Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AIME A Problems/Problem 3

Revision as of 14:28, 24 July 2006 by 4everwise (talk | contribs)

Problem

Let $\displaystyle P$ be the product of the first $\displaystyle 100$ positive odd integers. Find the largest integer $\displaystyle k$ such that $\displaystyle P$ is divisible by $\displaystyle 3^k .$

Solution

Note that the product of the first $\displaystyle 100$ positive odd integers can be written as $\displaystyle 1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{200!}{2(100)!}.$

Hence, we seek the number of threes in $\displaystyle 200!$ decreased by the number of threes in $\displaystyle 100!.$

There are

$\displaystyle \lfloor \frac{200}{3}\rfloor+\lfloor\frac{200}{9}\rfloor+\lfloor \frac{200}{27}\rfloor+\lfloor\frac{200}{81}\rfloor =66+22+7+2=97$

threes in $\displaystyle 200!$ and

$\displaystyle \lfloor \frac{100}{3}\rfloor+\lfloor\frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor\frac{100}{81}\rfloor=33+11+3+1=48$

threes in $\displaystyle 100!$

Therefore, we have a total of $\displaystyle 97-48=049$ threes.


See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_A_Problems/Problem_3&oldid=8312"