2005 AMC 10B Problems/Problem 24

Revision as of 17:06, 13 September 2017 by Reedmj (talk | contribs) (Solution)


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 $x + y + m$?

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


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. $99$ factorizes into $3^2 \cdot 11$, so $11|(a-b)(a+b)$. 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 $x+y+m$ $=$ $\boxed{\textbf{(E) }154}$.

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS