Problem 7 · 1992 AJHSME
Hard
Counting & Probability
digit-sumparity
The digit-sum of 998 is 9 + 9 + 8 = 26. How many 3-digit whole numbers, whose digit-sum is 26, are even?
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Answer: A — 1.
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Hint 1 of 3
The biggest a 3-digit digit-sum can be is 9 + 9 + 9 = 27. A sum of 26 is only 1 short of that ceiling — so how far from all-9s can the digits stray?
Still stuck? Show hint 2 →
Hint 2 of 3
When a digit-sum is pinned near its maximum, the digits are forced to be almost as large as possible — that drastically limits the possibilities, so you can just list them.
Still stuck? Show hint 3 →
Hint 3 of 3
For a number to be even, only the LAST digit matters — it has to be even.
Show solution
Approach: the near-maximum sum forces the digits; then check the last one
- The largest possible digit-sum is 9 + 9 + 9 = 27. To get 26 we must drop exactly 1 from that, so one digit becomes 8 and the rest stay 9: the digits are 9, 9, 8.
- Arranged as a 3-digit number, that gives only 998, 989, 899.
- Even numbers must end in an even digit. Of the three, only 998 ends in 8 (even), so the count is 1.
- Why this transfers: when a sum sits right against its maximum, treat it as "how much must I subtract from the all-max case?" A tiny deficit means only a handful of arrangements — cheap enough to list and check by hand.
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