IIT JEE , AIEEE Forum - Mathematics , Algebra

ridhima

total nno. of divisors of n= 3⁵2⁵5¹⁰7⁹that are of the form 4

+2

1 is equal to

a 240

b 30

c 120

d 15

ridhima

hey experts?/////

whr r u guys

nadeemoidu

I'm getting 660 but its not there in the choices.

coool_shetty

u r rite buddy , even i got 660 ...choices are obviously wrong!!!

iitkgp_bipin

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.

ridhima

3^x= (4-1)^x +(-1)^x

5^y = (4+1)^y + 1

7^z = (8-1)^z + (-1)^z

so any positive integral power of 5 will be in the form of 4x +1 . even power of 3 and 7 will be in the form of 4x+1 and odd power in the form of 4x-1 .

thus required no. of divisors is 8.(3.5 +3.5) = 240

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