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Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 1"

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== Solution ==
 
== Solution ==
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Let <math>\lfloor x \rfloor</math> 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 <math>\lfloor \frac{85}{3} \rfloor</math>. But that doesn't count the 2 3's in 9, so we need to add that to <math>\lfloor \frac{85}{9}</math> and <math>\lfloor \frac{85}{27} \rfloor</math> and <math>\lfloor \frac{85}{81}  \rfloor</math>, giving us 41.
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Let <math>\lfloor x \rfloor</math> 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 <math>\lfloor \frac{85}{3} \rfloor</math>. But that doesn't count the 2 powers of 3 in 9, so we need to add that to <math>\lfloor \frac{85}{9} \rfloor</math> and <math>\lfloor \frac{85}{27} \rfloor</math> and <math>\lfloor \frac{85}{81}  \rfloor</math>, giving us <math>28+9+3+1=41.</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:16, 5 April 2020

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 (Problems • Answer Key • Resources)
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
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