18. What is the largest integer $k$ whose square $k^{2}$ is a factor of $10!$ ?
$$
\text { [10! }=10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \text {.] }
$$

A 6
B 256
C 360
D 720
E 5040
Solution
D
Commentary
The most straightforward way to answer this question would be to first evaluate 10!, and then try out the squares of the given options, starting with the largest, until you find a square that is a factor of 10 !

If you do the multiplication you will find that $10!=3628800$. If you now test the options, you will find that $5040^{2}$ is not a factor of 10 !, but that $720^{2}$ is.

However, this approach has several disadvantages. In the SMC calculators are not allowed, and this method requires a lot of arithmetic. Outside the context of SMC, we do not have only five options to test. We would have to start with the largest integer, $k$, such that $k^{2} \leq 10$ !, and check whether $k^{2}$ is a factor of 10 !. If not we would need to check whether $(k-1)^{2}$ is factor of 10 !, and so on, until we find the largest integer whose square is a factor of 10 ! Also, this method gives no insight into the problem and if 10 were replaced by a much larger number, it would not be feasible to use this method.

A better method is to work with the prime factorization of 10 !, using the fact that when we factorize a square into primes, each prime occurs to an even power. So we begin by factorizing 10 ! into primes, and then we look for the highest product of even powers of the prime factors that is a factor of 10 !

We express 10 ! as a product of prime numbers as follows.
$$
\begin{aligned}
10! & =10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\
& =(2 \times 5) \times 3^{2} \times 2^{3} \times 7 \times(2 \times 3) \times 5 \times 2^{2} \times 3 \times 2 \\
& =2^{8} \times 3^{4} \times 5^{2} \times 7 \\
& =\left(2^{4} \times 3^{2} \times 5\right)^{2} \times 7 .
\end{aligned}
$$

We deduce that the square of $2^{4} \times 3^{2} \times 5$ is a factor of 10 ! but there is no larger integer $k$ such that $k^{2}$ is a factor of 10 ! So the largest integer $k$ whose square $k^{2}$ is a factor of 10 ! is $2^{4} \times 3^{2} \times 5=16 \times 9 \times 5=720$.

For investigation
18.1 Which is the largest integer $k$ whose square, $k^{2}$, is a factor of $11!$ ?
18.2 Which is the largest integer $k$ whose square, $k^{2}$, is a factor of 12 !?
18.3 Which is the least integer, $n$, such that $126^{2}$ is a factor of $n$ !?