https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Shiamk&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T14:11:04ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886132019 Mock AMC 10B Problems/Problem 212023-02-04T20:13:50Z<p>Shiamk: /* Solution 1: Author:(Shiva Kannan) */</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row. You will see at this point, that the first two elements of the third row determine the full arrangement. There are four possible ways to construct those two elements(Check different possibilities to arrive at the conclusion). Hence, the total number of ways is <math>24*9*4 = \boxed{\text{(C)} 864}</math>.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886122019 Mock AMC 10B Problems/Problem 212023-02-04T20:13:42Z<p>Shiamk: /* Solution 1: Author:(Shiva Kannan) */</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row. You will see at this point, that the first two elements of the third row determine the full arrangement. There are four possible ways to construct those two elements(Check different possibilities to arrive at the conclusion). Hence, the total number of ways is <math>24*9*4 = \boxed{\text{(C)} \864}</math>.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886112019 Mock AMC 10B Problems/Problem 212023-02-04T20:12:51Z<p>Shiamk: /* Solution 1: Author:(Shiva Kannan) */</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row. You will see at this point, that the first two elements of the third row determine the full arrangement. There are four possible ways to construct those two elements(Check different possibilities to arrive at the conclusion). Hence, the total number of ways is <math>24*9*4 = \qquad\textbf{(C)}\ 864</math>.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886102019 Mock AMC 10B Problems/Problem 212023-02-04T20:12:03Z<p>Shiamk: /* Solution 1: Author:(Shiva Kannan) */</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row. You will see at this point, that the first two elements of the third row determine the full arrangement. There are four possible ways to construct those two elements(Check different possibilities to arrive at the conclusion). Hence, the total number of ways is <math>24*9*4 = 864</math>.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886092019 Mock AMC 10B Problems/Problem 212023-02-04T20:09:15Z<p>Shiamk: /* Solution 1: Author:(Shiva Kannan) */</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. Number of arrangements in which there is a <math>1</math> under the <math>1</math> in the first row is <math>3! = 6</math>, as the remaining three numbers of the row can be arranged in <math>6</math> ways. By symmetry, this is the same for all <math>4</math> numbers, so the total number of cases in which one value is below itself is <math>24</math>. But we have to correct for overcounting. Number of arrangements in which both <math>1</math> and <math>2</math> are below themselves is <math>2! = 2</math>, as there are <math>2</math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which <math>2</math> values are below themselves is <math>2* {4 \choose 2} = 12</math>. For <math>3</math> numbers being under themselves, we can place the remaining number in <math>1</math> way, and there are three possible sets of <math>3</math> numbers from the four to pick, so the number of cases in which <math>3</math> numbers are below themselves is <math>4</math>. For all four numbers being under themselves, there is only <math>1</math> case. <math>24 - 24 + 12 - 4 + 1 = 9</math>, which yields <math>9</math> ways to construct the second row.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2019_Mock_AMC_10B_Problems/Problem_21&diff=1886082019 Mock AMC 10B Problems/Problem 212023-02-04T20:08:59Z<p>Shiamk: Created page with "==Solution 1: Author:(Shiva Kannan)== There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we..."</p>
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<div>==Solution 1: Author:(Shiva Kannan)==<br />
There are <math>4! = 24</math> ways to place one of each <math>1, 2, 3,</math> and <math>4</math> in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is <math>24</math>. We can use PIE. <math>Number of arrangements in which there is a </math>1<math> under the </math>1<math> in the first row is </math>3! = 6<math>, as the remaining three numbers of the row can be arranged in </math>6<math> ways. By symmetry, this is the same for all </math>4<math> numbers, so the total number of cases in which one value is below itself is </math>24<math>. But we have to correct for overcounting. Number of arrangements in which both </math>1<math> and </math>2<math> are below themselves is </math>2! = 2<math>, as there are </math>2<math> ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which </math>2<math> values are below themselves is </math>2* {4 \choose 2} = 12<math>. For </math>3<math> numbers being under themselves, we can place the remaining number in </math>1<math> way, and there are three possible sets of </math>3<math> numbers from the four to pick, so the number of cases in which </math>3<math> numbers are below themselves is </math>4<math>. For all four numbers being under themselves, there is only </math>1<math> case. </math>24 - 24 + 12 - 4 + 1 = 9<math>, which yields </math>9$ ways to construct the second row.</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1870402021 Fall AMC 10B Problems/Problem 232023-01-18T19:51:20Z<p>Shiamk: /* Solution 3 (Clear & Elementary Casework) */</p>
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<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Clear & Elementary Casework) - Author: Shiva Kannan ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^{10} = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2</math> (Choices of Colors) <math>\cdot 1</math> (Configuration for particular color existing on all 5 exterior edges) <math>= 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color. <br />
<br />
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.<br />
<br />
Subcase <math>1</math>: The case in which the two exterior sides colored differently from the other three are adjacent. <br />
<br />
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields <math>0</math> ways.<br />
<br />
Subcase <math>2</math>: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are <math>5</math> arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields <math>2*5=10</math> ways. <br />
<br />
In total, there are <math>10+2=12</math> ways, such that no triangle has all three sides in the same color. This yields <math>1024-12=1012</math> ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: <math>\frac{1012}{2^{10}} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
-Original Solution By Shiva Kannan<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==Video Solution by WhyMath==<br />
https://youtu.be/RqKYqCaZBlo<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868332020 AMC 10B Problems/Problem 222023-01-16T17:49:25Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
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<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</cmath><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868322020 AMC 10B Problems/Problem 222023-01-16T17:48:12Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</math> <math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868312020 AMC 10B Problems/Problem 222023-01-16T17:47:37Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math><br />
<br />
<br />
= <br />
<br />
\frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</math> <math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868302020 AMC 10B Problems/Problem 222023-01-16T17:46:33Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</math> <math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868292020 AMC 10B Problems/Problem 222023-01-16T17:46:14Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</math> <math></math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.<math><br />
<br />
Clearly, </math>201 < 2^{201} + 2^{51} + 1<math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is </math>\boxed{\textbf{(D) } 201}$.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868282020 AMC 10B Problems/Problem 222023-01-16T17:45:42Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =<br />
<br />
2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868272020 AMC 10B Problems/Problem 222023-01-16T17:45:14Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math><br />
<br />
= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> =<br />
<br />
2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} <math></math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.<math><br />
<br />
Clearly, </math>201 < 2^{201} + 2^{51} + 1<math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is </math>\boxed{\textbf{(D) } 201}$.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868262020 AMC 10B Problems/Problem 222023-01-16T17:44:55Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math><br />
<br />
= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>=<br />
<br />
2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><math>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868252020 AMC 10B Problems/Problem 222023-01-16T17:44:06Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math><br />
<br />
= <br />
<br />
\frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>=<br />
<br />
2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =<br />
<br />
<br />
2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<math>=<br />
<br />
2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} <br />
<br />
= <br />
<br />
2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}<br />
<br />
<br />
= <br />
<br />
<br />
2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868242020 AMC 10B Problems/Problem 222023-01-16T17:43:19Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math><br />
= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} =<br />
2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<math>=<br />
2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}<br />
= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868232020 AMC 10B Problems/Problem 222023-01-16T17:41:14Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<math>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868222020 AMC 10B Problems/Problem 222023-01-16T17:40:19Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<math>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868212020 AMC 10B Problems/Problem 222023-01-16T17:39:49Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<math>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1} = 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</math><br />
<br />
Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868202020 AMC 10B Problems/Problem 222023-01-16T17:35:34Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1} </math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868182020 AMC 10B Problems/Problem 222023-01-16T17:33:58Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1} = 2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51}}{2^{101}+2^{51}+1} </math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868172020 AMC 10B Problems/Problem 222023-01-16T17:32:30Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1} = 2^{101} - </math>\frac{2^{152}+2^{102}+2^{51}}{2^{101}+2^{51}+1}$<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868162020 AMC 10B Problems/Problem 222023-01-16T17:30:52Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math> <math>= 2^{101} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1} = 2^{101} - </math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868152020 AMC 10B Problems/Problem 222023-01-16T17:29:32Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868142020 AMC 10B Problems/Problem 222023-01-16T17:29:00Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </math> <math>-</math> <math>\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868132020 AMC 10B Problems/Problem 222023-01-16T17:28:14Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<math> \frac{2^{202}+202}{2^{101}+2^{51}+1} </math> <math>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </math> - <math>\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868122020 AMC 10B Problems/Problem 222023-01-16T17:27:44Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </cmath> - <cmath>\frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868112020 AMC 10B Problems/Problem 222023-01-16T17:27:30Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </cmath> - <cmath>\ frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868102020 AMC 10B Problems/Problem 222023-01-16T17:27:09Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </cmath> - <math>\ frac{2^{152} + 2^{101} + 202}{2^{101}+2^{51}+1}</math><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868092020 AMC 10B Problems/Problem 222023-01-16T17:26:03Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} </cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868072020 AMC 10B Problems/Problem 222023-01-16T17:25:05Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868062020 AMC 10B Problems/Problem 222023-01-16T17:24:36Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} = \frac{2^{202}+2^{152}+2^{101}}{2^{101}+2^{51}+1}} - </cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868052020 AMC 10B Problems/Problem 222023-01-16T17:24:21Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} = frac{2^{202}+2^{152}+2^{101}}{2^{101}+2^{51}+1}} - </cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868042020 AMC 10B Problems/Problem 222023-01-16T17:22:45Z<p>Shiamk: /* Solution 6(Author: Shiva Kannan) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
<cmath> \frac{2^202+202}{2^101+2^51+1} =</cmath><br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_22&diff=1868022020 AMC 10B Problems/Problem 222023-01-16T17:21:53Z<p>Shiamk: /* Solution 5 (Modular Arithmetic) */</p>
<hr />
<div>==Problem==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
==Solution 1==<br />
<br />
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.<br />
<br />
We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath><br />
<br />
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have<br />
<br />
<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath><br />
<br />
Rearranging, we can see that this is exactly what we need:<br />
<br />
<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath><br />
<br />
So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath><br />
<br />
Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88<br />
<br />
==Solution 2 (MAA Original Solution)==<br />
<br />
<cmath><br />
\begin{align*}<br />
2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\<br />
&= (2^{101} + 1)^2 - 2^{102} + 201\\<br />
&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.<br />
\end{align*}<br />
</cmath><br />
<br />
Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)<br />
<br />
==Solution 3 ==<br />
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>.<br />
Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>.<br />
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math><br />
<br />
==Solution 4 ==<br />
We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
==Solution 5 (Modular Arithmetic)==<br />
Let <math>n=2^{101}+2^{51}+1</math>. Then, <br />
<br />
<math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> <br />
<br />
<math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> <br />
<br />
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. <br />
<br />
Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>.<br />
<br />
~ Leo.Euler<br />
<br />
~ (edited by asops)<br />
<br />
==Solution 6(Author: Shiva Kannan)==<br />
<br />
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:<br />
<br />
$ \frac{2^{202}+202}{2^{101}+2^{51}+1} =<br />
<br />
==Video Solutions==<br />
<br />
===Video Solution 1 by Mathematical Dexterity (2 min)===<br />
https://www.youtube.com/watch?v=lLWURnmpPQA<br />
<br />
===Video Solution 2 by The Beauty Of Math===<br />
https://youtu.be/gPqd-yKQdFg<br />
<br />
===Video Solution 3===<br />
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx<br />
<br />
===Video Solution 4 Using Sophie Germain's Identity===<br />
https://youtu.be/ba6w1OhXqOQ?t=5155<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also== <br />
<br />
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_24&diff=1863772022 AMC 10A Problems/Problem 242023-01-12T01:57:10Z<p>Shiamk: /* Solution 4 (Answer Choices) */</p>
<hr />
<div>{{duplicate|[[2022 AMC 10A Problems/Problem 24|2022 AMC 10A #24]] and [[2022 AMC 12A Problems/Problem 24|2022 AMC 12A #24]]}}<br />
<br />
==Problem==<br />
<br />
How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math> are there such that for each <math>j \in \{1,2,3,4\}</math>, at least <math>j</math> of the digits are less than <math>j</math>? (For example, <math>02214</math> satisfies this condition<br />
because it contains at least <math>1</math> digit less than <math>1</math>, at least <math>2</math> digits less than <math>2</math>, at least <math>3</math> digits less<br />
than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it<br />
does not contain at least <math>2</math> digits less than <math>2</math>.)<br />
<br />
<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math><br />
<br />
== Solution 1 (Parking Functions) == <br />
For some <math>n</math>, let there be <math>n+1</math> parking spaces counterclockwise in a circle. Consider a string of <math>n</math> integers <math>c_1c_2 \ldots c_n</math> each between <math>0</math> and <math>n</math>, and let <math>n</math> cars come into this circle so that the <math>i</math>th car tries to park at spot <math>c_i</math>, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.<br />
<br />
Then the strings of <math>n</math> numbers between <math>0</math> and <math>n-1</math> that contain at least <math>k</math> integers <math><k</math> for <math>1 \leq k \leq n</math> are exactly the set of strings that leave spot <math>n</math> empty. Also note for any string <math>c_1c_2 \ldots c_n</math>, we can add <math>1</math> to each <math>c_i</math> (mod <math>n+1</math>) to shift the empty spot counterclockwise, meaning for each string there exists exactly one <math>j</math> with <math>0 \leq j \leq n</math> so that <math>(c_1+j)(c_2+j) \ldots (c_n+j)</math> leaves spot <math>n</math> empty. This gives there are <math>\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}</math> such strings. <br />
<br />
Plugging in <math>n = 5</math> gives <math>\boxed{\textbf{(E) }1296}</math> such strings.<br />
<br />
~oh54321<br />
<br />
==Solution 2 (Casework)==<br />
<br />
Note that a valid string must have at least one <math>0.</math><br />
<br />
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:<br />
<ol style="margin-left: 1.5em;"><br />
<li>The string has <math>1</math> different digit.</li><p><br />
The only possibility is <math>00000.</math> <p><br />
<b>There is <math>\boldsymbol{1}</math> string in this case.</b><br />
<li>The string has <math>2</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c||c}<br />
& & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{01} & \boldsymbol{02} & \boldsymbol{03} & \boldsymbol{04} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & \\ [-1.5ex]<br />
& 00001 & 00002 & 00003 & 00004 & \hspace{2mm}4\cdot\frac{5!}{4!1!}=20 \\ [2ex] <br />
& 00011 & 00022 & 00033 & & \hspace{2mm}3\cdot\frac{5!}{3!2!}=30 \\ [2ex] <br />
& 00111 & 00222 & & & \hspace{2mm}2\cdot\frac{5!}{2!3!}=20 \\ [2ex] <br />
& 01111 & & & & 1\cdot\frac{5!}{1!4!}=5 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{20+30+20+5=75}</math> strings in this case.</b><br />
<li>The string has <math>3</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c|c|c||c}<br />
& & & & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{012} & \boldsymbol{013} & \boldsymbol{014} & \boldsymbol{023} & \boldsymbol{024} & \boldsymbol{034} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & & & \\ [-1.5ex]<br />
& 00012 & 00013 & 00014 & 00023 & 00024 & 00034 & \hspace{2mm}6\cdot\frac{5!}{3!1!1!}=120 \\ [2ex] <br />
& 00112 & 00113 & 00114 & 00223 & 00224 & & \hspace{2mm}5\cdot\frac{5!}{2!2!1!}=150 \\ [2ex] <br />
& 00122 & 00133 & & 00233 & & & 3\cdot\frac{5!}{2!1!2!}=90 \\ [2ex] <br />
& 01112 & 01113 & 01114 & & & & 3\cdot\frac{5!}{1!3!1!}=60 \\ [2ex]<br />
& 01122 & 01133 & & & & & 2\cdot\frac{5!}{1!2!2!}=60 \\ [2ex]<br />
& 01222 & & & & & & 1\cdot\frac{5!}{1!1!3!}=20 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{120+150+90+60+60+20=500}</math> strings in this case.</b><br />
<li>The string has <math>4</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c}<br />
& & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{0123} & \boldsymbol{0124} & \boldsymbol{0134} & \boldsymbol{0234} \\ [0.5ex]<br />
\hline<br />
& & & & \\ [-1.5ex]<br />
& 00123 & 00124 & 00134 & 00234 \\ [2ex] <br />
& 01123 & 01124 & 01134 & \\ [2ex] <br />
& 01223 & 01224 & & \\ [2ex] <br />
& 01233 & & & \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{10\cdot\frac{5!}{2!1!1!1!}=600}</math> strings in this case.</b><br />
<li>The string has <math>5</math> different digits.</li><p><br />
<b>There are <math>\boldsymbol{5!=120}</math> strings in this case.</b><br />
</ol><br />
Together, the answer is <math>1+75+500+600+120=\boxed{\textbf{(E) }1296}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Recursive Equations Approach)==<br />
<br />
Denote by <math>N \left( p, q \right)</math> the number of <math>p</math>-digit strings formed by using numbers <math>0, 1, \cdots, q</math>, where for each <math>j \in \{1,2, \cdots , q\}</math>, at least <math>j</math> of the digits are less than <math>j</math>.<br />
<br />
We have the following recursive equation:<br />
<cmath><br />
N \left( p, q \right)<br />
= \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1<br />
</cmath><br />
and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.<br />
<br />
By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 1 \right)<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} \\<br />
& = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\<br />
& = 2^p - 1 .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 2</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 2 \right)<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right)<br />
- \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right)<br />
- p \\<br />
& = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\<br />
& = 3^p - 2^p - p .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 3</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 3 \right)<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right)<br />
- \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i}<br />
- \binom{p}{i} \left( p - i \right) \right)<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p<br />
- \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1}<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 4</math> and <math>p = 5</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( 5 , 4 \right)<br />
& = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\<br />
& = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\<br />
& = \boxed{\textbf{(E) }1296} .<br />
\end{align*}<br />
</cmath><br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 4 (Answer Choices)==<br />
Let the set of all valid sequences be <math>S</math>. <br />
Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences<br />
<cmath><br />
\begin{align*}<br />
\{a_2,a_3,a_4,a_5,a_1\}\\<br />
\{a_3,a_4,a_5,a_1,a_2\}\\<br />
\{a_4,a_5,a_1,a_2,a_3\}\\<br />
\{a_5,a_1,a_2,a_3,a_4\}<br />
\end{align*}<br />
</cmath><br />
must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which yields <math>\boxed{\textbf{(E) }1296}</math> by inspection.<br />
<br />
~Tau<br />
<br />
<br />
==Solution 5 (Casework on Number of 0s in the String) - Proposed by: Shiva Kannan==<br />
<br />
Case <math>1</math>: String contains <math>5</math> <math>0s</math> <br />
The only string that works is <math>00000</math> which yields <math>1</math> possible string that satisfies the given conditions<br />
Case <math>2</math>:<br />
<br />
==Video Solution==<br />
https://youtu.be/130OKAfG_-o<br />
<br />
~MathProblemSolvingSkills.com<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/mj78e_LnkX0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
== Video Solution By OmegaLearn using Complementary Counting == <br />
<br />
https://youtu.be/jWoxFT8hRn8<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_24&diff=1863762022 AMC 10A Problems/Problem 242023-01-12T01:56:37Z<p>Shiamk: /* Solution 4 (Answer Choices) */</p>
<hr />
<div>{{duplicate|[[2022 AMC 10A Problems/Problem 24|2022 AMC 10A #24]] and [[2022 AMC 12A Problems/Problem 24|2022 AMC 12A #24]]}}<br />
<br />
==Problem==<br />
<br />
How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math> are there such that for each <math>j \in \{1,2,3,4\}</math>, at least <math>j</math> of the digits are less than <math>j</math>? (For example, <math>02214</math> satisfies this condition<br />
because it contains at least <math>1</math> digit less than <math>1</math>, at least <math>2</math> digits less than <math>2</math>, at least <math>3</math> digits less<br />
than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it<br />
does not contain at least <math>2</math> digits less than <math>2</math>.)<br />
<br />
<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math><br />
<br />
== Solution 1 (Parking Functions) == <br />
For some <math>n</math>, let there be <math>n+1</math> parking spaces counterclockwise in a circle. Consider a string of <math>n</math> integers <math>c_1c_2 \ldots c_n</math> each between <math>0</math> and <math>n</math>, and let <math>n</math> cars come into this circle so that the <math>i</math>th car tries to park at spot <math>c_i</math>, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.<br />
<br />
Then the strings of <math>n</math> numbers between <math>0</math> and <math>n-1</math> that contain at least <math>k</math> integers <math><k</math> for <math>1 \leq k \leq n</math> are exactly the set of strings that leave spot <math>n</math> empty. Also note for any string <math>c_1c_2 \ldots c_n</math>, we can add <math>1</math> to each <math>c_i</math> (mod <math>n+1</math>) to shift the empty spot counterclockwise, meaning for each string there exists exactly one <math>j</math> with <math>0 \leq j \leq n</math> so that <math>(c_1+j)(c_2+j) \ldots (c_n+j)</math> leaves spot <math>n</math> empty. This gives there are <math>\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}</math> such strings. <br />
<br />
Plugging in <math>n = 5</math> gives <math>\boxed{\textbf{(E) }1296}</math> such strings.<br />
<br />
~oh54321<br />
<br />
==Solution 2 (Casework)==<br />
<br />
Note that a valid string must have at least one <math>0.</math><br />
<br />
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:<br />
<ol style="margin-left: 1.5em;"><br />
<li>The string has <math>1</math> different digit.</li><p><br />
The only possibility is <math>00000.</math> <p><br />
<b>There is <math>\boldsymbol{1}</math> string in this case.</b><br />
<li>The string has <math>2</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c||c}<br />
& & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{01} & \boldsymbol{02} & \boldsymbol{03} & \boldsymbol{04} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & \\ [-1.5ex]<br />
& 00001 & 00002 & 00003 & 00004 & \hspace{2mm}4\cdot\frac{5!}{4!1!}=20 \\ [2ex] <br />
& 00011 & 00022 & 00033 & & \hspace{2mm}3\cdot\frac{5!}{3!2!}=30 \\ [2ex] <br />
& 00111 & 00222 & & & \hspace{2mm}2\cdot\frac{5!}{2!3!}=20 \\ [2ex] <br />
& 01111 & & & & 1\cdot\frac{5!}{1!4!}=5 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{20+30+20+5=75}</math> strings in this case.</b><br />
<li>The string has <math>3</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c|c|c||c}<br />
& & & & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{012} & \boldsymbol{013} & \boldsymbol{014} & \boldsymbol{023} & \boldsymbol{024} & \boldsymbol{034} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & & & \\ [-1.5ex]<br />
& 00012 & 00013 & 00014 & 00023 & 00024 & 00034 & \hspace{2mm}6\cdot\frac{5!}{3!1!1!}=120 \\ [2ex] <br />
& 00112 & 00113 & 00114 & 00223 & 00224 & & \hspace{2mm}5\cdot\frac{5!}{2!2!1!}=150 \\ [2ex] <br />
& 00122 & 00133 & & 00233 & & & 3\cdot\frac{5!}{2!1!2!}=90 \\ [2ex] <br />
& 01112 & 01113 & 01114 & & & & 3\cdot\frac{5!}{1!3!1!}=60 \\ [2ex]<br />
& 01122 & 01133 & & & & & 2\cdot\frac{5!}{1!2!2!}=60 \\ [2ex]<br />
& 01222 & & & & & & 1\cdot\frac{5!}{1!1!3!}=20 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{120+150+90+60+60+20=500}</math> strings in this case.</b><br />
<li>The string has <math>4</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c}<br />
& & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{0123} & \boldsymbol{0124} & \boldsymbol{0134} & \boldsymbol{0234} \\ [0.5ex]<br />
\hline<br />
& & & & \\ [-1.5ex]<br />
& 00123 & 00124 & 00134 & 00234 \\ [2ex] <br />
& 01123 & 01124 & 01134 & \\ [2ex] <br />
& 01223 & 01224 & & \\ [2ex] <br />
& 01233 & & & \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{10\cdot\frac{5!}{2!1!1!1!}=600}</math> strings in this case.</b><br />
<li>The string has <math>5</math> different digits.</li><p><br />
<b>There are <math>\boldsymbol{5!=120}</math> strings in this case.</b><br />
</ol><br />
Together, the answer is <math>1+75+500+600+120=\boxed{\textbf{(E) }1296}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Recursive Equations Approach)==<br />
<br />
Denote by <math>N \left( p, q \right)</math> the number of <math>p</math>-digit strings formed by using numbers <math>0, 1, \cdots, q</math>, where for each <math>j \in \{1,2, \cdots , q\}</math>, at least <math>j</math> of the digits are less than <math>j</math>.<br />
<br />
We have the following recursive equation:<br />
<cmath><br />
N \left( p, q \right)<br />
= \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1<br />
</cmath><br />
and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.<br />
<br />
By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 1 \right)<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} \\<br />
& = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\<br />
& = 2^p - 1 .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 2</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 2 \right)<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right)<br />
- \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right)<br />
- p \\<br />
& = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\<br />
& = 3^p - 2^p - p .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 3</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 3 \right)<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right)<br />
- \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i}<br />
- \binom{p}{i} \left( p - i \right) \right)<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p<br />
- \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1}<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 4</math> and <math>p = 5</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( 5 , 4 \right)<br />
& = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\<br />
& = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\<br />
& = \boxed{\textbf{(E) }1296} .<br />
\end{align*}<br />
</cmath><br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 4 (Answer Choices)==<br />
Let the set of all valid sequences be <math>S</math>. <br />
Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences<br />
<cmath><br />
\begin{align*}<br />
\{a_2,a_3,a_4,a_5,a_1\}\\<br />
\{a_3,a_4,a_5,a_1,a_2\}\\<br />
\{a_4,a_5,a_1,a_2,a_3\}\\<br />
\{a_5,a_1,a_2,a_3,a_4\}<br />
\end{align*}<br />
</cmath><br />
must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which yields <math>\boxed{\textbf{(E) }1296}</math> by inspection.<br />
<br />
~Tau<br />
<br />
<br />
==Solution 5 (Casework on Number of 0s in the String)== (Proposed by: Shiva Kannan)<br />
<br />
Case <math>1</math>: String contains <math>5</math> <math>0s</math> <br />
The only string that works is <math>00000</math> which yields <math>1</math> possible string that satisfies the given conditions<br />
Case <math>2</math>:<br />
<br />
==Video Solution==<br />
https://youtu.be/130OKAfG_-o<br />
<br />
~MathProblemSolvingSkills.com<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/mj78e_LnkX0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
== Video Solution By OmegaLearn using Complementary Counting == <br />
<br />
https://youtu.be/jWoxFT8hRn8<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_24&diff=1863752022 AMC 10A Problems/Problem 242023-01-12T01:55:38Z<p>Shiamk: /* Solution 3 (Recursive Equations Approach) */</p>
<hr />
<div>{{duplicate|[[2022 AMC 10A Problems/Problem 24|2022 AMC 10A #24]] and [[2022 AMC 12A Problems/Problem 24|2022 AMC 12A #24]]}}<br />
<br />
==Problem==<br />
<br />
How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math> are there such that for each <math>j \in \{1,2,3,4\}</math>, at least <math>j</math> of the digits are less than <math>j</math>? (For example, <math>02214</math> satisfies this condition<br />
because it contains at least <math>1</math> digit less than <math>1</math>, at least <math>2</math> digits less than <math>2</math>, at least <math>3</math> digits less<br />
than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it<br />
does not contain at least <math>2</math> digits less than <math>2</math>.)<br />
<br />
<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math><br />
<br />
== Solution 1 (Parking Functions) == <br />
For some <math>n</math>, let there be <math>n+1</math> parking spaces counterclockwise in a circle. Consider a string of <math>n</math> integers <math>c_1c_2 \ldots c_n</math> each between <math>0</math> and <math>n</math>, and let <math>n</math> cars come into this circle so that the <math>i</math>th car tries to park at spot <math>c_i</math>, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.<br />
<br />
Then the strings of <math>n</math> numbers between <math>0</math> and <math>n-1</math> that contain at least <math>k</math> integers <math><k</math> for <math>1 \leq k \leq n</math> are exactly the set of strings that leave spot <math>n</math> empty. Also note for any string <math>c_1c_2 \ldots c_n</math>, we can add <math>1</math> to each <math>c_i</math> (mod <math>n+1</math>) to shift the empty spot counterclockwise, meaning for each string there exists exactly one <math>j</math> with <math>0 \leq j \leq n</math> so that <math>(c_1+j)(c_2+j) \ldots (c_n+j)</math> leaves spot <math>n</math> empty. This gives there are <math>\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}</math> such strings. <br />
<br />
Plugging in <math>n = 5</math> gives <math>\boxed{\textbf{(E) }1296}</math> such strings.<br />
<br />
~oh54321<br />
<br />
==Solution 2 (Casework)==<br />
<br />
Note that a valid string must have at least one <math>0.</math><br />
<br />
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:<br />
<ol style="margin-left: 1.5em;"><br />
<li>The string has <math>1</math> different digit.</li><p><br />
The only possibility is <math>00000.</math> <p><br />
<b>There is <math>\boldsymbol{1}</math> string in this case.</b><br />
<li>The string has <math>2</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c||c}<br />
& & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{01} & \boldsymbol{02} & \boldsymbol{03} & \boldsymbol{04} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & \\ [-1.5ex]<br />
& 00001 & 00002 & 00003 & 00004 & \hspace{2mm}4\cdot\frac{5!}{4!1!}=20 \\ [2ex] <br />
& 00011 & 00022 & 00033 & & \hspace{2mm}3\cdot\frac{5!}{3!2!}=30 \\ [2ex] <br />
& 00111 & 00222 & & & \hspace{2mm}2\cdot\frac{5!}{2!3!}=20 \\ [2ex] <br />
& 01111 & & & & 1\cdot\frac{5!}{1!4!}=5 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{20+30+20+5=75}</math> strings in this case.</b><br />
<li>The string has <math>3</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c|c|c||c}<br />
& & & & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{012} & \boldsymbol{013} & \boldsymbol{014} & \boldsymbol{023} & \boldsymbol{024} & \boldsymbol{034} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & & & \\ [-1.5ex]<br />
& 00012 & 00013 & 00014 & 00023 & 00024 & 00034 & \hspace{2mm}6\cdot\frac{5!}{3!1!1!}=120 \\ [2ex] <br />
& 00112 & 00113 & 00114 & 00223 & 00224 & & \hspace{2mm}5\cdot\frac{5!}{2!2!1!}=150 \\ [2ex] <br />
& 00122 & 00133 & & 00233 & & & 3\cdot\frac{5!}{2!1!2!}=90 \\ [2ex] <br />
& 01112 & 01113 & 01114 & & & & 3\cdot\frac{5!}{1!3!1!}=60 \\ [2ex]<br />
& 01122 & 01133 & & & & & 2\cdot\frac{5!}{1!2!2!}=60 \\ [2ex]<br />
& 01222 & & & & & & 1\cdot\frac{5!}{1!1!3!}=20 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{120+150+90+60+60+20=500}</math> strings in this case.</b><br />
<li>The string has <math>4</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c}<br />
& & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{0123} & \boldsymbol{0124} & \boldsymbol{0134} & \boldsymbol{0234} \\ [0.5ex]<br />
\hline<br />
& & & & \\ [-1.5ex]<br />
& 00123 & 00124 & 00134 & 00234 \\ [2ex] <br />
& 01123 & 01124 & 01134 & \\ [2ex] <br />
& 01223 & 01224 & & \\ [2ex] <br />
& 01233 & & & \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{10\cdot\frac{5!}{2!1!1!1!}=600}</math> strings in this case.</b><br />
<li>The string has <math>5</math> different digits.</li><p><br />
<b>There are <math>\boldsymbol{5!=120}</math> strings in this case.</b><br />
</ol><br />
Together, the answer is <math>1+75+500+600+120=\boxed{\textbf{(E) }1296}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Recursive Equations Approach)==<br />
<br />
Denote by <math>N \left( p, q \right)</math> the number of <math>p</math>-digit strings formed by using numbers <math>0, 1, \cdots, q</math>, where for each <math>j \in \{1,2, \cdots , q\}</math>, at least <math>j</math> of the digits are less than <math>j</math>.<br />
<br />
We have the following recursive equation:<br />
<cmath><br />
N \left( p, q \right)<br />
= \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1<br />
</cmath><br />
and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.<br />
<br />
By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 1 \right)<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} \\<br />
& = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\<br />
& = 2^p - 1 .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 2</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 2 \right)<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right)<br />
- \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right)<br />
- p \\<br />
& = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\<br />
& = 3^p - 2^p - p .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 3</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 3 \right)<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right)<br />
- \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i}<br />
- \binom{p}{i} \left( p - i \right) \right)<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p<br />
- \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1}<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 4</math> and <math>p = 5</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( 5 , 4 \right)<br />
& = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\<br />
& = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\<br />
& = \boxed{\textbf{(E) }1296} .<br />
\end{align*}<br />
</cmath><br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
==Solution 4 (Answer Choices)==<br />
Let the set of all valid sequences be <math>S</math>. <br />
Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences<br />
<cmath><br />
\begin{align*}<br />
\{a_2,a_3,a_4,a_5,a_1\}\\<br />
\{a_3,a_4,a_5,a_1,a_2\}\\<br />
\{a_4,a_5,a_1,a_2,a_3\}\\<br />
\{a_5,a_1,a_2,a_3,a_4\}<br />
\end{align*}<br />
</cmath><br />
must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which yields <math>\boxed{\textbf{(E) }1296}</math> by inspection.<br />
<br />
~Tau<br />
<br />
==Video Solution==<br />
https://youtu.be/130OKAfG_-o<br />
<br />
~MathProblemSolvingSkills.com<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/mj78e_LnkX0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
== Video Solution By OmegaLearn using Complementary Counting == <br />
<br />
https://youtu.be/jWoxFT8hRn8<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_24&diff=1863742022 AMC 10A Problems/Problem 242023-01-12T01:55:11Z<p>Shiamk: /* Solution 3 (Recursive Equations Approach) */</p>
<hr />
<div>{{duplicate|[[2022 AMC 10A Problems/Problem 24|2022 AMC 10A #24]] and [[2022 AMC 12A Problems/Problem 24|2022 AMC 12A #24]]}}<br />
<br />
==Problem==<br />
<br />
How many strings of length <math>5</math> formed from the digits <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math> are there such that for each <math>j \in \{1,2,3,4\}</math>, at least <math>j</math> of the digits are less than <math>j</math>? (For example, <math>02214</math> satisfies this condition<br />
because it contains at least <math>1</math> digit less than <math>1</math>, at least <math>2</math> digits less than <math>2</math>, at least <math>3</math> digits less<br />
than <math>3</math>, and at least <math>4</math> digits less than <math>4</math>. The string <math>23404</math> does not satisfy the condition because it<br />
does not contain at least <math>2</math> digits less than <math>2</math>.)<br />
<br />
<math>\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296</math><br />
<br />
== Solution 1 (Parking Functions) == <br />
For some <math>n</math>, let there be <math>n+1</math> parking spaces counterclockwise in a circle. Consider a string of <math>n</math> integers <math>c_1c_2 \ldots c_n</math> each between <math>0</math> and <math>n</math>, and let <math>n</math> cars come into this circle so that the <math>i</math>th car tries to park at spot <math>c_i</math>, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.<br />
<br />
Then the strings of <math>n</math> numbers between <math>0</math> and <math>n-1</math> that contain at least <math>k</math> integers <math><k</math> for <math>1 \leq k \leq n</math> are exactly the set of strings that leave spot <math>n</math> empty. Also note for any string <math>c_1c_2 \ldots c_n</math>, we can add <math>1</math> to each <math>c_i</math> (mod <math>n+1</math>) to shift the empty spot counterclockwise, meaning for each string there exists exactly one <math>j</math> with <math>0 \leq j \leq n</math> so that <math>(c_1+j)(c_2+j) \ldots (c_n+j)</math> leaves spot <math>n</math> empty. This gives there are <math>\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}</math> such strings. <br />
<br />
Plugging in <math>n = 5</math> gives <math>\boxed{\textbf{(E) }1296}</math> such strings.<br />
<br />
~oh54321<br />
<br />
==Solution 2 (Casework)==<br />
<br />
Note that a valid string must have at least one <math>0.</math><br />
<br />
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:<br />
<ol style="margin-left: 1.5em;"><br />
<li>The string has <math>1</math> different digit.</li><p><br />
The only possibility is <math>00000.</math> <p><br />
<b>There is <math>\boldsymbol{1}</math> string in this case.</b><br />
<li>The string has <math>2</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c||c}<br />
& & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{01} & \boldsymbol{02} & \boldsymbol{03} & \boldsymbol{04} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & \\ [-1.5ex]<br />
& 00001 & 00002 & 00003 & 00004 & \hspace{2mm}4\cdot\frac{5!}{4!1!}=20 \\ [2ex] <br />
& 00011 & 00022 & 00033 & & \hspace{2mm}3\cdot\frac{5!}{3!2!}=30 \\ [2ex] <br />
& 00111 & 00222 & & & \hspace{2mm}2\cdot\frac{5!}{2!3!}=20 \\ [2ex] <br />
& 01111 & & & & 1\cdot\frac{5!}{1!4!}=5 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{20+30+20+5=75}</math> strings in this case.</b><br />
<li>The string has <math>3</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c|c|c||c}<br />
& & & & & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{012} & \boldsymbol{013} & \boldsymbol{014} & \boldsymbol{023} & \boldsymbol{024} & \boldsymbol{034} & \textbf{Row's Count} \\ [0.5ex]<br />
\hline<br />
& & & & & & & \\ [-1.5ex]<br />
& 00012 & 00013 & 00014 & 00023 & 00024 & 00034 & \hspace{2mm}6\cdot\frac{5!}{3!1!1!}=120 \\ [2ex] <br />
& 00112 & 00113 & 00114 & 00223 & 00224 & & \hspace{2mm}5\cdot\frac{5!}{2!2!1!}=150 \\ [2ex] <br />
& 00122 & 00133 & & 00233 & & & 3\cdot\frac{5!}{2!1!2!}=90 \\ [2ex] <br />
& 01112 & 01113 & 01114 & & & & 3\cdot\frac{5!}{1!3!1!}=60 \\ [2ex]<br />
& 01122 & 01133 & & & & & 2\cdot\frac{5!}{1!2!2!}=60 \\ [2ex]<br />
& 01222 & & & & & & 1\cdot\frac{5!}{1!1!3!}=20 \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{120+150+90+60+60+20=500}</math> strings in this case.</b><br />
<li>The string has <math>4</math> different digits.</li><p><br />
We have the following table:<br />
<cmath>\begin{array}{c||c|c|c|c}<br />
& & & & \\ [-2.5ex]<br />
\textbf{Digits} & \boldsymbol{0123} & \boldsymbol{0124} & \boldsymbol{0134} & \boldsymbol{0234} \\ [0.5ex]<br />
\hline<br />
& & & & \\ [-1.5ex]<br />
& 00123 & 00124 & 00134 & 00234 \\ [2ex] <br />
& 01123 & 01124 & 01134 & \\ [2ex] <br />
& 01223 & 01224 & & \\ [2ex] <br />
& 01233 & & & \\ [0.75ex]<br />
\end{array}</cmath><br />
<b>There are <math>\boldsymbol{10\cdot\frac{5!}{2!1!1!1!}=600}</math> strings in this case.</b><br />
<li>The string has <math>5</math> different digits.</li><p><br />
<b>There are <math>\boldsymbol{5!=120}</math> strings in this case.</b><br />
</ol><br />
Together, the answer is <math>1+75+500+600+120=\boxed{\textbf{(E) }1296}.</math><br />
<br />
~MRENTHUSIASM<br />
<br />
==Solution 3 (Recursive Equations Approach)==<br />
<br />
Denote by <math>N \left( p, q \right)</math> the number of <math>p</math>-digit strings formed by using numbers <math>0, 1, \cdots, q</math>, where for each <math>j \in \{1,2, \cdots , q\}</math>, at least <math>j</math> of the digits are less than <math>j</math>.<br />
<br />
We have the following recursive equation:<br />
<cmath><br />
N \left( p, q \right)<br />
= \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1<br />
</cmath><br />
and the boundary condition <math>N \left( p, 0 \right) = 1</math> for any <math>p \geq 0</math>.<br />
<br />
By solving this recursive equation, for <math>q = 1</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 1 \right)<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\<br />
& = \sum_{i = 0}^{p - 1} \binom{p}{i} \\<br />
& = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\<br />
& = 2^p - 1 .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 2</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 2 \right)<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\<br />
& = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right)<br />
- \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right)<br />
- p \\<br />
& = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\<br />
& = 3^p - 2^p - p .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 3</math> and <math>p \geq q</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( p , 3 \right)<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\<br />
& = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right)<br />
- \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\<br />
& = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i}<br />
- \binom{p}{i} \left( p - i \right) \right)<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p<br />
- \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1}<br />
- \frac{3}{2} p \left( p - 1 \right) \\<br />
& = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) .<br />
\end{align*}<br />
</cmath><br />
<br />
For <math>q = 4</math> and <math>p = 5</math>, we get<br />
<cmath><br />
\begin{align*}<br />
N \left( 5 , 4 \right)<br />
& = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\<br />
& = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\<br />
& = \boxed{\textbf{(E) }1296} .<br />
\end{align*}<br />
</cmath><br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
Solution 4 - Casework on Number of 0s (Shiva Kannan)<br />
<br />
Case <math>1</math>: String contains <math>5</math> <math>0s</math> <br />
The only string that works is <math>00000</math> which yields <math>1</math> possible string that satisfies the given conditions<br />
Case <math>2</math>:<br />
<br />
==Solution 4 (Answer Choices)==<br />
Let the set of all valid sequences be <math>S</math>. <br />
Notice that for any sequence <math>\{a_1,a_2,a_3,a_4,a_5\}</math> in <math>S</math>, the sequences<br />
<cmath><br />
\begin{align*}<br />
\{a_2,a_3,a_4,a_5,a_1\}\\<br />
\{a_3,a_4,a_5,a_1,a_2\}\\<br />
\{a_4,a_5,a_1,a_2,a_3\}\\<br />
\{a_5,a_1,a_2,a_3,a_4\}<br />
\end{align*}<br />
</cmath><br />
must also belong in <math>S</math>. However, one must consider the edge case all 5 elements are the same (only <math>\{0,0,0,0,0\}</math>), in which case all sequences listed are equivalent. Then <math>\lvert S \rvert \equiv 1 \pmod 5</math>, which yields <math>\boxed{\textbf{(E) }1296}</math> by inspection.<br />
<br />
~Tau<br />
<br />
==Video Solution==<br />
https://youtu.be/130OKAfG_-o<br />
<br />
~MathProblemSolvingSkills.com<br />
<br />
==Video Solution==<br />
<br />
https://youtu.be/mj78e_LnkX0<br />
<br />
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)<br />
<br />
== Video Solution By OmegaLearn using Complementary Counting == <br />
<br />
https://youtu.be/jWoxFT8hRn8<br />
<br />
~ pi_is_3.14<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2022|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=AMC_12C_2020_Problems&diff=183158AMC 12C 2020 Problems2022-11-27T19:02:14Z<p>Shiamk: /* Problem 11 */</p>
<hr />
<div>1 plus 1<br />
<br />
==Problem 1<br />
<br />
A plane flies at a speed of <math>590</math> miles/hour. How many miles in two hours<br />
<br />
==Problem 3==<br />
<br />
In a bag are <math>7</math> marbles consisting of <math>3</math> blue marbles and <math>4</math> red marbles. If each marble is pulled out <math>1</math> at a time, what is the probability that the <math>6th</math> marble pulled out red?<br />
<br />
<br />
<math>\textbf{(A)}\ 0 \qquad\textbf{(B)}\ \frac{1}{8} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ \frac{4}{7} \qquad\textbf{(E)}\ 1</math><br />
<br />
<br />
==Problem 4==<br />
<br />
A spaceship flies in space at a speed of <math>s</math> miles/hour and the spaceship is paid <math>d</math> dollars for each <math>100</math> miles traveled. It’s only expense is fuel in which it pays <math>\frac{d}{2}</math> dollars per gallon, while going at a rate of <math>h</math> hours per gallon. Traveling <math>3s</math> miles, how much money would the spaceship have gained?<br />
<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ \ 21 \qquad\textbf{(C)}\ \ 22 \qquad\textbf{(D)}\ \ 23 \qquad\textbf{(E)}\ 24</math><br />
<br />
==Problem 5==<br />
<br />
Let <math>R(x)</math> be a function satisfying <math>R(m + n) = R(m)R(n)</math> for all real numbers <math>n</math> and <math>m</math>. Let <math>R(1) = \frac{1}{2}.</math> What is <math>R(1) + R(2) + R(3) + … + R(1000)</math>?<br />
<br />
==Problem 6==<br />
<br />
How many increasing(lower to higher numbered) subsets of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}</math> contain no <math>2</math> consecutive prime numbers?<br />
<br />
<br />
==Problem 7==<br />
<br />
A Regular Octagon has an area of <math>18 + 18\sqrt {2}</math>. What is the sum of the lengths of the diagonals of the octagon?<br />
<br />
==Problem 8==<br />
<br />
What is the value of <math>sin(1^\circ)sin(3^\circ)sin(5^\circ)…sin(179^\circ) - sin(181^\circ)sin(182^\circ)…sin(359^\circ)</math>?<br />
<br />
==Problem 9==<br />
<br />
Let <math>E(x)</math> denote the sum of the even digits of a positive integer and let <math>O(x)</math> denote the sum of the odd digits of a positive integer. For some positive integer <math>N</math>, <math>3E(3N)</math> = <math>4O(4N)</math>. What is the product of the digits of the least possible such <math>N</math>?<br />
<br />
==Problem 10==<br />
<br />
In how many ways can <math>n</math> candy canes and <math>n + 1</math> lollipops be split between <math>n - 4</math> children if each child must receive atleast <math>1</math> candy but no child receives both types?<br />
<br />
==Problem 11==<br />
<br />
Let <math>ABCD</math> be an isosceles trapezoid with <math>\overline{AB}</math> being parallel to <math>\overline{CD}</math> and <math>\overline{AB} = 5</math>, <math>\overline{CD} = 15</math>, and <math>\angle ADC = 60^\circ</math>. If <math>E</math> is the intersection of <math>\overline{AC}</math> and <math>\overline{BD}</math>, and <math>\omega</math> is the circumcenter of <math>\bigtriangleup ABC</math>, what is the length of <math>\overline{E\omega}</math>? Source: JHMMC 2019<br />
<br />
<br />
<math>\textbf{(A)} \frac {31}{12}\sqrt{2} \qquad \textbf{(B)} \frac {35}{12}\sqrt{3} \qquad \textbf{(C)} \frac {37}{12}\sqrt{5} \qquad \textbf{(D)} \frac {39}{12}\sqrt{7} \qquad \textbf{(E)} \frac {41}{12}\sqrt{11} \qquad </math><br />
<br />
==Problem 12==<br />
<br />
For some positive integer <math>k</math>, let <math>k</math> satisfy the equation<br />
<br />
<math>log(k - 2)! + log(k - 1)! + 2 = 2 log(k!)</math>.<br />
What is the sum of the digits of <math>k</math>?<br />
<br />
==Problem 13==<br />
<br />
An alien walks horizontally on the real number line starting at the origin. On each move, the alien can walk <math>1</math> or <math>2</math> numbers the right or left of it. What is the expected distance from the alien to the origin after <math>10</math> moves?<br />
<br />
==Problem 14==<br />
<br />
Let <math>K</math> be the set of solutions to the equation <math>(x + i)^{10} = 1</math> on the complex plane, where <math>i = \sqrt -1</math>. <math>2</math> points from <math>K</math> are chosen, such that a circle <math>\Omega</math> passes through both points. What is the least possible area of <math>\Omega</math>?<br />
<br />
==Problem 15==<br />
<br />
Let <math>N = 10^{10^{100…^{10000…(100 zeroes)}}}</math>. What is the remainder when <math>N</math> is divided by <math>629</math>?<br />
<br />
<br />
<br />
==Problem 16==<br />
<br />
Let <math>V</math> and <math>F</math> be the vertex and focus of the Parabola <math>P(x) = \frac{1}{8} x</math> respectively. For a point <math>G</math> lying on the directrix of <math>P(x)</math>, and a point <math>H</math> lying on <math>P(x)</math>, <math>\overline {GH} = 10</math> and Quadrilateral <math>VFGH</math> is cyclic. If <math>VFGH</math> has integral side lengths, what is the minimum possible area of <math>VFGH</math>?<br />
<br />
==Problem 17==<br />
<br />
Let <math>H(n)</math> denote the <math>2nd</math> nonzero digit from the right in the base - <math>10</math> expansion of <math>(2n + 1)!</math>, for example, <math>H(2) = 1</math>. What is the sum of the digits of <math>\prod_{k = 1}^{2020}H(k)</math>?<br />
<br />
==Problem 18==<br />
<br />
<math>\bigtriangleup ABC</math> lays flat on the ground and has side lengths <math>\overline{AB} = 3, \overline{BC} = 4</math>, and <math>\overline{AC} = 5</math>. Vertex <math>A</math> is then lifted up creating an elevation angle with the triangle and the ground of <math>60^{\circ}</math>. A wooden pole is dropped from <math>A</math> perpendicular to the ground, making an altitude of a <math>3</math> Dimensional figure. Ropes are connected from the foot of the pole, <math>D</math>, to form <math>2</math> other segments, <math>\overline{BD}</math> and <math>\overline{CD}</math>. What is the volume of <math>ABCD</math>?<br />
<br />
<br />
<br />
<math>\textbf{(A) } 180\sqrt{3} \qquad \textbf{(B) } 15 + 180\sqrt{3} \qquad \textbf{(C) } 20 + 180\sqrt{5} \qquad \textbf{(D) } 28 + 180\sqrt{5} \qquad \textbf{(E) } 440\sqrt{2}</math><br />
<br />
==Problem 19==<br />
<br />
<br />
Let <math>P(x)</math> be a cubic polynomial with integral coefficients and roots <math>\cos \frac{\pi}{13}</math>, <math>\cos \frac{5\pi}{13}</math>, and <math>\cos \frac{7\pi}{13}</math>. What is the least possible sum of the coefficients of <math>P(x)</math>?<br />
<br />
==Problem 20==<br />
<br />
What is the maximum value of <math>\sum_{k = 1}^{6}(2^{x} + 3^{x})</math> as <math>x</math> varies through all real numbers to the nearest integer?<br />
<br />
<br />
<math>\textbf{(A)}\ -3\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ 0\qquad\textbf{(E)}\ 1</math><br />
<br />
==Problem 21==<br />
<br />
Let <math>\lfloor x \rfloor</math> denote the greatest integer less than or equal to <math>x</math>. How many positive integers <math>x < 2020</math>, satisfy the equation<br />
<math>\frac{x^{4} + 2020}{108} = \lfloor \sqrt (x^{2} - x)\rfloor</math>?<br />
<br />
<br />
==Problem 22==<br />
<br />
A convex hexagon <math>ABCDEF</math> is inscribed in a circle. <math>\overline {AB}</math> <math>=</math> <math>\overline {BC}</math> <math>=</math> <math>\overline {AD}</math> <math>=</math> <math>2</math>. <math>\overline {DE}</math> <math>=</math> <math>\overline {CF}</math> <math>=</math> <math>\overline {EF}</math> <math>=</math> <math>4</math>. The measure of <math>\overline {DC}</math> can be written as <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m + n</math>?<br />
<br />
<br />
<br />
==Problem 23==<br />
<br />
==Problem 24==<br />
<br />
A sequence <math>(a_n)</math> is defined as <math>a_1 = \frac{1}{\sqrt{3}}</math>, <math>a_2 = \sqrt{3}</math>, and for all <math>n > 1</math>, <br />
<br />
<math>a_{n + 1} = \frac{2a_n-1}{1 - a_n^2}</math><br />
<br />
What is <math>\lfloor \ a_{2020}\rfloor</math>?<br />
<br />
<br />
<br />
==Problem 25==<br />
<br />
Let <math>P(x) = x^{2020} + 2x^{2019} + 3x^{2018} + … + 2019x^{2} + 2020x + 2021</math> and let <math>Q(x) = x^{4} + 2x^{3} + 3x^{2} + 4x + 5</math>. Let <math>U</math> be the sum of the <math>kth</math> power of the roots of <math>P(Q(x))</math>. It is given that the least positive integer <math>y</math>, such that <math>3^{y} > U</math> is <math>2021</math>. What is <math>k</math>?</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1754042021 Fall AMC 10B Problems/Problem 232022-06-28T00:17:55Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^{10} = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color. <br />
<br />
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.<br />
<br />
Subcase <math>1</math>: The case in which the two exterior sides colored differently from the other three are adjacent. <br />
<br />
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields <math>0</math> ways.<br />
<br />
Subcase <math>2</math>: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are <math>5</math> arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields <math>2*5=10</math> ways. <br />
<br />
In total, there are <math>10+2=12</math> ways, such that no triangle has all three sides in the same color. This yields <math>1024-12=1012</math> ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: <math>\frac{1012}{2^{10}} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
-Original Solution By Shiva Kannan<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1754032021 Fall AMC 10B Problems/Problem 232022-06-28T00:17:13Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^{10} = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color. <br />
<br />
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.<br />
<br />
Subcase <math>1</math>: The case in which the two exterior sides colored differently from the other three are adjacent. <br />
<br />
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields <math>0</math> ways.<br />
<br />
Subcase <math>2</math>: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are <math>5</math> arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields <math>2*5=10</math> ways. <br />
<br />
In total, there are <math>10+2=12</math> ways, such that no triangle has all three sides in the same color. This yields <math>1024-12=1012</math> ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: <math>\frac{1012}{2^{10}} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1754022021 Fall AMC 10B Problems/Problem 232022-06-28T00:16:22Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^{10} = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color. <br />
<br />
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.<br />
<br />
Subcase <math>1</math>: The case in which the two exterior sides colored differently from the other three are adjacent. <br />
<br />
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields <math>0</math> ways.<br />
<br />
Subcase <math>2</math>: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are <math>5</math> arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields <math>2*5=10</math> ways. <br />
<br />
In total, there are <math>10+2=12</math> ways, such that no triangle has all three sides in the same color. This yields <math>1024-12=1012</math> ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: <math>\frac{1012}{2^{10}}</math> = <math>\frac{253}{256}</math>.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1754012021 Fall AMC 10B Problems/Problem 232022-06-28T00:15:19Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^{10} = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color. <br />
<br />
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.<br />
<br />
Subcase <math>1</math>: The case in which the two exterior sides colored differently from the other three are adjacent. <br />
<br />
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields <math>0</math> ways.<br />
<br />
Subcase <math>2</math>: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are <math>5</math> arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields <math>2*5=10</math> ways. <br />
<br />
In total, there are <math>10+2=12</math> ways, such that no triangle has all three sides in the same color. This yields <math>1024-12=1012</math> ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: <math>frac{1012}{2^{10}}</math> = <math>frac{253}{256}</math>.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1753952021 Fall AMC 10B Problems/Problem 232022-06-27T20:19:41Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^(10) = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1753942021 Fall AMC 10B Problems/Problem 232022-06-27T20:19:13Z<p>Shiamk: /* Solution 4 (Clear & Elementary Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^10 = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
*Exterior sides are the five sides of the regular pentagon<br />
*Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
Case <math>3</math>: Three of the five exterior sides in one color and the remaining two in another color.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1753842021 Fall AMC 10B Problems/Problem 232022-06-27T18:45:01Z<p>Shiamk: /* Solution 4 (Clear & Concise Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Elementary Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is <math>2^10 = 1024</math>. We will be counting the number of arrangements in which there are no triangles with all <math>3</math> sides the same color, and we can subtract from the total(complementary counting). <br />
<br />
Case <math>1</math>: The five exterior sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways. <br />
<br />
Case <math>2</math>: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.<br />
<br />
There are <math>2</math> ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the <math>2</math> ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives <math>2</math> choices for each of <math>10</math> segments yielding <math>2^10</math>. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are <math>5</math> ways to arrange the one blue and four reds or one red and four blues on the <math>5</math> exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields <math>0</math> ways.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1753832021 Fall AMC 10B Problems/Problem 232022-06-27T17:59:44Z<p>Shiamk: /* Solution 4 (Clear & Concise Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Concise Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. <br />
<br />
Case 1: The five sides are all blue or all red.<br />
<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamkhttps://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10B_Problems/Problem_23&diff=1753822021 Fall AMC 10B Problems/Problem 232022-06-27T17:59:08Z<p>Shiamk: /* Solution 4 (Clear & Concise Casework) */</p>
<hr />
<div>==Problem==<br />
<br />
Each of the <math>5{ }</math> sides and the <math>5{ }</math> diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?<br />
<br />
<math>(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1</math><br />
<br />
==Solution 1==<br />
<br />
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear.<br />
After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.<br />
<br />
<math>\textbf{Case 1}</math>: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. <br />
<br />
<math>\textbf{Case 2}</math>: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.<br />
<br />
<math>\textbf{Case 3}</math>: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are <math>{4\choose2}</math> ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of <math>6\cdot 2 = 12</math> ways to color the pentagon so that no such triangle has the same color for all of its sides. <br />
<br />
These are all the cases, and there are a total of <math>2^{10}</math> ways to color the pentagon. Therefore the answer is <math>1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}</math><br />
<br />
~KingRavi<br />
<br />
== Solution 2 (Ramsey's Theorem)==<br />
<br />
This problem is related to a special case of [https://en.wikipedia.org/wiki/Ramsey%27s_theorem Ramsey's Theorem], [https://en.wikipedia.org/wiki/Ramsey%27s_theorem#R(3,_3)_=_6 R(3, 3) = 6]. Suppose we color every edge of a <math>6</math> vertex complete graph <math>(K_6)</math> with <math>2</math> colors, there must exist a <math>3</math> vertex complete graph <math>(K_3)</math> with all it's edges in the same color. That is, <math>K_6</math> with edges in <math>2</math> colors contains a monochromatic <math>K_3</math>. For <math>K_5</math> with edges in <math>2</math> colors, a monochromatic <math>K_3</math> does not always exist.<br />
<br />
This is a problem about the probability of a monochromatic <math>K_3</math> exist in a <math>5</math> vertex complete graph <math>K_5</math> with edges in <math>2</math> colors.<br />
<br />
Choose a vertex, it has <math>4</math> edges. <br />
<br />
<math>\textbf{Case 1}</math>: When <math>3</math> or more edges are the same color, there must exist a monochromatic <math>K_3</math>. Suppose the color is red, as shown below. <br />
<br />
[[File:K 5 3 colors.jpg | 500px]]<br />
<br />
There is only <math>1</math> way to color all the edges in the same color. There is <math>\binom{4}{3} = 4</math> ways to color <math>3</math> edges in the same color. There are <math>2</math> colors. The probability of <math>3</math> or more edges the same color is <math>\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}</math>. So the probability of containing a monochromatic <math>K_3</math> is <math>\frac{5}{8}</math>.<br />
<br />
<math>\textbf{Case 2}</math>: When <math>2</math> edges are the same color, graphs that does not contain a monochromatic <math>K_3</math> can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic <math>K_3</math>.<br />
<br />
[[File:K 5 2 colors.jpg | 500px]]<br />
<br />
There are <math>\binom{4}{2} = 6</math> ways to choose <math>2</math> edges with the same color. For the other <math>4</math> vertices there are <math>\binom{4}{2} = 6</math> edges among them, there are <math>2^6 = 64</math> ways to color the edges. There are only <math>2</math> cases without a monochromatic <math>K_3</math>. <br />
<br />
So the probability without monochromatic <math>K_3</math> is <math>\frac{2}{64} = \frac{1}{32}</math>.<br />
<br />
The probability with monochromatic <math>K_3</math> is <math>1 - \frac{1}{32} = \frac{31}{32}</math>.<br />
<br />
From case 1 and case 2, the probability with monochromatic <math>K_3</math> is <math>\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}</math><br />
<br />
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]<br />
<br />
== Solution 3 (Extremely Educated Guess)==<br />
<br />
Thinking about it, the probability there is a triangle in the pentagon with all its sides of the same color is extremely likely. It probably isn't 1 because MAA would never make it so easy. Also, it can't be 2/3 because the total amount of ways to color the diagonals is 2^10, so the answer should be in the form m/(2^k), not m/(3). Now we are left with B, C, and D. The biggest number is 253/256 or D, so it is probably the answer. <math>\fbox{D}</math><br />
~heheman<br />
<br />
<br />
== Solution 4 (Clear & Concise Casework) ==<br />
<br />
We can break this problem down into cases based on the distribution of the two colors throughout the <math>5</math> sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. <br />
<br />
Case 1: The five sides are all blue or all red<br />
There are <math>2</math> options for the color given in the title of this case, and for each color, only one option is given for the exterior(All <math>5</math> sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the <math>5</math> interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the <math>5</math> exterior sides are blue, all of the <math>5</math> interior diagonals must be red, and if all of the <math>5</math> exterior sides are red, all of the <math>5</math> interior diagonals must be blue. This gives us a total of <math>2(Choices of Colors) * 1(Configuration for particular color existing on all 5 exterior edges) = 2</math> ways.<br />
<br />
==Video Solution by Interstigation==<br />
https://www.youtube.com/watch?v=Iaz1dT_XBHU<br />
<br />
~Interstigation<br />
<br />
==See Also==<br />
{{AMC10 box|year=2021 Fall|ab=B|num-a=24|num-b=22}}<br />
{{MAA Notice}}</div>Shiamk