https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Whalefin&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T11:07:41ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_23&diff=983642014 AMC 8 Problems/Problem 232018-10-29T01:04:11Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
Three members of the Euclid Middle School girls' softball team had the following conversation.<br />
<br />
Ashley: I just realized that our uniform numbers are all <math>2</math>-digit primes.<br />
<br />
Brittany : And the sum of your two uniform numbers is the date of my birthday earlier this month.<br />
<br />
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.<br />
<br />
Ashley: And the sum of your two uniform numbers is today's date.<br />
<br />
What number does Caitlin wear?<br />
<br />
<math>\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23</math><br />
<br />
==Solution==<br />
The maximum amount of days any given month can have is <math>31</math>, and the smallest two digit primes are <math>11, 13,</math> and <math>17</math>. There are a few different sums that can be deduced from the following numbers, which are <math>24, 30,</math> and <math>28</math>, all of which represent the three days. Therefore, since Brittany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to <math>24</math>. Similarly, Caitlin says that the other two people's uniform numbers is later, so the sum must add up to <math>30</math>. This leaves <math>28</math> as today's date. From this, Caitlin was referring to the uniform wearers <math>13</math> and <math>17</math>, telling us that her number is <math>11</math>, giving our solution as <math>\boxed{(A)=11}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_16&diff=983632014 AMC 8 Problems/Problem 162018-10-29T01:00:24Z<p>Whalefin: /* See Also */</p>
<hr />
<div>==Problem==<br />
The "Middle School Eight" basketball conference has <math>8</math> teams. Every season, each team plays every other conference team twice (home and away), and each team also plays <math>4</math> games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?<br />
<br />
<math>\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160</math><br />
==Solution==<br />
<br />
Within the conference, there are 8 teams, so there are <math>\dbinom{8}{2}=28</math> pairings of teams, and each pair must play two games, for a total of <math>28\cdot 2=56</math> games within the conference.<br />
<br />
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of <math>4\cdot 8 =32</math> games outside the conference.<br />
<br />
Therefore, the total number of games is <math>56+32 = \boxed{88}</math>, so <math>\text{(B)}</math> is our answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_16&diff=983622014 AMC 8 Problems/Problem 162018-10-29T00:59:32Z<p>Whalefin: /* See Also */</p>
<hr />
<div>==Problem==<br />
The "Middle School Eight" basketball conference has <math>8</math> teams. Every season, each team plays every other conference team twice (home and away), and each team also plays <math>4</math> games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?<br />
<br />
<math>\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160</math><br />
==Solution==<br />
<br />
Within the conference, there are 8 teams, so there are <math>\dbinom{8}{2}=28</math> pairings of teams, and each pair must play two games, for a total of <math>28\cdot 2=56</math> games within the conference.<br />
<br />
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of <math>4\cdot 8 =32</math> games outside the conference.<br />
<br />
Therefore, the total number of games is <math>56+32 = \boxed{88}</math>, so <math>\text{(B)}</math> is our answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2014|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_18&diff=981562010 AMC 8 Problems/Problem 182018-10-17T01:04:41Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
A decorative window is made up of a rectangle with semicircles at either end. The ratio of <math>AD</math> to <math>AB</math> is <math>3:2</math>. And <math>AB</math> is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?<br />
<asy><br />
import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0));<br />
dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf);<br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy><br />
<math> \textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi </math><br />
<br />
==Solution==<br />
We can set a proportion:<br />
<br />
<cmath>\dfrac{AD}{AB}=\dfrac{3}{2}</cmath><br />
<br />
We substitute <math>AB</math> with 30 and solve for AD.<br />
<br />
<cmath>\dfrac{AD}{30}=\dfrac{3}{2}</cmath><br />
<br />
<cmath>AD=45</cmath><br />
<br />
We calculate the combined area of semicircle by putting together semicircle <math>AB</math> and <math>CD</math> to get a circle with radius <math>15</math>. Thus, the area is <math>225\pi</math>. The area of the rectangle is <math>30\cdot 45=1350</math>. We calculate the ratio:<br />
<br />
<cmath>\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{\textbf{(C)}\ 6:\pi}</cmath><br />
<br />
Note that we could have solved this without the measurement of <math>30</math> inches.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2010|num-b=17|num-a=19}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_10&diff=974292012 AMC 10B Problems/Problem 102018-08-28T02:26:18Z<p>Whalefin: /* Solution */</p>
<hr />
<div>== Problem 10 ==<br />
How many ordered pairs of positive integers (M,N) satisfy the equation <math>\frac {M}{6}</math> = <math>\frac{6}{N}</math><br />
<br />
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10 </math><br />
<br />
[[2012 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Solution ==<br />
<br />
<math>\frac {M}{6}</math> = <math>\frac{6}{N}</math><br />
<br />
is a ratio; therefore, you can cross-multiply.<br />
<br />
<math>MN=36</math><br />
<br />
Now you find all the factors of 36:<br />
<br />
<math>1\times36=36</math><br />
<br />
<math>2\times18=36</math><br />
<br />
<math>3\times12=36</math><br />
<br />
<math>4\times9=36</math><br />
<br />
<math>6\times6=36</math>.<br />
<br />
Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.<br />
<br />
<math>4\cdot 2+1=9</math> <br />
<br />
<math>\boxed{\textbf{(D)}\ 9}</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2012|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_9&diff=968822006 AMC 8 Problems/Problem 92018-08-09T03:56:31Z<p>Whalefin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
What is the product of <math> \frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005} </math> ?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1002\qquad\textbf{(C)}\ 1003\qquad\textbf{(D)}\ 2005\qquad\textbf{(E)}\ 2006 </math><br />
<br />
== Solution ==<br />
After looking at the problem, we immediately notice that terms cancel out, leaving us with<br />
: <math> \frac{2006}{2}=\boxed{\textbf{(C)}\ 1003} </math>. And that is the answer.<br />
<br />
==See Also==<br />
{{AMC8 box|year=2006|num-b=8|num-a=10}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=Modular_arithmetic/Introduction&diff=92396Modular arithmetic/Introduction2018-02-27T01:29:51Z<p>Whalefin: /* Problem */</p>
<hr />
<div>'''[[Modular arithmetic]]''' is a special type of arithmetic that involves only [[integers]]. This goal of this article is to explain the basics of modular arithmetic while presenting a progression of more difficult and more interesting problems that are easily solved using modular arithmetic.<br />
<br />
<br />
<br />
== Motivation ==<br />
Let's use a clock as an example, except let's replace the <math>12</math> at the top of the clock with a <math>0</math>. <br />
<br />
<asy>picture pic;<br />
path a;<br />
a = circle((0,0), 100);<br />
draw (a);<br />
draw((0,0), linewidth(4));<br />
<br />
pair b;<br />
b = (0,100);<br />
<br />
for (int i = 0; i < 12; ++i)<br />
{<br />
label (pic, (string) i, b);<br />
b = rotate(-30,(0,0)) * b;<br />
}<br />
pic = scale(0.8) * pic;<br />
add(pic);<br />
<br />
draw ((0,0) -- 50*dir(90));<br />
draw ((0,0) -- 70*dir(90));</asy><br />
<br />
Starting at noon, the hour hand points in order to the following:<br />
<br />
<math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math><br />
<br />
<br />
This is the way in which we count in '''modulo 12'''. When we add <math>1</math> to <math>11</math>, we arrive back at <math>0</math>. The same is true in any other [[modulus]] (modular arithmetic system). In modulo <math>5</math>, we [[counting | count]]<br />
<br />
<br />
<math>0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, \ldots</math><br />
<br />
<br />
We can also count backwards in modulo 5. Any time we subtract 1 from 0, we get 4. So, the integers from <math>-12</math> to <math>0</math>, when written in modulo 5, are<br />
<br />
<br />
<math>3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0,</math><br />
<br />
<br />
where <math>-12</math> is the same as <math>3</math> in modulo 5. Because all integers can be expressed as <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math> in modulo 5, we give these integers their own name: the '''[[residue class]]es''' modulo 5. In general, for a natural number <math>n</math> that is greater than 1, the modulo <math>n</math> residues are the integers that are [[whole number | whole numbers]] less than <math>n</math>:<br />
<br />
<br />
<math>0, 1, 2, \ldots, n-1. </math><br />
<br />
<br />
This just relates each integer to its [[remainder]] from the [[Division Theorem]]. While this may not seem all that useful at first, counting in this way can help us solve an enormous array of [[number theory]] problems much more easily!<br />
<br />
==Residue==<br />
We say that <math>a</math> is the modulo-<math>m</math> <b>residue</b> of <math>n</math> when <math>n\equiv a\pmod m</math>, and <math>0\le a<m</math>.<br />
<br />
== Congruence ==<br />
There is a mathematical way of saying that all of the integers are the same as one of the modulo 5 residues. For instance, we say that 7 and 2 are '''[[congruent]]''' modulo 5. We write this using the symbol <math>\equiv</math>: In other words, this means in base 5, these integers have the same residue modulo 5:<br />
<br />
<center><math>2\equiv 7\equiv 12 \pmod{5}. </math></center><br />
<br />
<br />
The '''(mod 5)''' part just tells us that we are working with the integers modulo 5. In modulo 5, two integers are congruent when their difference is a [[multiple]] of 5. <br />
In general, two integers <math>a</math> and <math>b</math> are congruent modulo <math>n</math> when <math>a - b</math> is a multiple of <math>n</math>. In other words, <math>a \equiv b \pmod{n}</math> when <math>\frac{a-b}{n}</math> is an integer. Otherwise, <math>a \not\equiv b \pmod{n}</math>, which means that <math>a</math> and <math>b</math> are '''not congruent''' modulo <math>n</math>.<br />
<br />
<br />
<br />
=== Examples ===<br />
* <math>31 \equiv 1 \pmod{10}</math> because <math>31 - 1 = 30</math> is a multiple of <math>10</math>.<br />
<br />
<br />
* <math>43 \equiv 22 \pmod{7}</math> because <math>\frac{43 - 22}{7} = \frac{21}{7} = 3</math>, which is an integer.<br />
<br />
<br />
* <math>8 \not\equiv -8 \pmod{3}</math> because <math>8 - (-8) = 16</math>, which is not a multiple of <math>3</math>.<br />
<br />
<br />
* <math>91 \not\equiv 18 \pmod{6}</math> because <math>\frac{91 - 18}{6} = \frac{73}{6}</math>, which is not an integer.<br />
<br />
<br />
<br />
=== Sample Problem ===<br />
Find the modulo <math>4</math> residue of <math>311</math>.<br />
==== Solution: ====<br />
Since <math>311 \div 4 = 77</math> R <math>3</math>, we know that <br />
<center><math>311 \equiv 3 \pmod{4}</math></center><br />
and <math>3</math> is the modulo <math>4</math> residue of <math>311</math>.<br />
<br />
==== Another Solution: ====<br />
Since <math> 311 = 300 + 11 </math>, we know that<br />
<center><math>311 \equiv 0+11 \pmod{4}</math></center><br />
We can now solve it easily<br />
<center><math>11 \equiv 3 \pmod{4}</math></center><br />
and <math>3</math> is the modulo <math>4</math> residue of <math>311</math><br />
<br />
== Making Computation Easier ==<br />
We don't always need to perform tedious computations to discover solutions to interesting problems. If all we need to know about are remainders when integers are divided by <math>n</math>, then we can work directly with those remainders in modulo <math>n</math>. This can be more easily understood with a few examples.<br />
<br />
=== Addition ===<br />
==== Problem ====<br />
Suppose we want to find the [[units digit]] of the following [[sum]]:<br />
<br />
<br />
<center><math>2403 + 791 + 688 + 4339.</math></center><br />
<br />
<br />
We could find their sum, which is <math>8221</math>, and note that the units digit is <math>1</math>. However, we could find the units digit with far less calculation.<br />
<br />
==== Solution ====<br />
We can simply add the units digits of the addends:<br />
<br />
<br />
<center><math>3 + 1 + 8 + 9 = 21.</math></center><br />
<br />
<br />
The units digit of this sum is <math>1</math>, which ''must'' be the same as the units digit of the four-digit sum we computed earlier.<br />
<br />
==== Why we only need to use remainders ====<br />
We can rewrite each of the integers in terms of multiples of <math>10</math> and remainders:<br><br />
<math>2403 = 240 \cdot 10 + 3</math><br><br />
<math>791 = 79 \cdot 10 + 1</math><br><br />
<math>688 = 68 \cdot 10 + 8</math><br><br />
<math>4339 = 433 \cdot 10 + 9</math>.<br><br />
When we add all four integers, we get<br />
<br />
<br />
<center><math> (240 \cdot 10 + 3) + (79 \cdot 10 + 1) + (68 \cdot 10 + 8) + (433 \cdot 10 + 9)</math></center><br />
<br />
<center><math>= (240 + 79 + 68 + 433) \cdot 10 + (3 + 1 + 8 + 9)</math></center><br />
<br />
<br />
At this point, we already see the units digits grouped apart and added to a multiple of <math>10</math> (which will not affect the units digit of the sum):<br />
<br />
<center><math>= 820 \cdot 10 + 21 = 8200 + 21 = 8221</math>.</center><br />
<br />
==== Solution using modular arithmetic ====<br />
Now let's look back at this solution, using modular arithmetic from the start. Note that<br><br />
<math>2403 \equiv 3 \pmod{10}</math><br><br />
<math>791 \equiv 1 \pmod{10}</math><br><br />
<math>688 \equiv 8 \pmod{10}</math><br><br />
<math>4339 \equiv 9 \pmod{10}</math><br><br />
Because we only need the modulo <math>10</math> residue of the sum, we add just the residues of the summands:<br />
<br />
<center><math>2403 + 791 + 688 + 4339 \equiv 3 + 1 + 8 + 9 \equiv 21 \equiv 1 \pmod{10},</math></center><br />
so the units digit of the sum is just <math>1</math>.<br />
<br />
==== Addition rule ====<br />
In general, when <math>a, b, c</math>, and <math>d</math> are integers and <math>m</math> is a positive integer such that<br><br />
<center><math>a \equiv c \pmod{m} </math></center><br />
<center><math>b \equiv d \pmod{m} </math></center><br />
the following is always true:<br />
<br />
<center><math>a + b \equiv c + d \pmod{m} </math>.</center><br />
And as we did in the problem above, we can apply more pairs of equivalent integers to both sides, just repeating this simple principle.<br />
<br />
====Proof of the addition rule====<br />
<br />
Let <math>a-c=m\cdot k</math>, and <math>b-d=m\cdot l</math> where <math>l</math> and <math>k</math> are integers. <br />
Adding the two equations we get:<br />
<cmath><br />
\begin{eqnarray*}<br />
mk+ml&=&(a-c)+(b-d)\\<br />
m(k+l)&=&(a+b)-(c+d)<br />
\end{eqnarray*}<br />
</cmath><br />
<br />
Which is equivalent to saying <math>a+b\equiv c+d\pmod{m}</math><br />
<br />
=== Subtraction ===<br />
The same shortcut that works with addition of remainders works also with subtraction.<br />
<br />
==== Problem ====<br />
Find the remainder when the difference between <math>60002</math> and <math>601</math> is divided by <math>6</math>.<br />
<br />
==== Solution ====<br />
Note that <math>60002 = 10000 \cdot 6 + 2</math> and <math>601 = 100 \cdot 6 + 1</math>. So,<br><br />
<math>60002 \equiv 2 \pmod{6}</math><br><br />
<math>601 \equiv 1 \pmod{6}</math><br><br />
Thus,<br />
<center><math>60002 - 601 \equiv 2 - 1 \equiv 1 \pmod{6}, </math></center><br />
so 1 is the remainder when the difference is divided by <math>6</math>. (Perform the subtraction yourself, divide by <math>6</math>, and see!)<br />
<br />
==== Subtraction rule ====<br />
When <math>a, b, c</math>, and <math>d</math> are integers and <math>m</math> is a positive integer such that<br><br />
<center><math>a \equiv c \pmod{m} </math></center><br />
<center><math>b \equiv d \pmod{m} </math></center><br />
the following is always true:<br />
<br />
<center><math>a - b \equiv c - d \pmod{m} </math></center>.<br />
<br />
<br />
=== Multiplication ===<br />
Modular arithmetic provides an even larger advantage when multiplying than when adding or subtracting. Let's take a look at a problem that demonstrates the point.<br />
<br />
==== Problem ====<br />
Jerry has <math>44</math> boxes of soda in his truck. The cans of soda in each box are packed oddly so that there are <math>113</math> cans of soda in each box. Jerry plans to pack the sodas into cases of <math>12</math> cans to sell. After making as many complete cases as possible, how many sodas will Jerry have leftover?<br />
<br />
==== Solution ====<br />
First, we note that this [[word problem]] is asking us to find the remainder when the product <math>44 \cdot 113</math> is divided by <math>12</math>.<br />
<br />
Now, we can write each <math>44</math> and <math>113</math> in terms of multiples of <math>12</math> and remainders:<br><br />
<math>44 = 3 \cdot 12 + 8</math><br><br />
<math>113 = 9 \cdot 12 + 5</math><br><br />
This gives us a nice way to view their product:<br />
<br />
<br />
<cmath>44 \cdot 113 = (3 \cdot 12 + 8)(9 \cdot 12 + 5)</cmath><br />
Using [[FOIL]], we get that this equals<br />
<cmath>(3 \cdot 9) \cdot 12^2 + (3 \cdot 5 + 8 \cdot 9) \cdot 12 + (8 \cdot 5)</cmath><br />
<br />
<br />
We can already see that each part of the product is a multiple of <math>12</math>, except the product of the remainders when each <math>44</math> and <math>113</math> are divided by 12. That part of the product is <math>8 \cdot 5 = 40</math>, which leaves a remainder of <math>4</math> when divided by <math>12</math>. So, Jerry has <math>4</math> sodas leftover after making as many cases of <math>12</math> as possible.<br />
<br />
==== Solution using modular arithmetic ====<br />
First, we note that<br><br />
<math>44 \equiv 8 \pmod{12}</math><br><br />
<math>113 \equiv 5 \pmod{12}</math><br><br />
Thus,<br />
<center><math>44 \cdot 113 \equiv 8 \cdot 5 \equiv 40 \equiv 4 \pmod{12},</math></center><br />
meaning there are <math>4</math> sodas leftover. Yeah, that was much easier.<br />
<br />
==== Multiplication rule ====<br />
When <math>a, b, c</math>, and <math>d</math> are integers and <math>m</math> is a positive integer such that<br><br />
<center><math>a \equiv c \pmod{m} </math></center><br />
<center><math>b \equiv d \pmod{m} </math></center><br />
The following is always true:<br />
<br />
<center><math>a \cdot b \equiv c \cdot d \pmod{m} </math>.</center><br />
<br />
<br />
=== Exponentiation ===<br />
Since [[exponentiation]] is just repeated multiplication, it makes sense that modular arithmetic would make many problems involving exponents easier. In fact, the advantage in computation is even larger and we explore it a great deal more in the [[intermediate modular arithmetic]] article.<br />
<br />
Note to everybody: Exponentiation is very useful as in the following problem:<br />
<br />
==== Problem #1====<br />
What is the last digit of <math>(...((7)^7)^7)...)^7</math> if there are 1000 7s as exponents and only one 7 in the middle?<br />
<br />
We could solve this problem using mods. This can also be stated as <math>7^{7^{1000}}</math>. After that, we see that 7 is congruent to -1 in mod 4, so we can use this fact to replace the 7s with -1s, because 7 has a pattern of repetitive period 4 for the units digit. <math>(-1)^{1000}</math> is simply 1, so therefore <math>7^1=7</math>, which really is the last digit.<br />
<br />
==== Problem #2====<br />
What are the tens and units digits of <math>7^{1942}</math>?<br />
<br />
We could (in theory) solve this problem by trying to compute <math>7^{1942}</math>, but this would be extremely time-consuming. Moreover, it would give us much more information than we need. Since we want only the tens and units digits of the number in question, it suffices to find the remainder when the number is divided by <math>100</math>. In other words, all of the information we need can be found using arithmetic mod <math>100</math>.<br />
<br />
We begin by writing down the first few powers of <math>7</math> mod <math>100</math>:<br />
<br />
<math>7, 49, 43, 1, 7, 49, 43, 1, \ldots</math><br />
<br />
A pattern emerges! We see that <math>7^4 = 2401 \equiv 1</math> (mod <math>100</math>). So for any positive integer <math>k</math>, we have <math>7^{4k} = (7^4)^k \equiv 1^k \equiv 1</math> (mod <math>100</math>). In particular, we can write<br />
<br />
<math>7^{1940} = 7^{4 \cdot 485} \equiv 1</math> (mod <math>100</math>).<br />
<br />
By the "multiplication" property above, then, it follows that<br />
<br />
<math>7^{1942} = 7^{1940} \cdot 7^2 \equiv 1 \cdot 7^2 \equiv 49</math> (mod <math>100</math>).<br />
<br />
Therefore, by the definition of congruence, <math>7^{1942}</math> differs from <math>49</math> by a multiple of <math>100</math>. Since both integers are positive, this means that they share the same tens and units digits. Those digits are <math>4</math> and <math>9</math>, respectively.<br />
<br />
==== Problem #3====<br />
<br />
Can you find a number that is both a multiple of <math>2</math> but not a multiple of <math>4</math> and a perfect square?<br />
<br />
No, you cannot. Rewriting the question, we see that it asks us to find an integer <math>n</math> that satisfies <math>4n+2=x^2</math>.<br />
<br />
Taking mod <math>4</math> on both sides, we find that <math>x^2\equiv 2\pmod{4}</math>. Now, all we are missing is proof that no matter what <math>x</math> is, <math>x^2</math> will never be a multiple of <math>4</math> plus <math>2</math>, so we work with cases:<br />
<br />
<math>x\equiv 0\pmod{4}\implies x^2\equiv 0\pmod{4}</math><br />
<br />
<math>x\equiv 1\pmod{4}\implies x^2\equiv 1\pmod{4}</math><br />
<br />
<math>x\equiv 2\pmod{4}\implies x^2\equiv 4\equiv 0\pmod{4}</math><br />
<br />
<math>x\equiv 3\pmod{4}\implies x^2\equiv 9\equiv 1\pmod{4}</math><br />
<br />
This assures us that it is impossible to find such a number.<br />
<br />
== Summary of Useful Facts ==<br />
<br />
Consider four integers <math>{a},{b},{c},{d}</math> and a positive integer <math>{m}</math> such that <math>a\equiv b\pmod {m}</math> and <math>c\equiv d\pmod {m}</math>. In modular arithmetic, the following [[identity | identities]] hold:<br />
<br />
* Addition: <math>a+c\equiv b+d\pmod {m}</math>.<br />
* Subtraction: <math>a-c\equiv b-d\pmod {m}</math>.<br />
* Multiplication: <math>ac\equiv bd\pmod {m}</math>.<br />
* Division: <math>\frac{a}{e}\equiv \frac{b}{e}\pmod {\frac{m}{\gcd(m,e)}}</math>, where <math>e</math> is a positive integer that divides <math>{a}</math> and <math>b</math>.<br />
* Exponentiation: <math>a^e\equiv b^e\pmod {m}</math> where <math>e</math> is a positive integer.<br />
<br />
<br />
<br />
== Applications of Modular Arithmetic ==<br />
Modular arithmetic is an extremely flexible problem solving tool. The following topics are just a few applications and extensions of its use:<br />
* [[Divisibility rules]]<br />
* [[Linear congruence | Linear congruences]]<br />
<br />
<br />
<br />
== Resources ==<br />
* The AoPS [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br />
* The AoPS [http://www.artofproblemsolving.com/School/courseinfo.php?course_id=intro:numbertheory Introduction to Number Theory Course]. Thousands of students have learned more about modular arithmetic and [[problem solving]] from this 12 week class.<br />
<br />
== See also ==<br />
* [[Intermediate modular arithmetic]]<br />
* [[Olympiad modular arithmetic]]<br />
<br />
<br />
[[Category:Introductory Mathematics Topics]]</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_6&diff=917322018 AMC 10B Problems/Problem 62018-02-17T01:40:50Z<p>Whalefin: /* Solution */</p>
<hr />
<div>A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?<br />
<br />
<math>\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math><br />
<br />
== Solution ==<br />
<br />
Notice that the only two ways such that more than <math>2</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Notice that only the first two draws are important, it doesn't matter what number we get third because no matter what combination of <math>3</math> numbers is picked, the sum will always be greater than 5. Also, note that it is necessary to draw a <math>1</math> in order to have 3 draws, otherwise <math>5</math> will be attainable in two or less draws. So the probability of getting a <math>1</math> is <math>\frac{1}{5}</math>. It is necessary to pull either a <math>2</math> or <math>3</math> on the next draw and the probability of that is <math>\frac{1}{2}</math>. But, the order of the draws can be switched so we get:<br />
<br />
<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math><br />
<br />
By: Soccer_JAMS<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_3&diff=899882016 AMC 10B Problems/Problem 32018-01-26T02:10:03Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>x=-2016</math>. What is the value of <math>\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x</math> ?<br />
<br />
<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math><br />
<br />
==Solution==<br />
Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math><br />
<br />
becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==Solution 2==<br />
<br />
Solution by e_power_pi_times_i<br />
<br />
Substitute <math>-y = x = -2016</math> into the equation. Now, it is <math>\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y</math>. Since <math>y = 2016</math>, it is a positive number, so <math>|y| = y</math>. Now the equation is <math>\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y</math>. This further simplifies to <math>2y-y+y = 2y</math>, so the answer is <math>\boxed{\textbf{(D)}\ 4032}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_3&diff=899872016 AMC 10B Problems/Problem 32018-01-26T02:09:38Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>x=-2016</math>. What is the value of <math>\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x</math> ?<br />
<br />
<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math><br />
<br />
==Solution==<br />
Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math><br />
<br />
becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>.<br />
<br />
==Solution 2==<br />
<br />
Solution by e_power_pi_times_i<br />
<br />
Substitute <math>-y = x = -2016</math> into the equation. Now, it is <math>\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y</math>. Since <math>y = 2016</math>, it is a positive number, so <math>|y| = y</math>. Now the equation is <math>\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y</math>. This further simplifies to <math>2y-y+y = 2y</math>, so the answer is <math>\boxed{\textbf{(D)}\ 4032}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_3&diff=899862016 AMC 10B Problems/Problem 32018-01-26T02:08:57Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>x=-2016</math>. What is the value of <math>\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x</math> ?<br />
<br />
<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math><br />
<br />
==Solution==<br />
Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math><br />
<br />
becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>.<br />
<br />
==Solution 2==<br />
<br />
Solution by e_power_pi_times_i<br />
<br />
Substitute <math>-y = x = -2016</math> into the equation. Now, it is <math>\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y</math>. Since <math>y = 2016</math>, it is a positive number, so <math>|y| = y</math>. Now the equation is <math>\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y</math>. This further simplifies to <math>2y-y+y = 2y</math>, so the answer is <math>\boxed{\textbf{(D)} 4032}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_2&diff=899852016 AMC 10B Problems/Problem 22018-01-26T02:08:16Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math><br />
<br />
==Solution 1==<br />
<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>.<br />
<br />
==Solution 2==<br />
We can replace <math>2</math> and <math>4</math> with <math>a</math> and <math>b</math> respectively. Then substituting with <math>n</math> and <math>m</math> we can get <math>\dfrac{a^3b^2}{b^3a^2}=\dfrac{a}{b}</math> and substitute to get <math>\dfrac{2}{4}=\boxed{\dfrac{1}{2}}</math> which is <math>\boxed{\textbf{(B)}}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_1&diff=899842016 AMC 10B Problems/Problem 12018-01-26T02:07:30Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \tfrac{1}{2}</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math><br />
<br />
==Solution==<br />
<br />
Factorizing the numerator, <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> then becomes <math>\frac{\frac{5}{2}}{a^{2}}</math> which is equal to <math>\frac{5}{2}\cdot 2^2</math> which is <math>\boxed{\textbf{(D) }10}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|before=-|num-a=2}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_11&diff=893902012 AMC 10A Problems/Problem 112018-01-03T22:28:24Z<p>Whalefin: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
Externally tangent circles with centers at points <math>A</math> and <math>B</math> have radii of lengths <math>5</math> and <math>3</math>, respectively. A line externally tangent to both circles intersects ray <math>AB</math> at point <math>C</math>. What is <math>BC</math>?<br />
<br />
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4 </math><br />
<br />
== Solution ==<br />
<br />
<center><asy><br />
unitsize(3.5mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375));<br />
path a=Circle(A,5);<br />
path b=Circle(B,3);<br />
draw(a); draw(b);<br />
draw(C--D);<br />
draw(A--C);<br />
draw(A--D);<br />
draw(B--E);<br />
<br />
pair[] ps={A,B,C,D,E};<br />
dot(ps);<br />
<br />
label("$A$",A,N); label("$B$",B,N); label("$C$",C,N);<br />
label("$D$",D,SE); label("$E$",E,SE);<br />
label("$5$",(A--D),SW);<br />
label("$3$",(B--E),SW);<br />
label("$8$",(A--B),N);<br />
label("$x$",(C--B),N);<br />
<br />
</asy></center><br />
<br />
Let <math>D</math> and <math>E</math> be the points of tangency on circles <math>A</math> and <math>B</math> with line <math>CD</math>. <math>AB=8</math>. Also, let <math>BC=x</math>. As <math>\angle ADC</math> and <math>\angle BEC</math> are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share <math>\angle ACD</math>, <math>\triangle ADC \sim \triangle BEC</math>. From this we can get a proportion.<br />
<br />
<math>\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}</math><br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2012|ab=A|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_19&diff=882182011 AMC 8 Problems/Problem 192017-11-12T03:56:42Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
How many rectangles are in this figure?<br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,I,J,K,L;<br />
A=(0,0);<br />
B=(20,0);<br />
C=(20,20);<br />
D=(0,20);<br />
draw(A--B--C--D--cycle);<br />
E=(-10,-5);<br />
F=(13,-5);<br />
G=(13,5);<br />
H=(-10,5);<br />
draw(E--F--G--H--cycle);<br />
I=(10,-20);<br />
J=(18,-20);<br />
K=(18,13);<br />
L=(10,13);<br />
draw(I--J--K--L--cycle);</asy><br />
<br />
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 </math><br />
<br />
==Solution==<br />
The figure can be divided into <math>7</math> sections. The number of rectangles with just one section is <math>3.</math> The number of rectangles with two sections is <math>5.</math> There are none with only three sections. The number of rectangles with four sections is <math>3.</math> <br />
<br />
<math>3+5+3=\boxed{\textbf{(D)}\ 11}</math><br />
<br />
==See Also==<br />
{{AMC8 box|year=2011|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_14&diff=871952008 AMC 10B Problems/Problem 142017-08-26T18:01:57Z<p>Whalefin: </p>
<hr />
<div>==Problem==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Triangle <math>OAB</math> has <math>O=(0,0)</math>, <math>B=(5,0)</math>, and <math>A</math> in the first quadrant. In addition, <math>\angle ABO=90^\circ</math> and <math>\angle AOB=30^\circ</math>. Suppose that <math>OA</math> is rotated <math>90^\circ</math> counterclockwise about <math>O</math>. What are the coordinates of the image of <math>A</math>? <!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
<math><br />
\mathrm{(A)}\ \left( - \frac {10}{3}\sqrt {3},5\right)<br />
\qquad<br />
\mathrm{(B)}\ \left( - \frac {5}{3}\sqrt {3},5\right)<br />
\qquad<br />
\mathrm{(C)}\ \left(\sqrt {3},5\right)<br />
\qquad<br />
\mathrm{(D)}\ \left(\frac {5}{3}\sqrt {3},5\right)<br />
\qquad<br />
\mathrm{(E)}\ \left(\frac {10}{3}\sqrt {3},5\right)<br />
</math><br />
<br />
==Solution 1==<br />
<br />
Since <math>\angle ABO=90^\circ</math>, and <math>\angle AOB=30^\circ</math>, we know that this triangle is one of the [[Special Right Triangles]].<br />
<br />
We also know that <math>A</math> is <math>(5,0)</math>, so <math>A</math> lies on the x-axis. Therefore, <math>OA = 5</math>.<br />
<br />
Then, since we know that this is a Special Right Triangle, we can use the proportion <cmath>\frac{5}{\sqrt 3}=\frac{x}{1}</cmath> to find <math>AB</math>.<br />
<br />
We find that <cmath>AB=\frac{5\sqrt 3}{3}</cmath><br />
<br />
That means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>.<br />
<br />
Rotate this triangle <math>90^\circ</math> counterclockwise around <math>O</math>, and you will find that <math>A</math> will end up in the second quadrant with the coordinates<br />
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) \text{or B.}}</math>.<br />
<br />
==Solution 2==<br />
<br />
As <math>\angle ABO=90^\circ</math> and <math>A</math> in the first quadrant, we know that the <math>x</math> coordinate of <math>A</math> is <math>5</math>. We now need to pick a positive <math>y</math> coordinate for <math>A</math> so that we'll have <math>\angle AOB=30^\circ</math>. <br />
<br />
By the [[Pythagorean theorem]] we have <math>AO^2 = AB^2 + BO^2 = AB^2 + 25</math>. <br />
<br />
By the definition of [[sine]], we have <math>\frac{AB}{AO} = \sin AOB = \sin 30^\circ = \frac 12</math>, hence <math>AO=2\cdot AB</math>.<br />
<br />
Substituting into the previous equation, we get <math>AB^2 = \frac{25}3</math>, hence <math>AB=\frac{5\sqrt 3}3</math>.<br />
<br />
This means that the coordinates of <math>A</math> are <math>\left(5,\frac{5\sqrt 3}3\right)</math>. <br />
<br />
After we rotate <math>A</math> <math>90^\circ</math> counterclockwise about <math>O</math>, it will get into the second quadrant and have the coordinates <br />
<math>\boxed{ \left( -\frac{5\sqrt 3}3, 5\right) }</math>.<br />
So the answer is <math>\boxed{\text{B}}</math>.<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_20&diff=866902010 AMC 10A Problems/Problem 202017-07-29T22:32:25Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
A fly trapped inside a cubical box with side length <math>1</math> meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?<br />
<br />
<math>\textbf{(A)}\ 4+4\sqrt{2} \qquad \textbf{(B)}\ 2+4\sqrt{2}+2\sqrt{3} \qquad \textbf{(C)}\ 2+3\sqrt{2}+3\sqrt{3} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad \textbf{(E)}\ 3\sqrt{2}+5\sqrt{3}</math><br />
<br />
==Solution 1==<br />
The distance of an interior diagonal in this cube is <math>\sqrt{3}</math> and the distance of a diagonal on one of the square faces is <math>\sqrt{2}</math>. It is not possible for the fly to travel any interior diagonal twice, as then it would visit a corner more than once. So, the final sum can have at most <math>4</math> as the coefficient of <math>\sqrt{3}</math>. The other 4 paths taken can be across a diagonal on one of the faces (try to find a such path!), so the maximum distance traveled is <math>\textbf{(D)}\ 4\sqrt{2}+4\sqrt{3}</math>.<br />
==Solution 2==<br />
<br />
The path of the fly consists of eight line segments, where each line segment goes from one corner to another corner. The distance of each such line segment is 1, <math>\sqrt{2}</math>, or <math>\sqrt{3}</math>.<br />
<br />
The only way to obtain a line segment of length <math>\sqrt{3}</math> is to go from one corner of the cube to the opposite corner. Since the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length <math>\sqrt{3}</math>. Hence, the length of the fly's path can be at most <math>\boxed{4 \sqrt{3} + 4 \sqrt{2}\text{, or D.}}</math>. This length can be achieved by taking the path<br />
<cmath>A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A.</cmath><br />
<br />
<asy><br />
import graph;<br />
<br />
unitsize(2 cm);<br />
<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
draw((1,0)--(1.3,0.3));<br />
draw((1,1)--(1.3,1.3));<br />
draw((0,1)--(0.3,1.3));<br />
draw((1.3,0.3)--(1.3,1.3)--(0.3,1.3));<br />
draw((0,0)--(0.3,0.3),dashed);<br />
draw((0.3,0.3)--(1.3,0.3),dashed);<br />
draw((0.3,0.3)--(0.3,1.3),dashed);<br />
<br />
label("$A$", (0,0), SW);<br />
label("$B$", (1,0), SE);<br />
label("$C$", (1.3,0.3), E);<br />
label("$D$", (0.3,0.3), SE);<br />
label("$E$", (0,1), W);<br />
label("$F$", (1,1), E);<br />
label("$G$", (1.3,1.3), NE);<br />
label("$H$", (0.3,1.3), N);<br />
</asy><br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|num-b=19|num-a=21|ab=A}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_20&diff=866892010 AMC 10A Problems/Problem 202017-07-29T22:31:56Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
A fly trapped inside a cubical box with side length <math>1</math> meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?<br />
<br />
<math>\textbf{(A)}\ 4+4\sqrt{2} \qquad \textbf{(B)}\ 2+4\sqrt{2}+2\sqrt{3} \qquad \textbf{(C)}\ 2+3\sqrt{2}+3\sqrt{3} \qquad \textbf{(D)}\ 4\sqrt{2}+4\sqrt{3} \qquad \textbf{(E)}\ 3\sqrt{2}+5\sqrt{3}</math><br />
<br />
==Solution 1==<br />
The distance of an interior diagonal in this cube is <math>\sqrt{3}</math> and the distance of a diagonal on one of the square faces is <math>\sqrt{2}</math>. It is not possible for the fly to travel any interior diagonal twice, as then it would visit a corner more than once. So, the final sum can have at most <math>4</math> as the coefficient of <math>\sqrt{3}</math>. The other 4 paths taken can be across a diagonal on one of the faces (try to find a such path!), so the maximum distance traveled is <math>\textbf{(D)}\ 4\sqrt{2}+4\sqrt{3}</math>.<br />
==Solution 2==<br />
<br />
The path of the fly consists of eight line segments, where each line segment goes from one corner to another corner. The distance of each such line segment is 1, <math>\sqrt{2}</math>, or <math>\sqrt{3}</math>.<br />
<br />
The only way to obtain a line segment of length <math>\sqrt{3}</math> is to go from one corner of the cube to the opposite corner. Since the fly visits each corner exactly once, it cannot traverse such a line segment twice. Also, the cube has exactly four such diagonals, so the path of the fly can contain at most four segments of length <math>\sqrt{3}</math>. Hence, the length of the fly's path can be at most <math>\boxed{4 \sqrt{3} + 4 \sqrt{2}\text{, or D.}}</math>. This length can be achieved by taking the path<br />
<cmath>A \rightarrow G \rightarrow B \rightarrow H \rightarrow C \rightarrow E \rightarrow D \rightarrow F \rightarrow A.</cmath><br />
Hence, the answer is <math>\text{(D)} 4 \sqrt{3} + 4 \sqrt{2}</math>.<br />
<br />
<asy><br />
import graph;<br />
<br />
unitsize(2 cm);<br />
<br />
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);<br />
draw((1,0)--(1.3,0.3));<br />
draw((1,1)--(1.3,1.3));<br />
draw((0,1)--(0.3,1.3));<br />
draw((1.3,0.3)--(1.3,1.3)--(0.3,1.3));<br />
draw((0,0)--(0.3,0.3),dashed);<br />
draw((0.3,0.3)--(1.3,0.3),dashed);<br />
draw((0.3,0.3)--(0.3,1.3),dashed);<br />
<br />
label("$A$", (0,0), SW);<br />
label("$B$", (1,0), SE);<br />
label("$C$", (1.3,0.3), E);<br />
label("$D$", (0.3,0.3), SE);<br />
label("$E$", (0,1), W);<br />
label("$F$", (1,1), E);<br />
label("$G$", (1.3,1.3), NE);<br />
label("$H$", (0.3,1.3), N);<br />
</asy><br />
<br />
== See also ==<br />
{{AMC10 box|year=2010|num-b=19|num-a=21|ab=A}}<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_21&diff=865172004 AMC 10A Problems/Problem 212017-07-23T16:51:55Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is <math>\frac{8}{13}</math> of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: <math>\pi</math> radians is <math>180</math> degrees.)<br />
<br />
<center><asy>fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7));<br />
fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7));<br />
fill((-20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7));<br />
fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7));<br />
fill((0,10)..(-10,0)--(10,0)..cycle,white);<br />
fill((0,-10)..(-10,0)--(10,0)..cycle,white);<br />
fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7));<br />
fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7));<br />
draw(Circle((0,0),10),linewidth(0.7));<br />
draw(Circle((0,0),20),linewidth(0.7));<br />
draw(Circle((0,0),30),linewidth(0.7));<br />
draw((-28,-21)--(28,21),linewidth(0.7));<br />
draw((-28,21)--(28,-21),linewidth(0.7));</asy></center><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4} </math><br />
<br />
==Solution 1==<br />
Let the area of the shaded region be <math>S</math>, the area of the unshaded region be <math>U</math>, and the acute angle that is formed by the two lines be <math>\theta</math>. We can set up two equations between <math>S</math> and <math>U</math>:<br />
<br />
<math>S+U=9\pi</math><br />
<br />
<math>S=\dfrac{8}{13}U</math><br />
<br />
Thus <math>\dfrac{21}{13}U=9\pi</math>, and <math>U=\dfrac{39\pi}{7}</math>, and thus <math>S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}</math>.<br />
<br />
Now we can make a formula for the area of the shaded region in terms of <math>\theta</math>:<br />
<br />
<math>\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math><br />
<br />
Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math><br />
<br />
==Solution 2==<br />
<br />
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of <math>\theta</math>.<br />
<br />
<math>\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi</math>.<br />
<br />
So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>.<br />
<br />
Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for theta!<br />
<br />
<br />
<math>3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta</math>.<br />
<br />
Simplifying, we get <math>63\theta=9\pi\implies \theta=\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2004|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_21&diff=865162004 AMC 10A Problems/Problem 212017-07-23T16:51:01Z<p>Whalefin: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is <math>\frac{8}{13}</math> of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: <math>\pi</math> radians is <math>180</math> degrees.)<br />
<br />
<center><asy>fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7));<br />
fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7));<br />
fill((-20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7));<br />
fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7));<br />
fill((0,10)..(-10,0)--(10,0)..cycle,white);<br />
fill((0,-10)..(-10,0)--(10,0)..cycle,white);<br />
fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7));<br />
fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7));<br />
draw(Circle((0,0),10),linewidth(0.7));<br />
draw(Circle((0,0),20),linewidth(0.7));<br />
draw(Circle((0,0),30),linewidth(0.7));<br />
draw((-28,-21)--(28,21),linewidth(0.7));<br />
draw((-28,21)--(28,-21),linewidth(0.7));</asy></center><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4} </math><br />
<br />
==Solution 1==<br />
Let the area of the shaded region be <math>S</math>, the area of the unshaded region be <math>U</math>, and the acute angle that is formed by the two lines be <math>\theta</math>. We can set up two equations between <math>S</math> and <math>U</math>:<br />
<br />
<math>S+U=9\pi</math><br />
<br />
<math>S=\dfrac{8}{13}U</math><br />
<br />
Thus <math>\dfrac{21}{13}U=9\pi</math>, and <math>U=\dfrac{39\pi}{7}</math>, and thus <math>S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}</math>.<br />
<br />
Now we can make a formula for the area of the shaded region in terms of <math>\theta</math>:<br />
<br />
<math>\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math><br />
<br />
Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math><br />
<br />
==Solution 2==<br />
<br />
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of <math>\theta</math>.<br />
<br />
<math>\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi</math>.<br />
<br />
So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>.<br />
<br />
Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for theta!<br />
<br />
<br />
<math>3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta</math>.<br />
<br />
Simplifying, we get <math>63\theta=9\pi\implies \theta=\boxed{(B)\frac{\pi}{7}}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2004|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10A_Problems/Problem_21&diff=865152004 AMC 10A Problems/Problem 212017-07-23T16:50:44Z<p>Whalefin: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is <math>\frac{8}{13}</math> of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: <math>\pi</math> radians is <math>180</math> degrees.)<br />
<br />
<center><asy>fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7));<br />
fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7));<br />
fill((-20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((20,0)..(0,20)--(0,-20)..cycle,white);<br />
fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7));<br />
fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7));<br />
fill((0,10)..(-10,0)--(10,0)..cycle,white);<br />
fill((0,-10)..(-10,0)--(10,0)..cycle,white);<br />
fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7));<br />
fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7));<br />
draw(Circle((0,0),10),linewidth(0.7));<br />
draw(Circle((0,0),20),linewidth(0.7));<br />
draw(Circle((0,0),30),linewidth(0.7));<br />
draw((-28,-21)--(28,21),linewidth(0.7));<br />
draw((-28,21)--(28,-21),linewidth(0.7));</asy></center><br />
<br />
<math> \mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4} </math><br />
<br />
==Solution 1==<br />
Let the area of the shaded region be <math>S</math>, the area of the unshaded region be <math>U</math>, and the acute angle that is formed by the two lines be <math>\theta</math>. We can set up two equations between <math>S</math> and <math>U</math>:<br />
<br />
<math>S+U=9\pi</math><br />
<br />
<math>S=\dfrac{8}{13}U</math><br />
<br />
Thus <math>\dfrac{21}{13}U=9\pi</math>, and <math>U=\dfrac{39\pi}{7}</math>, and thus <math>S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}</math>.<br />
<br />
Now we can make a formula for the area of the shaded region in terms of <math>\theta</math>:<br />
<br />
<math>\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}</math><br />
<br />
Thus <math>3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math><br />
<br />
==Solution 2==<br />
<br />
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of <math>\theta</math>.<br />
<br />
<math>\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi</math>.<br />
<br />
So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>.<br />
<br />
Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for theta!<br />
<br />
<br />
<math>3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta</math>.<br />
<br />
Simplifying, we get <math>63\theta=9\pi\implies \theta=\boxed{\frac{\pi}{7}}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2004|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_18&diff=862431983 AHSME Problems/Problem 182017-07-01T18:56:07Z<p>Whalefin: </p>
<hr />
<div>Problem:<br />
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,<br />
<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath><br />
For all real <math>x</math>, <math>f(x^2 - 1)</math> is<br />
<br />
(A) <math>x^4 + 5x^2 + 1</math> (B) <math>x^4 + x^2 - 3</math> (C) <math>x^4 - 5x^2 + 1</math> (D) <math>x^4 + x^2 + 3</math> (E) none of these<br />
<br />
Solution:<br />
<br />
Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as<br />
<math>f(y) &= x^4 + 5x^2 + 3 \\<br />
&= (x^2)^2 + 5x^2 + 3 \\<br />
&= (y - 1)^2 + 5(y - 1) + 3 \\<br />
&= y^2 - 2y + 1 + 5y - 5 + 3 \\<br />
&= y^2 + 3y - 1.</math><br />
Then substituting <math>x^2 - 1</math>, we get<br />
<math>f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\<br />
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\<br />
&= \boxed{x^4 + x^2 - 3}.</math><br />
The answer is (B).</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_18&diff=862421983 AHSME Problems/Problem 182017-07-01T18:55:39Z<p>Whalefin: </p>
<hr />
<div>Problem:<br />
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,<br />
<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath><br />
For all real <math>x</math>, <math>f(x^2 - 1)</math> is<br />
<br />
(A) <math>x^4 + 5x^2 + 1</math> (B) <math>x^4 + x^2 - 3</math> (C) <math>x^4 - 5x^2 + 1</math> (D) <math>x^4 + x^2 + 3</math> (E) none of these<br />
<br />
Solution:<br />
<br />
Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as<br />
<math>\begin{align*}<br />
f(y) &= x^4 + 5x^2 + 3 \\<br />
&= (x^2)^2 + 5x^2 + 3 \\<br />
&= (y - 1)^2 + 5(y - 1) + 3 \\<br />
&= y^2 - 2y + 1 + 5y - 5 + 3 \\<br />
&= y^2 + 3y - 1.<br />
\end{align*}</math><br />
Then substituting <math>x^2 - 1</math>, we get<br />
<math>\begin{align*}<br />
f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\<br />
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\<br />
&= \boxed{x^4 + x^2 - 3}.<br />
\end{align*}</math><br />
The answer is (B).</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_18&diff=862411983 AHSME Problems/Problem 182017-07-01T18:55:13Z<p>Whalefin: </p>
<hr />
<div>Problem:<br />
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,<br />
<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath><br />
For all real <math>x</math>, <math>f(x^2 - 1)</math> is<br />
<br />
(A) <math>x^4 + 5x^2 + 1</math> (B) <math>x^4 + x^2 - 3</math> (C) <math>x^4 - 5x^2 + 1</math> (D) <math>x^4 + x^2 + 3</math> (E) none of these<br />
<br />
Solution:<br />
<br />
Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as<br />
\begin{align*}<br />
f(y) &= x^4 + 5x^2 + 3 \\<br />
&= (x^2)^2 + 5x^2 + 3 \\<br />
&= (y - 1)^2 + 5(y - 1) + 3 \\<br />
&= y^2 - 2y + 1 + 5y - 5 + 3 \\<br />
&= y^2 + 3y - 1.<br />
\end{align*}<br />
Then substituting <math>x^2 - 1</math>, we get<br />
\begin{align*}<br />
f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\<br />
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\<br />
&= \boxed{x^4 + x^2 - 3}.<br />
\end{align*}<br />
The answer is (B).</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_21&diff=850662006 AMC 10A Problems/Problem 212017-03-29T23:48:15Z<p>Whalefin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? <br />
<br />
<math>\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad</math><br />
<br />
== Solution ==<br />
Since we are asked for the number of positive 4-digit [[integer]]s with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. <br />
<br />
The total number of 4-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have 10 choices for each digit except the first (which can't be 0).<br />
<br />
Similarly, the total number of 4-digit integers without any 2 or 3 is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>.<br />
<br />
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their [[decimal representation]] is <math>9000-3584=\boxed{5416} \Longrightarrow \mathrm{(E)} </math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Whalefinhttps://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_21&diff=850652006 AMC 10A Problems/Problem 212017-03-29T23:46:44Z<p>Whalefin: /* Solution */</p>
<hr />
<div>== Problem ==<br />
How many four-[[digit]] [[positive integer]]s have at least one digit that is a 2 or a 3? <br />
<br />
<math>\mathrm{(A) \ } 2439\qquad\mathrm{(B) \ } 4096\qquad\mathrm{(C) \ } 4903\qquad\mathrm{(D) \ } 4904\qquad\mathrm{(E) \ } 5416\qquad</math><br />
<br />
== Solution ==<br />
Since we are asked for the number of positive 4-digit [[integer]]s with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits. <br />
<br />
The total number of 4-digit integers is <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>, since we have 10 choices for each digit except the first (which can't be 0).<br />
<br />
Similarly, the total number of 4-digit integers without any 2 or 3 is <math>7 \cdot 8 \cdot 8 \cdot 8 ={3584}</math>.<br />
<br />
Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their [[decimal representation]] is <math>9000-3584=\boxed5416 \Longrightarrow \mathrm{(E)} </math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2006|ab=A|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Whalefin