Question 53

Given that $$2^{20} + 1$$ is completely divisible by a whole number, which of the following is completely divisible by the same number?

Solution

Let $$2^{20} + 1$$ is completely divisible by a whole number W

$$=$$> $$2^{20} + 1$$ = kW (where k is a constant)

$$\therefore\ $$ $$\left(2^{20}\right)^3+1^3=\left(2^{20}+1\right)\left(2^{40}+2^{20}+1\right)$$

$$=$$> $$2^{60}+1=\left(kW\right)\left(2^{40}+2^{20}+1\right)$$

$$=$$> $$2^{60}+1$$ is completely divisible by whole number W

Hence, the correct answer is Option D

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