Problem of the Week
Problem E and Solution
Sixty-Four!

Problem

The product $64\times 63\times 62\times \cdots \times 3\times 2\times 1$ can be written as $64!$ and called "$64$ factorial".

In general, the product of the positive integers $1$ to $m$ is $$m!=m\times (m-1)\times (m-2)\times \cdots \times 3\times 2\times 1$$

If $64!$ is divisible by ${2025}^n$, determine the largest positive integer value of $n$.

Solution

Let $P=64!$. The prime factorization of $2025$ is $3^4\times 5^2.$ We must determine how many times the factors of $3$ and $5$ are repeated in the factorization of $P.$

First we count the number of factors of $3$ in $P$ by looking at the multiples of $3$ from $1$ to $64.$ They are $3$, $6$, $9$, $\dots$, $57$, $60$, and $63.$ Each of these $21$ numbers contains a factor of $3.$

Now, each multiple of $9$ from $1$ to $64$ will contain a second factor of $3$. These multiples of $9$ are $9$, $18$, $27$, $36$, $45$, $54$, and $63.$ Each of these $7$ numbers contains two factors of $3.$

Now, each multiple of $27$ from $1$ to $64$ will contain a third factor of $3$. These multiples of $27$ are $27$ and $54.$ Each of these $2$ numbers contains three factors of $3.$

There are no higher powers of $3$ less than $64.$ Thus, $P$ has $21+7+2=30$ factors of $3$, and so the largest power of $3$ that $P$ is divisible by is $3^{30}.$

Next we count the number of factors of $5$ in $P$ by looking at the multiples of $5$ from $1$ to $64.$ They are $5$, $10$, $15$, $\dots$, $50$, $55$, and $60.$ Each of these $12$ numbers contains a factor of $5.$

Now, each multiple of $25$ from $1$ to $64$ will contain a second factor of $5.$ These multiples of $25$ are $25$ and $50.$ Each of these $2$ numbers contains two factors of $5.$

There are no higher powers of $5$ less than $64.$ Thus, $P$ has $12+2=14$ factors of $5$, and so the largest power of $5$ that $P$ is divisible by is $5^{14}.$

Thus, $P$ is divisible by $3^{30}\times 5^{14}.$ $$3^{30}\times 5^{14}=3^{28}\times 3^2\times 5^{14}={(3^4\times 5^2)}^7\times 3^2={2025}^7\times 3^2$$ Thus, $P$ is divisible by ${2025}^7$, and $7$ is the largest value of $n.$