IIT JEE , AIEEE Forum - Mathematics , Algebra - Number of divisers of the form 4n+2(n>=0)of the int. 240 is a)4 b)8 c)10 d)3

little_genius

4n+2= 2*(2n+1)
so the power of 2 is only 1.. (since 2n+1 is odd)

240= 2^4 * 5* 3
every divisor is of the form 2^a *5^b*3^c .. ( 0<=a<=4, 0<=b<= 1, 0<=c<=1..)
we have to keep the power of 2 as 1. i.e a=1.
therefore the no of divisors . total no of ways in which we can choose b and c = 2*2=4 :)

Ask & Discuss Questions with Community & Experts