2011 UNCO Math Contest II Problems/Problem 1

Problem

The largest integer $n$ so that $3^n$ evenly divides $9! = 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9$ is $n = 4$ . Determine the largest integer $n$ so that $3^n$ evenly divides $85! = 1\cdot 2\cdot 3\cdot 4\cdots 84\cdot 85$ .

Solution

Let $\lfloor x \rfloor$ indicate the largest integer less than or equal to x. To solve this problem, we simply need to find the powers of 3 that go into 85!. Thus we get $\lfloor \frac{85}{3} \rfloor$ . But that doesn't count the 2 powers of 3 in 9, so we need to add that to $\lfloor \frac{85}{9} \rfloor$ and $\lfloor \frac{85}{27} \rfloor$ and $\lfloor \frac{85}{81} \rfloor$ , giving us $28+9+3+1=41.$

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
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2011 UNCO Math Contest II Problems/Problem 1

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