Subject: General Aptitude · Chapter: Number System · Exam: 2010 · Marks: · Difficulty:
The number ( $2^{48}-1$ ) is exactly divisible by two numbers between 60 and 70. The numbers are
(a)63 and 65
(b)63 and 67
(c)61 and 65
(d)65 and 67
Answer
Answer (as printed): A
Explanation
$\left(2^{48}-1\right)=\left[\left(2^{6}\right)^{8}-1\right]=(64)^{8}-1$. When $n$ is even, $\left(x^{n}-a^{n}\right)$ is completely divisible by both $(x-a)$ and $(x+a)$. $\therefore\left(64^{8}-1^{8}\right)$ is divisible by both $(64-1)$ and $(64+1)$ $\Rightarrow\left(2^{48}-1\right)$ is divisible by both 63 and 65.
Explanation as extracted from the printed page; notation may be imperfect.