2006 AIME II Problems/Problem 3

Problem

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

Solution

Note that the product of the first $100$ positive odd integers can be written as $1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}$

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

There are

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

threes in $200!$ and

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

threes in $100!$

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

For more information, see also prime factorizations of a factorial.

Solution 2

We count the multiples of $3^k$ below 200 and subtract the count of multiples of $2\cdot 3^k$ :

$\left\lfloor \frac{200}{3}\right\rfloor - \left\lfloor \frac{200}{6}\right\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \frac{200}{54}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor - \left\lfloor \frac{200}{162}\right\rfloor$ $= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.$

2006 AIME II (Problems • Answer Key • Resources)
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2006 AIME II Problems/Problem 3

Contents

Problem

Solution

Solution 2

See also