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Sum of all the odd divisors of 720 is...

Sum of all the odd divisors of 720 is

A

76

B

78

C

80

D

84

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The correct Answer is:
B
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2^m3^n5^p is a divisor of 2^10 cdot 3^8 cdot 5^7 if m, n and p are all whole number such that 0 le m le 10, 0 le n le 8 and 0 le p le 7 . also, (2^0 + 2^1 +.........+2^10)(3^0 + 3^1 +.......+3^8)(5^0+5^1+....+5^7) = sum of all the divisors of 2^10 cdot 3^8 cdot 5^7 The sum of odd divisors (ne 1) of 10800 is

2^m3^n5^p is a divisor of 2^10 cdot 3^8 cdot 5^7 if m, n and p are all whole number such that 0 le m le 10, 0 le n le 8 and 0 le p le 7 . also, (2^0 + 2^1 +.........+2^10)(3^0 + 3^1 +.......+3^8)(5^0+5^1+....+5^7) = sum of all the divisors of 2^10 cdot 3^8 cdot 5^7 The number of proper divisors of 16200 is

Product of all the even divisors of N = 1000 , is

The number of even divisors of 10800 is

A positive integer n is of the form n=2^(alpha)3^(beta) , where alpha ge 1 , beta ge 1 . If n has 12 positive divisors and 2n has 15 positive divisors, then the number of positive divisors of 3n is

Find the total number of proper factors of the number 35700. Also find (1)sum of all these factors (2)sum of the odd proper divisors (3)the number of proper divisors divisible by 10 and the sum of these divisors.

Find the number of divisors of 720. How many of these are even? Also find the sum of divisors.

In an A.P. of 99 terms, the sum of all the odd-numbered terms is 2550. Then find the sum of all the 99 terms of the A.P.

An A.P. consists of n terms. If the sum of its first three terms is x and the sum of the last three terms is y, then show that, the sum of all the terms of the A.P. is (n)/(6)(x+y) .

The sum of all the solutions of cottheta=sin2theta(theta!=npi, n integer) , 0lt=thetalt=pi, is

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