Home
Class 11
MATHS
Number of divisors of the form 4n+2(n ge...

Number of divisors of the form `4n+2(n ge 0)` of integer 240 is :

A

4

B

8

C

10

D

3

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

If N = 10800, find the(i) the number of divisors of the form 4m +2,(ii) the number of divisors which are multiple of 10(iii) the number of divisors which are multiple of 15.

Total number of 480 that are of the form 4n+2, n ge 0 , is equal to

Statement-1: the number of even divisors of the number N=12600 is 54. Statement-2: 0,2,4,6,8, . . . Are even integers.

Total number of divisors of n = 3^5. 5^7. 7^9 that are in the form of 4lambda + 1; lamda >=0 is equal to

For some integer n every odd integer is of the form

Find the number of divisors of the number N=2^3 .3^5 .5^7 .7^9 which are perfect squares.

Let f(n) denotes the number of different ways, the positive integer n ca be expressed as the sum of the 1's and 2's. for example, f(4)=5. i.e., 4=1+1+1+1 =1+1+2=1+2+1=2+1+1=2+2 Q. The number of solutions of the equation f(n)=n , where n in N is

Number of odd proper divisors of 3^(p).6^(m).21^(n) is :

Let n be 4-digit integer in which all the digits are different. If x is the number of odd integers and y is the number of even integers, then

Total number of divisors of N=2^(5)*3^(4)*5^(10)*7^(6) that are of the form 4n+2,n ge 1 , is equal to