If m and n are positive integers, is n^m n divisible by 6?

\( n^m - n = n(n^{m-1} - 1)\)

Let's start with statement 2 as its simpler of the two

Statement 2

n = 2

Case 1 : m = 3

2 (\(2^2 \)- 1) = 2 * 3 = 6 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

Case 2 : m = 4

2 (\(2^3 \)- 1) = 2 * 7 = 14 ⇒ Is \(n^m − n\) divisible by 6 ? -- No

Statement 2 is not sufficient, hence we can rule out B and D.

Statement 1

m = 3

\(\frac{n}{6}\) can leave a remainder between 0 and 5, both inclusive.

Let's see in each case what's the remainder when \(n^3 - n\) is divided by 6

When Rem(\(\frac{n}{m}\)) = 0

Rem(\(\frac{n^3 - n }{ 6}\)) = 0 - 0 = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

When Rem(\(\frac{n}{m}\)) = 1

Rem(\(\frac{n^3 - n }{ 6}\)) = 1 - 1 = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

When Rem(\(\frac{n}{m}\)) = 2

Rem(\(\frac{n^3 - n }{ 6}\)) = 8 - 2 = 6

Rem(\(\frac{6}{6}\)) = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

When Rem(\(\frac{n}{m}\)) = 3

Rem(\(\frac{n^3 - n }{ 6}\)) = 27 - 2 = 24

Rem(\(\frac{24}{6}\)) = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

When Rem(\(\frac{n}{m}\)) = 4

Rem(\(\frac{n^3 - n }{ 6}\)) = 64 - 4 = 60

Rem(\(\frac{60}{6}\)) = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

When Rem(\(\frac{n}{m}\)) = 5

Rem(\(\frac{n^3 - n }{ 6}\)) = 125 - 5 = 120

Rem(\(\frac{120}{6}\)) = 0 ⇒ Is \(n^m − n\) divisible by 6 ? -- Yes

Statement 1 is sufficient

Option A