The number of positive even divisors of 6300 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) First,find the prime factorization of $6300$: $6300 = 63 	imes 100 = (9 	imes 7) 	imes (10^2) = (3^2 	imes 7^1) 	imes (2^2 	imes 5^2) = 2^2

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) First,find the prime factorization of $6300$: $6300 = 63 	imes 100 = (9 	imes 7) 	imes (10^2) = (3^2 	imes 7^1) 	imes (2^2 	imes 5^2) = 2^2 	imes 3^2 	imes 5^2 	imes 7^1$. Total number of divisors is given by the product of (exponent $+ 1$) for each prime factor: Total divisors $= (2+1)(2+1)(2+1)(1+1) = 3 	imes 3 	imes 3 	imes 2 = 54$. Odd divisors are formed by the prime factors excluding $2$: Odd divisors $= (2+1)(2+1)(1+1) = 3 	imes 3 	imes 2 = 18$. Therefore,the number of even divisors is the total number of divisors minus the number of odd divisors: Even divisors $= 54 - 18 = 36$.

The number of positive even divisors of $6300$ is

Similar Questions

The sum of the divisors of $9000$ is . . . . . . .

The coefficient of $x^{1012}$ in the expansion of $(1 + x^n + x^{253})^{10}$,where $n \leq 22$ is any positive integer,is

If $33!$ is divisible by $2^n$,where $n \in N$,then the sum of all possible values of $n$ is:

If $2^{n}$ divides $16!$ and $2^{n+1}$ does not divide $16!$,then $n=$

If the integer represented by $100!$ has $K$ consecutive zeroes at the end,then $K=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module