Sufficiently divisible

Number Theory Level 2

$1 \times 2\times 3\times \cdots \times n$

I've calculated the number above, but I seem to have forgotten the value of $n$ .
I recall that the above number is divisible 17, 18 and 19.
If my recollection is correct, is this number also divisible by 20?

Yes No

1 solution

Sabhrant Sachan
Nov 27, 2016

Let $P_n = 1\times 2 \times 3 \times 4 \cdots \times n \\ P_n = n!$

Since $P_n$ is divisible by $19$ , $18$ and $17$ .

$n \ge19$ , since $19$ and $17$ are prime numbers , which must be present in the factorial .

$P_{n} = 1\times 2\times 3 \times \color{#3D99F6}{4 \times 5} \color{#333333}{\times \cdots \cdots}$

It is clearly divisible by $20$

No, n must not necessary be equal to 19, it can be any number greater or equal to 19.

Pi Han Goh - 4 years, 6 months ago

I agree , modifying my solution.

Sabhrant Sachan - 4 years, 6 months ago

Sufficiently divisible

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