Difference between revisions of "2005 AMC 10B Problems/Problem 24"

Revision as of 11:36, 5 January 2017

Problem

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ . What is $x + y + m$ ?

$\mathrm{(A)} 88 \qquad \mathrm{(B)} 112 \qquad \mathrm{(C)} 116 \qquad \mathrm{(D)} 144 \qquad \mathrm{(E)} 154$

Solution

Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero. Then $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . Since this must be a perfect square, all the exponents in its prime factorization must be even. However, $99$ factorizes into $3^2 \cdot 11$ . Therefore, $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum of $a+b$ is $9+8=17$ , so $a+b=11$ . Then we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{13}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=65+56+33=154\ \mathbf{(E)}$ .

Solution 2

The first steps are the same as above. Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ . This is where the solution diverges.

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 $3^2\cdot 11$ . In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even. Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals 11 while $(a-b)$ is a factor of nine (1 or 9). The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ . The sum of the numbers is $\boxed{\textbf{(E) }154}$

Solution 3

Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$ . Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$ Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$ . A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $\boxed{\textbf{(E) }154}$ .

@@ Line 15: / Line 15: @@
 == Solution 3 ==
-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>
+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>.
-This is quite easy for a Problem <math>24</math>.
-*Solution by <math>AOPS12142015</math>
 == See Also ==

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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