Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
One less than (49) ${ }^{15}$ is exactly divisible by
(a)8
(b)14
(c)50
(d)51
Answer
Answer (as printed): A
Explanation
$\left(49^{15}-1\right)=\left(7^{2}\right)^{15}-1=7^{30}-1$. Now, when $n$ is even, $\left(x^{n}-a^{n}\right)$ is divisible by both $(x-a)$ and $(x+a)$. $\therefore\left(7^{30}-1\right)$ is divisible by both $(7-1)$ and $(7+1)$ i.e. by both 6 and 8. Thus, $\left[(49)^{15}-1\right]$ is divisible by both 6 and 8 .
Explanation as extracted from the printed page; notation may be imperfect.