JEE Advanced Mathematics question

MathematicsPermutations and CombinationsJEE Advanced

The sum of all the proper divisors of 9900 is

Answer: (C)

Solution

. Sum of all divisors:

σ(9900) = (1 + 2 + + 3 + + 5 + + 11) = 7 × 13 × 31

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Question ID VK-107781