AMC 8 2023 Problem 16: Number Theory Solution

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Answer: C — 133 Ps, 134 Qs, 133 Rs.

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Hint 1 of 2

20 isn't a multiple of 3, so the three letters can't come out perfectly equal — 400 = 3 × 133 + 1 means exactly one letter gets one extra. The real work is finding which letter.

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Hint 2 of 2

Trick to balance the board: pretend you add a 21st column that's a copy of column 3. Each row of P,Q,R repeating across 21 cells holds exactly 7 of each — perfectly even. Then just subtract back off the column you added.

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Approach: pad to a clean width, then subtract the extra column

Since 20 isn't divisible by 3, the counts can't be equal: 400 = 3 × 133 + 1, so two letters tie at 133 and one gets 134. The whole question is which letter wins the leftover.
Make the width friendly: imagine appending a 21st column identical to the 3rd column. Now every row spans 21 = 3 × 7 cells, so each row has exactly 7 P, 7 Q, 7 R. Over 20 rows that's 140 of each — perfectly balanced.
Now undo it: the column you added (a copy of column 3, which reads P, R, Q, P, R, … down 20 cells) contains 7 Ps, 6 Qs, 7 Rs. Subtract from 140 each: 140−7 = 133 Ps, 140−6 = 134 Qs, 140−7 = 133 Rs.
So the table has 133 Ps, 134 Qs, 133 Rs. This transfers: when a repeating pattern doesn't fit evenly, pad it to a clean multiple where counting is trivial, then subtract the padding.

Another way — tile with 1×3 triominoes (MAA):

Any 1×3 straight tile laid on the pattern covers exactly one P, one Q, and one R. So wherever the board tiles cleanly, the three letters stay perfectly balanced.
Split the 20×20 board into an 18×18 square, a 2×18 strip, an 18×2 strip (all divisible by 3, so balanced), plus a leftover 2×2 corner.
That corner reads Q R / P Q — one extra Q. So the whole board has one more Q than P or R: 133 Ps, 134 Qs, 133 Rs.