The number of divisors of the form `4K + 2 , K ge 0` of the integers 240 is

The number of divisors of the form (4n+2) of the integer 240 is

If n=10800, then find the a. total number of divisors of n. (b) The numberof even divisors. (c) The number of divisors of the form 4m+2. (d) The number of divisors which are multiples of 15.

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.

Show that there are an infinite number of primes of the form 4k + 1 and of the form 4k+ 3.

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

The total number of divisor of 480 that are of the form 4n+2,ngeq0, is equal to a. 2 b. 3 c. 4 d. none of these

If N denotes the set of all positive integers and if f : N -> N is defined by f(n)= the sum of positive divisors of n then f(2^k*3) , where k is a positive integer is

If sintheta=K, -1<=K<=1 then number of values of theta for same value of K in [0,2pi]

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

The number of values of k for which the system of the equations (k+1)x+8y=4k and k x+(k+3)y=3k-1 has infinitely many solutions is 0 b. 1 c. 2 d. infinite