# The sum of the Natural Numbers which are Divisors of -100 is

- 116
- 117
- 216
- 217

The sum of the natural numbers which are Divisors of 100 is** 217.**

**Formula** –

(P^{a}) = (P^{a + 1} – 1) / (P – 1)

Where, P = Divisors and a = power of prime number.

s(100) = s(2^{2} x 5^{5}). As 2^{2} and 5^{5} are relatively prime,

s(2^{2}) = 2^{3} – 1 = 7 and s(5^{5}) = (5^{3} – 1) / 4 = 31.

Therefore, s(100) = 7 x 31 = 217. Ans.

(P^{a}) = (P^{a + 1} – 1) / (P – 1)

Where, P = Divisors and a = power of prime number.

s(100) = s(2^{2} x 5^{5}). As 2^{2} and 5^{5} are relatively prime,

s(2^{2}) = 2^{3} – 1 = 7 and s(5^{5}) = (5^{3} – 1) / 4 = 31.

Therefore, s(100) = 7 x 31 = 217. Ans.