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The number $(2^{48} - 1)$ is exactly divisible by two numbers between 60 and 70. Which are they?

Competition practiceNumber System, HCF & LCM

A61 and 67
B65 and 67
C63 and 65
D63 and 67

Step-by-Step Solution

2^{48} - 1 = (2^{24} - 1) (2^{24} + 1)

and

2^{24} - 1 = (2^{6})^{4} - 1

is divisible by

2^{6} - 1 = 63

and

2^{6} + 1 = 65

.

Answer: C) 63 and 65

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