2012/014) The no. of integers that are both multiples of 2002^2004 and factors of 2002^2006 is

we know 2002^2006/2002^2004 = 2002^2

the no. of integers that are both multiples of 2002^2004 and factors of 2002^2006 is number of inetgers factor of 2002^2

now 2002 =2 * 7 * 11 * 13

so 2002^2 = 2^2 * 7^2 * 11^2 * 13 ^2
number of factors = (2+1)(2+1)(2+1)(2+1) = 3^ 4 = 81

( rationale:
Sure, there is! It just depends on how many prime factors your number has, and how many times each one appears.

Suppose your number factors as a product p^a * q^b * .... *r^k. Then to get any factor you want you should take _up to_ a copies of p, _up to_ b copies of q, etc. and multiply them all together. The number of ways you can choose _up to_ a copies of p is (a+1) since you could choose 0 copies, 1 copy, 2 copies, ..., a copies. Likewise there are (b+1) ways to choose how many q's to include, etc.

So the number of factors for your number would be (a+1)*(b+1)*...(k+1).)