Problem 24 · 2006 AMC 8
Hard
Number Theory
factor-1010cryptarithm
In the multiplication problem below, A, B, C, D are different digits. ABA × CD = CDCD. What is A + B?
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Answer: A — 1.
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Hint 1 of 2
Look at the answer CDCD: it's the two-digit block CD written twice. Ask what number you multiply a two-digit block by to repeat it — like how 37 × 1001 = 37037.
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Hint 2 of 2
Repeating a 2-digit block means multiplying by 101 (since CD·101 = CD00 + CD = CDCD). Compare that to the given ABA × CD and the value of ABA falls right out.
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Approach: recognize the repeat as multiplying by 101
- Writing the block CD twice is exactly CD × 101, because CD × 100 shifts it left two places and adding CD fills the bottom: CDCD.
- But the problem says ABA × CD = CDCD = 101 × CD. Cancelling the CD on both sides forces ABA = 101.
- Reading off the digits: A = 1, B = 0, so A + B = 1.
- Worth keeping: repeating an n-digit block multiplies it by a "1 0…0 1" number — 11 for 1 digit, 101 for 2 digits, 1001 for 3. Recognizing these place-value patterns cracks most cryptarithms without trial and error.
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