$20!$ Has nos that are multiples of $2,3,4$ and so on. However, the total number of integers is large. So, please help me.
$\endgroup$ 22 Answers
$\begingroup$Since $20!=2^{18}\cdot3^{8}\cdot5^{4}\cdot7^{2}\cdot11^{1}\cdot13^{1}\cdot17^{1}\cdot19^{1}$:
- $2$ can appear in every divisor between $0$ and $18$ times, i.e., $19$ combinations
- $3$ can appear in every divisor between $0$ and $8$ times, i.e., $9$ combinations
- $5$ can appear in every divisor between $0$ and $4$ times, i.e., $5$ combinations
- $7$ can appear in every divisor between $0$ and $2$ times, i.e., $3$ combinations
- $11$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations
- $13$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations
- $17$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations
- $19$ can appear in every divisor between $0$ and $1$ times, i.e., $2$ combinations
Therefore, the number of divisors of $20!$ is $19\cdot9\cdot5\cdot3\cdot2\cdot2\cdot2\cdot2=41040$.
$\endgroup$ 2 $\begingroup$Hint:
Use Legendre's formula:
For each prime $p\le n$, the exponent of $p$ in the prime decomposition of $n!$ is $$v_p(n!)=\biggl\lfloor\frac{n}{p}\biggr\rfloor+\biggl\lfloor\frac{n}{p^2}\biggr\rfloor+\biggl\lfloor\frac{n}{p^3}\biggr\rfloor+\dotsm$$
The number of prime divisors of $n!$ is then $$\prod_{\substack{ p\;\text{prime}\\p\le n}}\bigl(v_p(n!)+1\bigr).$$
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