Question

Find the no of divisors of 2^2*3^3*5^3*7^5 of the form 4n+1 when n belongs to w

Answer 1

Step-by-step explanation:

number of divisors are given by the product of the powers of prime number thus

(2+1)(3+1)(3+1)(5+1)

which is

3 * 4 * 4 * 6 = 288

let 288 = 4n + 1

4n = 287

n = 71.75

thus

the answer is 4(71.75)+1