Problem 23 · 1997 AJHSME
Stretch
Number Theory
boundingcasework
Some positive integers have both properties: (I) the sum of the squares of their digits is 50, and (II) each digit is larger than the one to its left. The product of the digits of the largest such integer is
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Answer: C — 36.
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Hint 1 of 2
'Each digit larger than the one before' secretly means the digits are all DIFFERENT — and the smallest possible distinct digits already pile up squares fast. First pin down how many digits can even fit under a square-sum of 50.
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Hint 2 of 2
Two moves: (1) bound the count of digits using the minimum square-sum, then (2) to make the NUMBER biggest, work from the largest possible final digit downward.
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Approach: bound the digit count, then settle the largest digit
- Strictly increasing ⇒ distinct digits. Five distinct increasing digits force squares ≥ 1+4+9+16+25 = 55 > 50, so there are at most 4 digits. (A 4-digit answer beats any 3-digit one, so aim for 4.)
- Test the last (largest) digit d: d = 7 gives 49, leaving only 1 for three more squares — impossible. d = 6 gives 36, leaving 14 = 1 + 4 + 9 for the smaller digits 1, 2, 3. That works: 1236, with 1 + 4 + 9 + 36 = 50.
- Smaller last digits (d = 5 etc.) can't reach a 4-digit set summing to 50, so 1236 is the largest valid integer.
- Digit product = 1 × 2 × 3 × 6 = 36.
- Why this transfers: 'largest integer with property P' problems split into bounding the length, then greedily fixing digits from the most significant (or here, the constraint-heaviest) end down.
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