Problem 23 · 1994 AJHSME
Stretch
Number Theory
place-valuemaximize
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Answer: D — Form YYZ.
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Hint 1 of 2
Turn the column-addition into one expression by place value: XXX is 111·X, YX is 10·Y + X, and X is X. Add them and the X's collect to 113·X, plus 10·Y.
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Hint 2 of 2
Two competing wants: make the sum big, but it must stay 3 digits (the question says so). X drives it hardest (the 113), so push X as high as it can go WITHOUT spilling to 4 digits, then push Y. Last step: read off the answer's digits and translate them back into X/Y/Z letters.
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Approach: express the sum, then maximize within three digits
- Sum = XXX + YX + X = 113·X + 10·Y. X matters most, but 113·9 = 1017 is four digits, so the biggest legal X is 8 → 113·8 = 904.
- Now max out Y: Y = 9 adds 90, giving 904 + 90 = 994. (Y can equal a different digit than X.)
- Read 994 in letters: with X = 8, Y = 9, Z = 4, the digits 9, 9, 4 are Y, Y, Z. So the form is YYZ.
- Why two-step greedy works: when one slot's coefficient (113) dwarfs the other's (10), maximize the heavyweight first, then the rest. The sneaky part is the question wants the FORM (pattern of letters), not the number — so always re-map the digits back to the symbols you were given.
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