Problem 11 · 2006 AMC 8
Medium
Number Theory
digit-sumcasework
How many two-digit numbers have digits whose sum is a perfect square?
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Answer: C — 17.
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Hint 1 of 2
Flip the question: instead of testing all 90 two-digit numbers, first ask which digit SUMS are even allowed. The smallest sum is 1 (from 10) and the largest is 18 (from 99), so only the perfect squares 1, 4, 9, 16 can occur.
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Hint 2 of 2
Now handle each target sum separately. The tens digit must be 1–9 (no leading zero), but the units digit may be 0–9 — that asymmetry is what makes the counts uneven, so watch the edges.
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Approach: first pin down the possible digit sums, then count each case
- A two-digit number's digit sum runs from 1 (for 10) up to 18 (for 99). The perfect squares in that window are 1, 4, 9, and 16 — only four cases to check.
- Sum = 1: just 10 ⇒ 1 number.
- Sum = 4: 13, 22, 31, 40 ⇒ 4 numbers.
- Sum = 9: 18, 27, 36, 45, 54, 63, 72, 81, 90 ⇒ 9 numbers.
- Sum = 16: 79, 88, 97 ⇒ 3 numbers (only these fit, since each digit caps at 9 — big sums leave very few splits).
- Total: 1 + 4 + 9 + 3 = 17.
- Why the counts shrink at the top: for sum 16 the digits are nearly maxed out (each at most 9), so few splits fit; for middling sums like 9 there's lots of freedom. Narrowing to the four legal sums first is what keeps this from being a 90-number slog.
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