2011 UNCO Math Contest II Problems/Problem 1

Revision as of 19:16, 5 April 2020 by Enderramsby (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $28+9+3+1=41.$

See Also

2011 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
Invalid username
Login to AoPS