# Difference between revisions of "2006 AIME A Problems/Problem 3"

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<math>\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97</math> | <math>\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97</math> | ||

+ | |||

threes in <math>\displaystyle 200!</math> and | threes in <math>\displaystyle 200!</math> and | ||

<math>\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48 </math> | <math>\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48 </math> |

## Revision as of 14:29, 24 July 2006

## Problem

Let be the product of the first positive odd integers. Find the largest integer such that is divisible by

## Solution

Note that the product of the first positive odd integers can be written as

Hence, we seek the number of threes in decreased by the number of threes in

There are

threes in and

threes in

Therefore, we have a total of threes.