\(\text {Proof the following}\)

\(\bullet\) Given any positive integer \(\displaystyle k\) ,prove that there are \(\displaystyle k\) consecutive integers that are all composite.

\(\text {Proof the following}\)

\(\bullet\) Given any positive integer \(\displaystyle k\) ,prove that there are \(\displaystyle k\) consecutive integers that are all composite.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestOne line proof. \( (k+1)! + i \) where \( 2 \leq i \leq k+1 \) – Siddhartha Srivastava · 2 years, 7 months ago

Log in to reply

@Siddhartha Srivastava – Anik Mandal · 2 years, 7 months ago

Sorry,I didn't get it.Please explain it clearlyLog in to reply

– Siddhartha Srivastava · 2 years, 7 months ago

Expanding on what Dinesh Chavan said. \( (k+1)! = (k+1)*k*(k-1)...*4*3*2 \). So for each \( 2 \leq i \leq k+1 \), we see that \( i|(k+1)! \). Also, it is trivially true that \( i|i \). Therefore, \( i| (k+1)! + i \). Since \( i \) divides the number and it is greater than one, the number must be composite.Log in to reply

– Dinesh Chavan · 2 years, 7 months ago

If u know factorial notation then notice that for each $i$ that same $i$ from factorial expansion will come out common and the expression will be composite.Log in to reply