Question

sum of natural no which are divisor by 100​

Answer 1

Answer:

As the prime factorization of 100 is 2²×5², the sum sigma(100) of all positive divisors is (1+2+2²)×(1+5+5²) = 7×31 = 217, because every divisor of 100 is a unique product of a divisor of 4 with a divisor of 25.

Step-by-step explanation:

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Answer 2

Answer:

The sum of the naturalnumbers which are divisors of-100 is - 217