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Q.

Consider the natural number $N = 3600$ . If the sum of the reciprocals of all the positive integral divisors of $N$ is $R$ then $[R]$ is (where $[R]$ is the greatest integer less than or equal to $R$ )

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answer is 3.

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Detailed Solution

$3600 = 2^{4} \times 3^{2} \times 5^{2} Sum of all factors = (2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4}) (3^{0} + 3^{1} + 3^{2}) (5^{0} + 5^{1} + 5^{2}) \begin{aligned} \Rightarrow (1 + 2 + 4 + 8 + 16) (1 + 3 + 9) (1 + 5 + 25) \\ \Rightarrow 31 \times 13 \times 31 \\ \Rightarrow 12493 \end{aligned}$

$R = \frac{S u m o f a l l d i v i s o r s}{g i v e n n u m b e r} = \frac{12493}{3600}$

$\Rightarrow 3.47$

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Consider the natural number N = 3600. If the sum of the reciprocals of all the positive integral divisors of N is R then R is (where R is the greatest integer less than or equal to R)