https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mistypond2018&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:14:42ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Sugar_rush&diff=148133User:Sugar rush2021-03-01T23:13:43Z<p>Mistypond2018: /* User Count */</p>
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<div>==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="120px”>19<br />
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<div style="border:2px solid black; background:#ccccff;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">sugar_rush is 14 years old.<br><br />
sugar_rush was born on January 12, 2007.<br />
<br />
sugar_rush is in 8th grade.<br />
<br />
sugar_rush is a competitive runner and pro at math.<br />
<br />
Got 96 on AMC 12A (although there are debates on whether it's 90 or 96) and 109.5 on AMC 10B<br />
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</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">Qualify for AIME and get at least 6 on 2021 AIME I (never qualified for AIME before)<br />
<br />
</div><br />
</div><br />
Also: here is this guy's youtube channel: https://www.youtube.com/c/PunxsutawnyPhil</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=Diophantine_equation&diff=125874Diophantine equation2020-06-19T01:07:48Z<p>Mistypond2018: /* Babylonian Method */</p>
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<div>A '''Diophantine equation''' is an [[equation]] relating [[integer]] (or sometimes [[natural number]] or [[whole number]]) quanitites.<br />
<br />
Finding the solution or solutions to a Diophantine equation is closely tied to [[modular arithmetic]] and [[number theory]]. Often, when a Diophantine equation has infinitely many solutions, [[parametric form]] is used to express the relation between the variables of the equation.<br />
<br />
Diophantine equations are named for the ancient Greek/Alexandrian mathematician Diophantus.<br />
<br />
== Linear Combination ==<br />
A Diophantine equation in the form <math>ax+by=c</math> is known as a linear combination. If two [[relatively prime]] integers <math>a</math> and <math>b</math> are written in this form with <math>c=1</math>, the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when <math>\gcd(a,b)=1</math>. If <math>\gcd(a,b)\nmid c</math>, then there are no solutions to the equation. To see why, consider the equation <math>3x-9y=3(x-3y)=17</math>. <math>3</math> is a divisor of the [[LHS]] (also notice that <math>x-3y</math> must always be an integer). However, <math>17</math> will never be a multiple of <math>3</math>, hence, no solutions exist. <br />
<br />
Now consider the case where <math>c=0</math>. Thus, <math>ax=-by</math>. If <math>a</math> and <math>b</math> are relatively prime, then all solutions are obviously in the form <math>(bk,-ak)</math> for all integers <math>k</math>. If they are not, we simply divide them by their greatest common divisor.<br />
<br />
== Pythagorean Triples ==<br />
{{main|Pythagorean triple}}<br />
A Pythagorean triple is a set of three [[integer]]s that satisfy the [[Pythagorean Theorem]], <math>a^2+b^2=c^2</math>. There are three main methods of finding Pythagorean triples:<br />
=== Method of Pythagoras ===<br />
If <math>m>1</math> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.<br />
<br />
=== Method of Plato ===<br />
If <math>n>1</math>, <math>2n, n^2 - 1, n^2 + 1</math> is a Pythagorean triple.<br />
<br />
=== Babylonian Method ===<br />
For any <math>m,n</math>, we have <math>m^2 - n^2, 2mn, m^2 + n^2</math> is a Pythagorean triple.<br />
<br />
== Sum of Fourth Powers ==<br />
A equation of form <math>x^4+y^4=z^2</math> has no [[integer]] solutions, as follows:<br />
<!-- taken from AoPS Vol. 2 --><br />
We assume that the equation does have integer solutions, and consider the solution which minimizes <math>z</math>. Let this solution be <math>(x_0,y_0,z_0)</math>. If <math>\gcd (x_0,y_0)\neq 1</math> then their [[GCD]] <math>d</math> must satsify <math>d^2|z</math>. The solution <math>\left(\frac{x_0}{d},\frac{y_0}{d},\frac{z_0}{d}\right)</math> would then be a solution less than <math>z_0</math>, which contradicts our assumption. Thus, this equation has no integer solutions. <br />
<br />
If <math>\gcd(x_0,y_0)=1</math>, we then proceed with casework, in <math>\mod 4</math>.<br />
<br />
Note that every square, and therefore every fourth power, is either <math>1</math> or <math>0\mod 4</math>. The proof of this is fairly simple, and you can show it yourself.<br />
<br />
<br />
Case 1: <math>x_0^4 \equiv y_0^4 \equiv 1\mod 4</math><br />
<br />
This would imply <math>z^2 \equiv 2\mod 4</math>, a contradiction. <br />
<br />
<br />
Case 2: <math>x_0^4 \equiv y_0^4 \equiv 0\mod 4</math><br />
<br />
This would imply <math>z^2 \equiv 0\mod 4</math>, a contradiction since we assumed <math>\gcd(x_0,y_0)=1</math>.<br />
<br />
<br />
Case 3: <math>x_0^4 \equiv 1\mod 4, y_0^4 \equiv 0\mod 4</math>, and <math>z_0^2 \equiv 1\mod 4</math><br />
<br />
We also know that squares are either <math> -1, 0 </math> or <math>1 \mod 5</math>. Thus, all fourth powers are either <math>0</math> or <math>1 \mod 5</math>. <br />
<br />
<br />
By similar approach, we show that:<br />
<br />
<br />
<math>x_0^4 \equiv 0\mod 5, y_0^4 \equiv 1\mod 5</math>, so <math>z_0^2 \equiv 1\mod 5</math>. <br />
<br />
This is a contradiction, as <math>z_0^2 \equiv 1\mod 4</math> implies <math>z_0</math> is odd, and <math>z_0^2 \equiv 1\mod 5</math> implies <math>z_0</math> is even. QED [Oops, this doesn't work. 21 (or <math>3^4 = 81</math>) are equal to <math>1\mod 5</math> and not even...]<br />
<br />
== Pell Equations ==<br />
{{main|Pell equation}}<br />
A Pell equation is a type of Diophantine equation in the form <math>x^2-Dy^2=\pm1</math> for [[natural number]] <math>D</math>. The solutions to the Pell equation when <math>D</math> is not a perfect square are connected to the continued [[fraction]] expansion of <math>\sqrt D</math>. If <math>a</math> is the [[period]] of the continued fraction and <math>C_k=P_k/Q_k</math> is the <math>k</math>th convergent, all solutions to the Pell equation are in the form <math>(P_{ia},Q_{ia})</math> for [[positive]] integer <math>i</math>.<br />
<br />
== Methods of Solving ==<br />
=== Coordinate Plane ===<br />
Note that any linear combination can be transformed into the linear equation <math>y=\frac{-b}{a}x+\frac{c}{a}</math>, which is just the slope-intercept equation for a line. The solutions to the diophantine equation correspond to [[lattice point]]s that lie on the line. For example, consider the equation <math>-3x+4y=4</math> or <math>y=\frac{3}{4}x+1</math>. One solution is (0,1). If you graph the line, it's easy to see that the line intersects a [[lattice point]] as x and y increase or decrease by the same multiple of <math>4</math> and <math>3</math>, respectively (wording?). Hence, the solutions to the equation may be written [[parametrically]] <math>x=4t, y=3t+1</math> (if we think of <math>(0,1)</math> as a "starting point"). <br />
=== Modular Arithmetic === <br />
Sometimes, [[modular arithmetic]] can be used to prove that no solutions to a given Diophantine equation exist. Specifically, if we show that the equation in question is never true mod <math>m</math>, for some integer <math>m</math>, then we have shown that the equation is false. However, this technique cannot be used to show that solutions to a Diophantine equation do exist.<br />
<br />
===Induction===<br />
Sometimes, when a few solutions have been found, [[induction]] can be used to find a family of solutions. Techniques such as [[infinite Descent]] can also show that no solutions to a particular equation exist, or that no solutions outside of a particular family exist.<br />
<br />
=== General Solutions ===<br />
<br />
It is natural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no.<br />
<br />
== Fermat's Last Theorem ==<br />
{{main|Fermat's Last Theorem}}<br />
<math>x^n+y^n=z^n</math> is known as [[Fermat's Last Theorem]] for the condition <math>n\geq3</math>. In the 1600s, Fermat, as he was working through a book on Diophantine Equations, wrote a comment in the margins to the effect of "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." Fermat actually made many conjectures and proposed plenty of "theorems," but wasn't one to write down the proofs or much other than scribbled comments. After he died, all his conjectures were re-proven (either false or true) except for Fermat's Last Theorem. After over 350 years of failing to be proven, the theorem was finally proven by [[Andrew Wiles]] after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof.<br />
<br />
<br />
== Problems ==<br />
=== Introductory ===<br />
* Two farmers agree that pigs are worth <math>300</math> dollars and that goats are worth <math>210</math> dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a <math>390</math> dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?<br />
<br />
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 90\qquad \mathrm{(E) \ } 210</math><br />
<br />
([[2006_AMC_10A_Problems/Problem_22|Source]])<br />
<br />
=== Intermediate ===<br />
*Let <math> P(x) </math> be a polynomial with integer coefficients that satisfies <math> P(17)=10 </math> and <math> P(24)=17. </math> Given that <math> P(n)=n+3 </math> has two distinct integer solutions <math> n_1 </math> and <math> n_2, </math> find the product <math> n_1\cdot n_2</math>. ([[2005 AIME II Problems/Problem 13|Source]])<br />
<br />
=== Olympiad ===<br />
*Determine the maximum value of <math>m^2 + n^2 </math>, where <math>m </math> and <math>n </math> are integers satisfying <math> m, n \in \{ 1,2, \ldots , 1981 \} </math> and <math>( n^2 - mn - m^2 )^2 = 1 </math>. ([[1981 IMO Problems/Problem 3|Source]])<br />
*Solve in integers the equation <math>x^3+x^2+x+1=y^2</math>.<br />
<br />
==References==<br />
*[http://www.ams.org/notices/199507/faltings.pdf Proof of Fermat's Last Theorem]<br />
<br />
==See also==<br />
* [[Number Theory]]<br />
* [[Pell equation]]<br />
<br />
[[Category:Number theory]]</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_24&diff=1007082005 AMC 10B Problems/Problem 242019-01-21T00:19:23Z<p>Mistypond2018: /* Solution */</p>
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<div>== Problem ==<br />
Let <math>x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits<br />
of <math>x</math>. The integers <math>x</math> and <math>y</math> satisfy <math>x^2 - y^2 = m^2</math> for some positive integer <math>m</math>.<br />
What is <math>x + y + m</math>?<br />
<br />
<math>\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154 </math><br />
<br />
== Solution 1 ==<br />
Let <math>x = 10a+b, y = 10b+a</math>. The given conditions imply <math>x>y</math>, which implies <math>a>b</math>, and they also imply that both <math>a</math> and <math>b</math> are nonzero. Then <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. Since this must be a perfect square, all the exponents in its prime factorization must be even. <math>99</math> factorizes into <math>3^2 \cdot 11</math>, so <math>11|(a-b)(a+b)</math>. However, the maximum value of <math>a-b</math> is <math>9-1=8</math>, so <math>11|a+b</math>. The maximum of <math>a+b</math> is <math>9+8=17</math>, so <math>a+b=11</math>. Then we have <math>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</math> and <math>4</math>. We know <math>a+b=11</math>, and, for <math>a-b=1</math>, adding equations to eliminate <math>b</math> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math>b</math> must be digits. Therefore, <math>(a,b) = (6,5)</math>, and <math>x+y+m=65+56+33=154\ \mathbf{(E)}</math>.<br />
<br />
== Solution 2 ==<br />
The first steps are the same as above. Let <math>x = 10a+b, y = 10b+a</math>, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting <math>(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. This is where the solution diverges. <br />
<br />
We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get <math>3^2\cdot 11</math>. In order to get a perfect square on the left side, <math>(a-b)(a+b)</math> must make both prime exponents even. Because the a and b are digits, a simple guess would be that <math>(a+b)</math> (the bigger number) equals 11 while <math>(a-b)</math> is a factor of nine (1 or 9). The correct guesses are <math>a = 6, b = 5</math> causing <math>x = 65, y = 56,</math> and <math>m = 33</math>. The sum of the numbers is <math>\boxed{\textbf{(E) }154}</math><br />
<br />
== Solution 3 ==<br />
Once again, the solution is quite similar as the above solutions. Since <math>x</math> and <math>y</math> are two digit integers, we can write <math>x = 10a+b, y = 10b+a</math> and because <math>x^2 - y^2 = (x-y)(x+y)</math>, substituting and factoring, we get <math>99(a+b)(a-b) = m^2</math>. Therefore, <math>(a+b)(a-b) = \frac{m^2}{99}</math> and <math>\frac{m^2}{99}</math> must be an integer. A quick strategy is to find the smallest such integer <math>m</math> such that <math>\frac{m^2}{99}</math> is an integer. We notice that 99 has a prime factorization of <math>3^2 \cdot 11.</math> Let <math>m^2 = n.</math> Since we need a perfect square and 3 is already squared, we just need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math>(x-y)(x+y) = 33^2</math>, which we can also write as <math>(x-y)(x+y) = 11^2 \cdot 3^2</math>. A very simple guess assumes that <math>x-y=3^2</math> and <math>x+y=11^2</math> since <math>x</math> and <math>y</math> are positive. Finally, we come to the conclusion that <math>x=65</math> and <math>y=56</math>, so <math>x+y+m</math> <math>=</math> <math>\boxed{\textbf{(E) }154}</math>.<br />
Note that all of the solutions used <math>a+b</math> or <math>a-b</math> as part of their solution.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_23&diff=1007072005 AMC 10B Problems/Problem 232019-01-21T00:17:55Z<p>Mistypond2018: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>?<br />
<br />
<math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math><br />
<br />
== Solution 1==<br />
Since the height of both trapezoids are equal, and the area of <math>ABEF</math> is twice the area of <math>FECD</math>, <br />
<br />
<math>\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)</math>.<br />
<br />
<math>\frac{AB+EF}{2}=DC+EF</math>, so<br />
<br />
<math>AB+EF=2DC+2EF</math>.<br />
<br />
<math>EF</math> is exactly halfway between <math>AB</math> and <math>DC</math>, so <math>EF=\frac{AB+DC}{2}</math>. <br />
<br />
<math>AB+\frac{AB+DC}{2}=2DC+AB+DC</math>, so<br />
<br />
<math>\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB</math>, and <br />
<br />
<math>\frac{1}{2}AB=\frac{5}{2}DC</math>.<br />
<br />
<math>\frac{AB}{DC} = \boxed{5}</math>.<br />
<br />
==Solution 2==<br />
<br />
Mark <math>DC=z</math>, <math>AB=x</math>, and <math>FE=y.</math><br />
Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</math>.<br />
<br />
Then, we have that <math>\frac{x+y}{2}\cdot h=2(\frac{y+z}{2} \cdot h)</math>.<br />
<br />
From this, we get that <math>x=2z+y</math>.<br />
<br />
We also get that <math>\frac{x+z}{2} \cdot h= 3(\frac{y+z}{2} \cdot h)</math>.<br />
<br />
Simplifying, we get that <math>2x=z+3y</math><br />
<br />
Notice that we want <math>\frac{AB}{DC}=\frac{x}{z}</math>.<br />
<br />
Dividing the first equation by <math>z</math>, we get that <math>\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})</math>.<br />
<br />
Dividing the second equation by <math>z</math>, we get that <math>2(\frac{x}{z})=1+3(\frac{y}{z})</math>.<br />
<br />
Now, when we subtract the top equation from the bottom, we get that <math>\frac{x}{z}=5</math><br />
<br />
Hence, the answer is <math>\boxed{5}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10B_Problems/Problem_21&diff=1007062005 AMC 10B Problems/Problem 212019-01-21T00:10:57Z<p>Mistypond2018: /* Solution (where the order of drawing slips matters) */</p>
<hr />
<div>== Problem ==<br />
Forty slips are placed into a hat, each bearing a number <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, or <math>10</math>, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let <math>p</math> be the probability that all four slips bear the same number. Let <math>q</math> be the probability that two of the slips bear a number <math>a</math> and the other two bear a number <math>b \neq a</math>. What is the value of <math>q/p</math>?<br />
<br />
<math>\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720 </math><br />
<br />
== Solution 1 (where the order of drawing slips matters) ==<br />
There are <math>10</math> ways to determine which number to pick. There are <math>4!</math> ways to then draw those four slips with that number, and <math>40 \cdot 39 \cdot 38 \cdot 37</math> total ways to draw four slips. Thus <math>p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}</math>. <br />
<br />
There are <math>{10 \choose 2} = 45</math> ways to determine which two numbers to pick for the second probability. There are <math>{4 \choose 2} = 6</math> ways to arrange the order which we draw the non-equal slips, and in each order there are <math>4 \times 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>. <br />
<br />
Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}</math>.<br />
<br />
==Solution 2 (order does not matter)==<br />
<br />
For probability <math>p</math>, there are <math>\binom{10}{1}=10</math> ways to choose the card you want to show up <math>4</math> times..<br />
<br />
Hence, the probability is <math>\frac{10}{\binom{40}{4}}</math>.<br />
<br />
For probability <math>q</math>, there are <math>\binom{10}{2}=45</math> ways to choose the <math>2</math> numbers you want to show up twice. There are <math>\binom{4}{2}\cdot\binom{4}{2}</math> to pick which cards you want out of the <math>4</math> of each.<br />
<br />
Hence, the probability is <math>\frac{45\cdot6\cdot6}{\binom{40}{4}}</math><br />
<br />
Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}</math>.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Combinatorics Problems]]<br />
{{MAA Notice}}</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_18&diff=1005772009 AMC 12A Problems/Problem 182019-01-18T02:27:39Z<p>Mistypond2018: /* Alternate Solution */</p>
<hr />
<div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}}<br />
<br />
== Problem ==<br />
For <math>k > 0</math>, let <math>I_k = 10\ldots 064</math>, where there are <math>k</math> zeros between the <math>1</math> and the <math>6</math>. Let <math>N(k)</math> be the number of factors of <math>2</math> in the prime factorization of <math>I_k</math>. What is the maximum value of <math>N(k)</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 />
== Solution ==<br />
<br />
The number <math>I_k</math> can be written as <math>10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6</math>.<br />
<br />
For <math>k\in\{1,2,3\}</math> we have <math>I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)</math>. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have <math>N(k)=k+2\leq 5</math>.<br />
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For <math>k>4</math> we have <math>I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)</math>. For <math>k>4</math> the value in the parentheses is odd, hence <math>N(k)=6</math>.<br />
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This leaves the case <math>k=4</math>. We have <math>I_4 = 2^6 \left( 5^6 + 1 \right)</math>. The value <math>5^6 + 1</math> is obviously even. And as <math>5\equiv 1 \pmod 4</math>, we have <math>5^6 \equiv 1 \pmod 4</math>, and therefore <math>5^6 + 1 \equiv 2 \pmod 4</math>. Hence the largest power of <math>2</math> that divides <math>5^6+1</math> is <math>2^1</math>, and this gives us the desired maximum of the function <math>N</math>: <math>N(4) = \boxed{7}</math>.<br />
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== Solution 2 ==<br />
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Notice that 2 is a prime factor of an integer <math>n</math> if and only if <math>n</math> is even. Therefore, given any sufficiently high positive integral value of <math>k</math>, dividing <math>I_k</math> by <math>2^6</math> yields a terminal digit of zero, and dividing by 2 again leaves us with <math>2^7 \cdot a = I_k</math> where <math>a</math> is an odd integer.<br />
Observe then that <math>\boxed{\textbf{(B)}7}</math> must be the maximum value for <math>N(k)</math> because whatever value we choose for <math>k</math>, <math>N(k)</math> must be less than or equal to <math>7</math>.<br />
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"Isn't this solution incomplete because we need to show that <math>N(k) = 7</math> can be reached?"<br />
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An example of 7 being reached is 1000064. 1000064 divided by <math>2^7=128</math> is 7813.<br />
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In fact, 1000064 is the ONLY <math>N(k)</math> that satisfies 7. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 (<math>2^7</math>), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.<br />
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== Solution 3 ==<br />
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As in the first solution, the number <math>I_k</math> can be written as <math>10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6</math>. Factor <math>2^6</math> out of the expression to get <math>I_k = 2^6(1+2^{k-4}5^{k+2})</math>.<br />
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You can now easily see that <math>I_k</math> is divisible by at least 6 factors of two, from <math>2^6</math>. Any other factors of two will come from the expression <math>(1+2^{k-4}5^{k+2})</math>. <br />
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Make the substitution: <math>x=k-4</math>.<br />
You now have <math>(1+2^{x}5^{x+6}) = (1+10^x 5^6)</math><br />
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<math>5^6=15625</math>, so the expression becomes <math>(1+15625\cdot10^x)</math> This is valid when <math>x>-4</math>. <br />
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Obviously, if <math>x</math> is negative, the expression will be fractional and not contain any extra factors of two.<br />
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If <math>x>0</math>, the value <math>15625\cdot10^x</math> will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.<br />
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If <math>x=0</math>, the expression evaluates to <math>15626</math>. Dividing out twos, you find that this value is only divisible by one factor of 2.<br />
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The six factors of two from earlier add to this factor of two to give <math>\textbf{(B)}\ 7\qquad</math><br />
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Solution written by coolak<br />
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== See Also ==<br />
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{{AMC12 box|year=2009|ab=A|num-b=17|num-a=19}}<br />
{{AMC10 box|year=2009|ab=A|num-b=24|after=Last Question}}<br />
{{MAA Notice}}</div>Mistypond2018https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_8&diff=1005762009 AMC 12A Problems/Problem 82019-01-18T02:25:34Z<p>Mistypond2018: /* Solution1 */</p>
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<div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #8]] and [[2009 AMC 10A Problems|2009 AMC 10A #14]]}}<br />
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== Problem ==<br />
Four congruent rectangles are placed as shown. The area of the outer square is <math>4</math> times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?<br />
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 + \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4</math><br />
[[Category: Introductory Geometry Problems]]<br />
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== Solution 1 ==<br />
<math>\boxed{(A)}</math><br />
The area of the outer square is <math>4</math> times that of the inner square.<br />
Therefore the side of the outer square is <math>\sqrt 4 = 2</math> times that of the inner square.<br />
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Then the shorter side of the rectangle is <math>1/4</math> of the side of the outer square, and the longer side of the rectangle is <math>3/4</math> of the side of the outer square, hence their ratio is <math>\boxed{3}</math>.<br />
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== Solution 2 ==<br />
Let the side length of the smaller square be <math>1</math>, and let the smaller side of the rectangles be <math>y</math>. Since the larger square's area is four times larger than the smaller square's, the larger square's side length is <math>2</math>. That too is then equivalent to <math>2y+1</math>, giving <math>y=1/2</math>. Then, the larger piece of the rectangles is <math>3/2</math>. <math>3/2/1/2=\boxed{3}</math>.<br />
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== Solution 3 ==<br />
Let the longer side length be <math>x</math>, and the shorter side be <math>a</math>.<br />
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We have that <math>(x+a)^2=4(x-a)^2\implies x+a=2(x-a)\implies x+a=2x-2a\implies x=3a \implies \frac{x}{a}=3</math><br />
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Hence, the answer is <math>\boxed{A}\implies 3</math><br />
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== See Also ==<br />
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{{AMC12 box|year=2009|ab=A|num-b=7|num-a=9}}<br />
{{AMC10 box|year=2009|ab=A|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mistypond2018