Problem 7 · 2017 AMC 8
Medium
Number Theory
factorizationdivisibility
Let Z be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of Z?
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Answer: A — 11.
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Hint 1 of 2
The block of 3 digits repeats. Ask what number you multiply abc by to shove it 3 places left and then add it back — that single multiplier carries every factor that must divide Z.
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Hint 2 of 2
Repeating a digit block is always multiplication by a repunit-like constant: abcabc = abc × 1001. Factor that constant to read off the guaranteed divisors.
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Approach: factor the repeat multiplier
- Repeating abc means: abc shifted up 3 digits (×1000) plus abc itself, so Z = abc × 1000 + abc = abc · 1001. That 1001 is the whole problem — it divides Z no matter what abc is.
- Factor it: 1001 = 7 · 11 · 13. So 7, 11, and 13 are all guaranteed factors — from the choices, 11 works.
- Worth memorizing: 1001 = 7·11·13, and likewise 11 = 11, 101 prime, 1111 = 11·101. Repeated-block numbers are a classic AMC divisibility setup — the answer is always inside the repeat constant.
Another way — test the example:
- The problem hands you 247247. Check the choices: 247247 ÷ 11 = 22477 exactly, so 11 divides this case. Since 11 must work for the given example, it's the safe pick — and the abc·1001 argument confirms it holds for every such number.
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