AMC 8 · 2006 · #11

Grade 3 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

digit-sumperfect-squaressystematic-enumerationdigit-constraints systematic-enumerationcasework ↑ Prerequisites: digit-sumperfect-squares

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Problem

Among all two-digit numbers (the integers from 10 to 99), count how many have a digit sum that is a perfect square.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Count Systematically

Tool #14 (Try the Extreme Case) shows that digit sums only run from 1 up to 18, so the perfect squares we need to hit are just 1, 4, 9, 16 — that's the whole list. Tool #7 (Break Into Subproblems) splits the question into four small counting tasks, one per square. Tool #13 (Count Systematically) then handles each subproblem the same way: list every pair (t, u) with t + u equal to the target, where t ∈ {1, …, 9} and u ∈ {0, …, 9}. Adding the four counts gives the answer.

1STEP 1

Digit sums run from 1 up to 9 + 9 = 18, so the perfect squares to hit are just 1, 4, 9, 16.

1 ≤ t + u ≤ 18 → t + u ∈ {1, 4, 9, 16}

2STEP 2

With t ≥ 1, the only pair summing to 1 is t = 1, u = 0, giving the number 10 — 1 number.

t + u = 1: (1, 0) → {10} → 1 number

3STEP 3

Pairs with t + u = 4 are (1,3), (2,2), (3,1), (4,0) → 13, 22, 31, 40, so 4 numbers.

t + u = 4: {13, 22, 31, 40} → 4 numbers

4STEP 4

For t + u = 9, each t from 1 to 9 gives a valid u = 9 - t, so 9 numbers (18, 27, …, 90).

t + u = 9: {18, 27, 36, 45, 54, 63, 72, 81, 90} → 9 numbers

5STEP 5

For t + u = 16, u = 16 - t ≤ 9 forces t ≥ 7: (7,9), (8,8), (9,7) → 79, 88, 97, so 3 numbers.

t + u = 16: {79, 88, 97} → 3 numbers

6STEP 6

Adding the four counts: 1 + 4 + 9 + 3 = 17 two-digit numbers, which is choice (C).

1 + 4 + 9 + 3 = 17 → (C)

Answer

Each case can be re-checked by reading the list out loud — 10; 13, 22, 31, 40; 18, 27, 36, 45, 54, 63, 72, 81, 90; 79, 88, 97 — and these are clearly different numbers, so no double-counting. The total 1 + 4 + 9 + 3 = 17 matches answer (C). It also makes sense that the middle square 9 produces the most numbers: 9 is right in the middle of the digit-sum range [1, 18], so neither the lower bound (t ≥ 1) nor the upper bound (u ≤ 9) trims any cases away.

💡Key takeaway

Digit sums for two-digit numbers only run from 1 to 18, so the perfect squares to chase are 1, 4, 9, 16. Count each case, add them up, and you get 17.

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