Problem 22 · 2010 AMC 8
Medium
Algebra & Patterns
place-value-difference
The hundreds digit of a three-digit number is 2 more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
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Answer: E — 8.
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Hint 1 of 3
When you reverse a 3-digit number, the tens digit doesn't move — only the hundreds and units swap. So in the subtraction, the tens completely cancel and won't affect the answer.
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Hint 2 of 3
Reversing always gives a difference of 99 × (hundreds digit − units digit). Here that gap is fixed at 2, so the difference is a single fixed number — no variables survive.
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Hint 3 of 3
So you don't need the actual digits: just compute 99 × 2 and read off its units digit.
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Approach: the tens digit cancels; difference is forced
- Write the number as 100·(hundreds) + 10·(tens) + (units). Reversing swaps hundreds and units, so when you subtract, the 10·(tens) terms cancel exactly.
- What's left is 99·(hundreds − units). The hundreds digit is 2 more than the units, so hundreds − units = 2, every time.
- Difference = 99 × 2 = 198, whose units digit is 8.
- Why this transfers: ‘a number minus its reversal’ is always a multiple of 99 (for 3 digits), and only the gap between the outer digits matters. Spotting that the middle digit cancels means you can answer without ever choosing specific digits.
Another way — try a concrete example:
- Pick any number fitting the rule, say hundreds 2 more than units: 301. Reverse to 103.
- 301 − 103 = 198 ⇒ units digit 8. Trying 412 − 214 = 198 too — the units digit is locked at 8 no matter what you choose, which is the whole point.
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