Problem 13 · 2013 AMC 8
Medium
Number Theory
place-value-differencedivisibility-by-9
When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?
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Answer: A — 45.
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Hint 1 of 2
Swapping the tens and units digit only moves value between the "tens place" and "ones place." A digit worth 10 in one spot is worth 1 in the other — so each unit you shuffle changes the number by 9. The difference can only be a multiple of 9.
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Hint 2 of 2
Reversing two adjacent digits always changes a number by a multiple of 9 (it's a place-value gap of 10 − 1). So scan the choices for the one divisible by 9 — no equation needed.
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Approach: digit swap forces a multiple-of-9 difference
- Write the swapped two digits as 10a + b before and 10b + a after. The difference is (10a + b) − (10b + a) = 9(a − b) — a multiple of 9, no matter what the digits are.
- So Clara's error in the sum must be divisible by 9.
- Check the choices for divisibility by 9 (digit-sum trick): only 45 has digit sum 9. So the answer is 45 = 9 × 5.
- You'll see this again: any "reversed digits" puzzle hinges on this multiple-of-9 fact — it's the same reason the digit-sum test for 9 works.
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