Problem 11 · 2016 AMC 8
Medium
Number Theory
place-valuecareful-counting
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is 132.
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Answer: B — 7 numbers.
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Hint 1 of 3
Reversing swaps the tens digit and the ones digit. So when you ADD a number to its reversal, each digit lands once in the tens place and once in the ones place — the two digits end up playing perfectly equal roles. Write both numbers with place value and watch that symmetry.
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Hint 2 of 3
That symmetry forces the sum to be 11 × (digit sum). So 132 = 11 × (something) instantly pins the digit sum, and the whole problem becomes "count the digit pairs that add to it."
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Hint 3 of 3
Counting trap: the tens digit can't be 0 (it must stay a two-digit number), so a pair like (0, 12) is out — and a digit can't exceed 9 anyway.
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Approach: place value turns number + reversal into 11 × (digit sum)
- Write the number as 10a + b and its reversal as 10b + a. Adding: (10a + b) + (10b + a) = 11a + 11b = 11(a + b).
- So 11(a + b) = 132, giving a + b = 12. The reversal trick converted the whole condition into one tidy digit-sum equation.
- Count digit pairs with a from 1–9 (leading digit, can't be 0) and b a single digit: (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3) — that's 7 numbers.
- Why this transfers: "number plus its reversal" is ALWAYS a multiple of 11, and "number minus its reversal" is always a multiple of 9 — remembering these two facts collapses a whole family of digit problems.
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