AMC 10 · 2005 · #19

Grade 8 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
difference-of-squaresdigit-decompositionperfect-squares convert-to-algebracasework ↑ Prerequisites: difference-of-squaresperfect-squaresdigit-decomposition
📏 Long solution 💡 3 insights
Problem
A two-digit number and its reversal have squares differing by a perfect square. Find the sum of the two numbers and that square's root.

Pick an answer.

(A)
88
(B)
112
(C)
116
(D)
144
(E)
154
How to solve
Strategy Introduce a Variable

Name the two digits so the reversed number becomes an expression. Turning the digit-swap into algebra reveals a clean factorization, and then divisibility rules narrow the digits down to a single possibility.

1STEP 1

Name the digits

Naming the digits writes both numbers with one pair of unknowns.

x = 10a + b, y = 10b + a
2STEP 2

Factor the difference of squares

The difference of squares factors into two neat multiples.

x² - y² = (x-y)(x+y) = 9(a-b) · 11(a+b) = 99(a-b)(a+b)
3STEP 3

Force a perfect square

An odd prime must appear twice, forcing the digit sum to 11.

99(a-b)(a+b) = 3² · 11·(a-b)(a+b) → a+b = 11
4STEP 4

Pin down the digits

Then the difference must be a square too, pinning the digits to 6 and 5.

1089(a-b)=m²; a+b=11, a-b=1 → a=6, b=5
5STEP 5

Add up x, y, and m

Adding everything gives 154, choice (E).

x=65, y=56, m=√(1089)=33 → 65+56+33 = 154
Answer
154
Check directly: 65² - 56² = 4225 - 3136 = 1089 = 33², a genuine perfect square, so m = 33 works. The digits 6 and 5 are valid, and 65 + 56 + 33 = 154 matches choice (E).
💡Key takeaway

Write a two-digit number as 10a + b, and a difference of reversed squares always becomes 99 times (a - b)(a + b) -- then the primes tell you which digits fit.

  • Name the digits
  • Factor the difference of squares
  • Force a perfect square
  • Pin down the digits
  • Add up x, y, and m

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