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$20!$ Has nos that are multiples of $2,3,4$ and so on. However, the total number of integers is large. So, please help me.

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    Hint : Look at the prime factorization of $20!$ and use the formula for the number of divisors. You should get $41040$ – 2017-01-03
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    What are all the prime factors in $20!$ ? – 2017-01-03

2 Answers 2

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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 $19\cdot9\cdot5\cdot3\cdot2\cdot2\cdot2\cdot2=41040$.

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    Thanks a lot for this.....I have now understood this problem very nicely. – 2017-01-04
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    @LokeshSangewar: You're welcome. Feel free to accept my answer by clicking on the V next to it :) – 2017-01-04
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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 $$\prod_{\substack{ p\;\text{prime}\\p\le n}}\bigl(v_p(n!)+1\bigr).$$