Problem 14 · 1989 AJHSME
Hard
Number Theory
minimize-differenceplace-value
When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?
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Show answer
Answer: C — 149.
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Hint 1 of 3
To make a difference small, push the top number down and the bottom number up — but not all digit slots matter equally. Which single slot moves the answer the most?
Still stuck? Show hint 2 →
Hint 2 of 3
Place digits where they have the most leverage first: the hundreds place of the top number, then the tens place of the bottom number, then fill in the rest.
Still stuck? Show hint 3 →
Hint 3 of 3
The top must be 3 digits, so its hundreds digit can't be helped much — make it the smallest, 2. Then give the bottom number the biggest tens digit you can, 9.
Show solution
Approach: place the high-leverage digits first
- A digit in the hundreds column is worth 100, in the tens column 10 — so the slots that swing the difference most are the top number's hundreds and the bottom number's tens. Lock those down first: top hundreds = smallest = 2; bottom tens = largest = 9.
- Now fill the leftover digits {4, 5, 6} to keep the gap tight: top becomes 245 (smallest with leading 2) and bottom becomes 96 (largest with leading 9). Difference: 245 − 96 = 149.
- Why this transfers: in build-the-number puzzles, assign digits to the highest place values first — that's where they count for the most. Greedily optimizing the biggest column, then the next, beats blindly trying combinations.
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