2005 AMC 10B Problem 24

P21-25~6 minPrint

Diophantine Equations Bases & Digits Number Properties Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Bound the quantity above/below or estimate to pin it down

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 $x + y + m$ ?

Show solution without answering

Key insight

With digits a and b, x squared minus y squared equals 99(a + b)(a - b). So a + b must be 11 and a - b a perfect square.

Solution

Reversed digits suggest writing the numbers with their digits. Let $x = 10a + b$ and $y = 10b + a$ , where $a$ and $b$ are digits. Since $m$ is positive, $x > y$ , so $a > b$ .

1. Factor. $x^2 - y^2 = (x - y)(x + y)$ (difference of squares). Now
- $x - y = 9a - 9b = 9(a - b)$ ,
- $x + y = 11a + 11b = 11(a + b)$ .

So $m^2 = 9 \cdot 11 \cdot (a - b)(a + b)$ .
2. What makes a perfect square? In a perfect square, every prime appears an even number of times. The $9 = 3 \cdot 3$ is fine, but $11$ appears only once so far. So $(a - b)(a + b)$ must supply another $11$ .
3. Where can the $11$ come from? $a - b$ is between $1$ and $8$ , so it cannot be a multiple of $11$ . So $a + b$ must be. Since $a + b$ is at most $9 + 8 = 17$ , we need $a + b = 11$ .
4. Now $m^2 = 9 \cdot 11 \cdot 11 \cdot (a - b) = 33^2 \cdot (a - b)$ . For this to be a perfect square, $a - b$ must be a perfect square: $1$ or $4$ .
5. Find the digits.
- $a + b = 11$ , $a - b = 1$ : adding gives $2a = 12$ , so $a = 6$ , $b = 5$ .
- $a + b = 11$ , $a - b = 4$ : adding gives $2a = 15$ , not a whole number. Rejected.
6. Answer. $x = 65$ , $y = 56$ , and $m^2 = 33^2 \cdot 1$ , so $m = 33$ . Check: $65^2 - 56^2 = 4225 - 3136 = 1089 = 33^2$ .
$x + y + m = 65 + 56 + 33 = 154.$

The answer is $\boxed{\textbf{(E)}\ 154}$ .

Why this works

For problems about reversing digits, write the numbers as $10a + b$ and $10b + a$ ; their sum is always a multiple of $11$ and their difference a multiple of $9$ . A perfect square needs each prime an even number of times, and the small size of digits leaves only a couple of options to check.

Alternative approach

From step 2, $m^2$ is a multiple of $99$ , so $m$ must be a multiple of both $3$ and $11$ , that is, a multiple of $33$ . Try $m = 33$ : we need $x^2 = y^2 + 1089$ with $x$ , $y$ reversed two-digit numbers. Checking $65$ and $56$ : $4225 - 3136 = 1089$ . It works, giving $65 + 56 + 33 = 154$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let $x = 10a + b$ and $y = 10b + a$ with digits $a, b$ . Then $x - y = 9(a - b)$ and $x + y = 11(a + b)$ , so
$x^2 - y^2 = (x - y)(x + y) = 99\,(a - b)(a + b) = 3^2 \cdot 11 \cdot (a - b)(a + b).$
For this to be a perfect square, the prime $11$ must appear to an even power, so $11$ divides $(a - b)(a + b)$ . Since $m > 0$ we need $a > b$ , so $1 \le a - b \le 8$ and $11 \nmid a - b$ . Hence $11 \mid a + b$ , and as $a + b \le 17$ , we get $a + b = 11$ .

Now $x^2 - y^2 = 3^2 \cdot 11^2 \cdot (a - b)$ , so $a - b$ must be a perfect square. With $a + b = 11$ odd, $a - b$ is odd, so $a - b = 1$ (the only odd square at most $8$ ). Solving $a + b = 11$ , $a - b = 1$ gives $a = 6$ , $b = 5$ .

Then $x = 65$ , $y = 56$ , and $m^2 = 99 \cdot 1 \cdot 11 = 1089 = 33^2$ , so $m = 33$ .

$x + y + m = 65 + 56 + 33 = 154.$

The answer is $\boxed{\textbf{(E)}\ 154}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 11: Special Factorizations