Waste less time on Facebook — follow Brilliant.
×

Proof Practice-I

\(\text {Proof the following}\)

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

Note by Anik Mandal
1 year, 11 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

One line proof. \( (k+1)! + i \) where \( 2 \leq i \leq k+1 \) Siddhartha Srivastava · 1 year, 11 months ago

Log in to reply

@Siddhartha Srivastava Sorry,I didn't get it.Please explain it clearly@Siddhartha Srivastava Anik Mandal · 1 year, 11 months ago

Log in to reply

@Anik Mandal 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. Siddhartha Srivastava · 1 year, 11 months ago

Log in to reply

@Anik Mandal 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. Dinesh Chavan · 1 year, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...