Problem 11 · 2019 Math Kangaroo
Hard
Number Theory
factorizationfactor-pairs
Which is the highest power of three that divides the number \(7! + 8! + 9!\)? (Recall that \(n! = n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\).)
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Answer: D — \(3^{6}\)
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Hint 1 of 2
Factor out the smallest factorial: \(7! + 8! + 9! = 7!\,(1 + 8 + 8\cdot 9)\).
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Hint 2 of 2
Count the factors of 3 in \(7!\) and in the bracket \(1 + 8 + 72 = 81\) separately, then add.
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Approach: factor out \(7!\) and count powers of 3
- \(7! + 8! + 9! = 7!\,(1 + 8 + 8\cdot 9) = 7!\,(1 + 8 + 72) = 7!\cdot 81\).
- Among 1…7 the multiples of 3 are 3 and 6, so \(7!\) contributes \(3^{2}\).
- \(81 = 3^{4}\), so the total power of 3 is \(2 + 4 = 6\).
- The highest power of three dividing the sum is \(3^{6}\) — answer (D).
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