AJHSME 1991 Problem 20: Number Theory Solution

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Answer: A — 1.

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Hint 1 of 3

The letter A appears in the hundreds place (in ABC), the tens place (in AB), and the ones place (in A). So A pulls a LOT of weight in the total of 300. Roughly how big can A be before three copies of it overshoot 300?

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Hint 2 of 3

Write what each letter is worth: ABC = 100A + 10B + C, AB = 10A + B, A = A. Add them: 111A + 11B + C = 300. Now the hundreds force A almost immediately.

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Hint 3 of 3

111A alone must be near 300 without passing it: A = 3 gives 333 (too big), A = 1 gives only 111 (can't reach 300 with the small leftovers). So A is pinned. Then solve 11B + C = whatever's left.

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Approach: expand by place value; the hundreds digit pins everything down

Translate the stacked sum into place value: ABC + AB + A = (100A + 10B + C) + (10A + B) + A = 111A + 11B + C = 300.
Look at A first: it's multiplied by 111. A = 3 gives 333 (over 300); A = 1 gives 111, and even with B = C = 9 you'd reach only 111 + 99 + 9 = 219 (short of 300). So A = 2 is forced.
That leaves 11B + C = 300 − 222 = 78. The biggest B can be is 7 (since 11 × 8 = 88 > 78), and 11 × 7 = 77 leaves C = 1.
Check: 271 + 27 + 2 = 300, and A, B, C = 2, 7, 1 are all different. So C = 1.
Why this transfers: in a cryptarithm, attack the leftmost (highest place-value) digit first — it carries the most weight and usually only one digit fits, which then unlocks the rest.

Another way — add column by column with carries:

Units column: A + B + C must end in 0 (the total's ones digit). Tens column: A + B plus any carry must also end in 0. Hundreds column: A plus the carry from the tens must equal 3.
If the tens carry 1 into the hundreds, then A + 1 = 3, so A = 2. Filling that back through the columns gives B = 7 and C = 1, and 271 + 27 + 2 = 300 checks out — so C = 1.