Question

The number $\left(49^{2}-4\right)\left(49^{3}-49\right)$ is divisible by

• A. $5 !$
• B. $6 !$
• C. $7 !$
• D. $9 !$

Answer 1

Answer: (A)

$\left(49^{2}-4\right)\left(49^{3}-49\right) =\left(49^{2}-2^{2}\right) 49 .\left(49^{2}-1\right)$
$=(49-2)(49+2) 49 .(49-1)(49+1)$
$=47.51 .49 .48 .50$
$=47.48 .49 .50 .51$
It is the product of $5$ consecutive integers and hence it is divisible by $5!$