How to find the $20$ consecutive composite numbers

Question

I have a little confusion over the following problem:

Find the $20$ consecutive composite numbers.

HINTS: $$\color {green} {\text{Numbers}\,\, 20!+2,20!+3,\cdots ,20!+21 \,\,\text{will do the trick. The following result by Euclid has been known for more than 2000 years.}}$$

But the solution is not very clear to me specially I do not understand why $k$ has been added to $20!$ where the numbers are of the form $20!+k,k=2....,21$.

Can someone explain it? Thanks and regards to all.

You could use $19!$ instead of $20!$ – Mark Bennet Aug 15 '13 at 09:35 — Mark Bennet, Aug 15 '13 at 09:35

score 1 · Answer 1 · answered Aug 15 '13 at 09:01

1

Observe that for each $k \in \{2,\ldots,20\}$, $k$ divides $20!$ and $k$ divides $k$, so $k$ divides their sum. Hence, $20!+k$ is composite.

For the case where $k=21$, simply observe that: $$ 20! = (20) \cdot \ldots \cdot (7) \cdot \ldots \cdot (3) \ldots \cdot (1) $$ Hence, since $21 \mid 20!$ and $21 \mid 21$, it follows that $20! + 21$ is also composite.

answered Aug 15 '13 at 09:01

Adriano

41,969

Thanks @Adriano for the nice explanation. – learner Aug 15 '13 at 09:07
Note also that $20!+22$ is composite – Mark Bennet Aug 15 '13 at 09:33

score 1 · Accepted Answer · answered Aug 15 '13 at 09:02

For every natural number $n$, if $k\le n$ is a natural number, then $k\mid n!+k$ (simply since $k\mid n!$). In particular, if $1<k\le n$, then $n!+k$ is not prime. It follows that all of $n!+2,n!+3,\cdots ,n!+n$ are composite. Thus, for every $n$ you can find a list of $n-1$ consecutive composite numbers.

This should explain how the result you quote was arrived at (you just need to notice that $20!+21$ is also composite.

How to find the $20$ consecutive composite numbers

2 Answers2