Question 10

N = $$2^{48} - 1$$ and N are exactly divisible by two numbers between 60 and 70. What is the sum of those two numbers?

Solution

$$2^{48}-1$$

$$=\left(2^{24}+1\right)\left(2^{24}-1\right)$$

$$=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{12}-1\right).$$

$$=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^6+1\right)\left(2^6-1\right).$$

So, Those two numbers are $$\left(2^6+1\right)\ and\ \ \left(2^6-1\right).$$

or, 65 and 63.

Sum of these numbers=65+63=128.

A is correct choice.

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