AMC 10 · 2011 · #21

Grade 9 number-theoryalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

mean-median-mode-rangedigit-decompositionperfect-squaresdifference-of-squares convert-to-algebra ↑ Prerequisites: mean-median-mode-range

📏 Long solution 💡 4 insights

Problem

Two whole numbers have a two-digit arithmetic mean, and their geometric mean is that number reversed. Find the gap between them.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

The words "reverse the digits" only become math once the digits have names, so Tool #4 (Introduce a Variable) is primary: call the two-digit mean 10a+b, and its reversal is 10b+a. Tool #15 (Organize Information in More Ways) is what makes the problem crack — instead of chasing x and y, rewrite (x-y)² as (x+y)²-4xy, which is built entirely from the two means, and then factor it so the prime 11 pops out. Tool #14 (Extreme Principle) supplies the only other real input: digits are boxed into 0 through 9, and those bounds are what force a single digit sum. Tool #3 (Eliminate Possibilities) clears the remaining candidates by parity. Finally Tool #11 (Work Backwards) rebuilds the actual pair x, y from the digits found, which is the step that proves such numbers exist at all rather than merely proving what they would have to be.

1STEP 1

Name the digits of the mean

Two digits describe both means.

x+y = 2(10a+b), xy = (10b+a)², a ＞ b

2STEP 2

Express the gap using only the means

The gap squared comes straight from the two means.

(x-y)² = 4(M-G)(M+G) = 396(a-b)(a+b)

3STEP 3

Integer gap means perfect square

A whole-number gap forces a perfect square.

|x-y| = 6√(11(a-b)(a+b)), 11 ∣ (a-b)(a+b)

4STEP 4

Digit limits force the digit sum

Digit limits pin the digit sum at 11.

1 ≤ a-b ≤ 9, 1 ≤ a+b ≤ 17 → a+b = 11 → |x-y| = 66√(a-b)

5STEP 5

Parity kills the other candidates

The remaining candidates fail on parity.

a+b = 11, a-b = 1 → a = 6, b = 5

6STEP 6

Rebuild x and y to confirm

Rebuilding the pair confirms 66, choice (D).

x = 98, y = 32: (98+32)/2 = 65, √(98 · 32) = 56, |x-y| = 66 → (D)

Answer

The found pair verifies exactly: (98+32)/2 = 65, √(98 · 32) = 56, and 56 is 65 reversed, so |x-y| = 66 is achieved and not just permitted. It is worth seeing which step carries the weight. From |x-y| = 6√(11(a-b)(a+b)) one is tempted to say "so the answer is a multiple of 6 and of 11, hence 66" — but that only works because the choice list happens to contain a single multiple of 66; the same reasoning is equally happy with 132 or 198. What actually eliminates those is arithmetic on digits: a+b ≤ 17 forces a+b=11, and then the parity of 11 forces a-b odd, killing a-b=4 (which would give 132), while a=10 is not a digit, killing a-b=9 (which would give 198). A sanity scan of all 90 two-digit means confirms 65 and 56 is the only pair that works.

💡Key takeaway

Write the two-digit mean as 10a+b and its reversal as 10b+a; their sum and difference carry the factors 11 and 9, and since digits stop at 9, the 11 has only one place to go.

Name the digits of the mean
Express the gap using only the means
Integer gap means perfect square
Digit limits force the digit sum
Parity kills the other candidates
Rebuild x and y to confirm