3. How many positive perfect cubes are divisors of the product $1!\cdot 2!\cdot 3!\cdots 10$ !?
Answer: 468
Solution: We have $N:=1!\cdot 2!\cdot 3!\cdots 10!=2^{38} 3^{17} 5^{7} 7^{4}$. Thus, a positive divisor of $N$ that is a perfect cube must be of the form $2^{3 a} 3^{3 b} 5^{3 c} 7^{3 d}$ for some nonnegative integers $a, b, c, d$. We see that $3 a \leq 38,3 b \leq 17,3 c \leq 7$ and $3 d \leq 4$. Thus, there are $\left\lceil\frac{38}{3}\right\rceil=13$ choices for $a,\left\lceil\frac{17}{3}\right\rceil=6$ choices for $b$, $\left\lceil\frac{7}{3}\right\rceil=3$ choices for $c$, and $\left\lceil\frac{4}{3}\right\rceil=2$ choices for $d$. Hence, there are $13 \cdot 6 \cdot 3 \times 2=468$ positive perfect cube divisors of $N$.