|
|
Divisors are in the form 4k+2 = 2(2k+1)
All the odd divisors when multiplied by 2 would give the required divisors, hence we have to find the no. of odd divisors (including 1).
Exponents of odd prime divisors are 5, 10, 9.
So no. of odd divisors (including 1) are (5+1)(10+1)(9+1) = 660
Total no. of divisors of the form 4k+2 is 660, none of the options given are correct.
|
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
1044
1
