# Problem Of The Week #451 January 18th 2021

Status
Not open for further replies.

Staff member

#### anemone

##### MHB POTW Director
Staff member
Congratulations to kaliprasad for his correct solution, which you can find below:

Let us factorise $30^{2003}$ and $20^{2000}$

$30^{2003}= 3^{2003} * 2 ^{2003} * 5^{2003}$

as 2,3,5 are pairwise co-primes and in factor each can come 0 to 2003 times that is 2004 ways
so number of factors = $(2003+1) * (2003 + 1) * (2003 +1)= 2004^3$

Now
$20^{2000}= 2^{2000} * (2 * 5) ^{2000}= 2^ {4000} * 5^{2000}$

$\gcd (30^{2003},20^{2000}) = 2^{2003} * 5 ^{2000}$

Any number that divides $30^{2003}$ and $20^{2000}$ must divide $\gcd (30^{2003},20^{2000})$

So number of numbers that divide $30^{2003}$ and $20^{2000}$ = $(2003+1)(2000+1) = 2004 * 2001$

So number of numbers that divide $30^{2003}$ and does not divide $20^{2000}$ = $2004^3 - 2004 * 2001 = 2004(2004^2-2021) = 8044086060$

8044086060 is the number of divisors of $30^{2003}$ that are not divisor of $20^{2000}$.

Status
Not open for further replies.