N = $$2^{48} - 1$$ and N are exactly divisible by two numbers between 60 and 70. What is the sum of those two numbers?
$$2^{48}-1$$
$$=\left(2^{24}+1\right)\left(2^{24}-1\right)$$
$$=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{12}-1\right).$$
$$=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^6+1\right)\left(2^6-1\right).$$
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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