Question

The product 1 × 2 × 3 × ………… × n is denoted by n!. For example 4! = 1 × 2 × 3 × 4 = 24. Let M = 1! × 2! × 3! × 4! × 5! × 6! × 7! × 8! × 9!. Let the number of factors of M that are perfect squares be N. what is N/8?

Answer 1

$\rm\large\underline{\text{Answer}}$

$84$

$\rm\large\underline{\text{Explanation}}$

$\boxed{\begin{aligned}M&=1! \times 2! \times 3! \times \cdots \times 9!\\\\&=1^{9} \times 2^{8} \times 3^{7} \times \cdots \times 9 \\\\&=(2^{8} \times 4^{6} \times 8^{2}) \times (3^{7} \times 9) \times 5^{5} \times (2 \times 3)^{4} \times 7^{3} \\\\&=2^{8+12+6+4} \times 3^{7+2+4} \times 5^{5} \times 7^{3}\ [ \because a^{m} \times a^{n} = a^{m+n}]\\\\&=2^{30} \times 3^{13} \times 5^{5} \times 7^{3} \end{aligned}}$

We can find the perfect squares by squaring every factor of,

$\cdots\longrightarrow 2^{15} \times 3^{6} \times 5^{2} \times 7$

So, the number of the factors is equivalent to the number.

The number of the factors

$\boxed{\begin{aligned}N&=(15+1) \times (6+1) \times (2+1) \times (1+1)\\\\&=16\times 7\times 3\times 2\\\\&=2^{5}\times 3\times 7\end{aligned}}$

Hence,

$\cdots \longrightarrow \boxed{ \dfrac{N}{8} = 2^{2} \times 3 \times 7 = \underline{ 84 } }$