Problem 1 · 1990 AJHSME
Medium
Number Theory
place-valueminimize
What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in this addition problem?
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Answer: C — 1047.
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Hint 1 of 2
A digit in the hundreds place is worth 100 of itself, but only 1 of itself in the units. So which digits do you most want to keep out of the hundreds?
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Hint 2 of 2
This is the place-value greedy rule: to make a sum small, feed your smallest digits to the most expensive seats (hundreds), and dump your biggest digits in the cheapest seats (units).
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Approach: feed small digits to the expensive place values
- Don't try arrangements at random — notice that each hundreds digit costs 100× its value, each tens digit 10×, each units digit 1×. So you want the *smallest* digits sitting where the multiplier is biggest.
- Smallest two digits (4, 5) go in the hundreds; next two (6, 7) in the tens; largest two (8, 9) in the units. The order *within* a place doesn't matter (4+5 hundreds is the same as 5+4).
- Sum = (4+5)×100 + (6+7)×10 + (8+9) = 900 + 130 + 17 = 1047.
- *Why this transfers:* whenever you're placing fixed digits to minimize (or maximize) a total, sort by the place's weight — smallest values into the heaviest places to minimize, the reverse to maximize. You'll meet this again in 'arrange the digits' problems.
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