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

Revision as of 14:17, 13 August 2015

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 int(85/3). But that doesn't count the 2 3's in 9, so we need to add that to int(85/9), int(85/27), and int(85/81), giving us 41.

@@ Line 5: / Line 5: @@
 == Solution ==
-Let <math>\floor x</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 int(85/3). But that doesn't count the 2 3's in 9, so we need to add that to int(85/9), int(85/27), and int(85/81), giving us 41.
+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 int(85/3). But that doesn't count the 2 3's in 9, so we need to add that to int(85/9), int(85/27), and int(85/81), giving us 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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Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 1"

Revision as of 14:17, 13 August 2015

Problem

Solution

See Also