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**Q.) Find the sum and no of factors of 1200 such that the factors are divisible by 15.**

Solution:

1200 = ${ 2 }^{ 4 }*{ 5 }^{ 2 }*3$

for the numbers to be divisible by 15, it should compulsorily have 3 and 5 in it.

sum of factors divisible by 15 = ${ (2 }^{ 0 }+{ 2 }^{ 1 }+{ 2 }^{ 2 }+{ 2 }^{ 3 }+{ 2 }^{ 4 })(5^{ 0 }+5^{ 1 })(3)$

no of such factors = (4+1)*(2)*(1) Ans