What is the GCD of 12^50 and 100!?

Find the prime factorization of these numbers and it's pretty easy to find the greatest common DIVISOR:

12^50 = (2^2 * 3)^50 = 2^100 * 3^50

Now - we can ignore all other prime factors of 100! except 2 and 3. How many factors of 2 and 3 does 100! have?

100! = 1*2*3...*100. This includes 50 factors of 2, 25 factors of 4 (2^2, an extra "2"), 12 factors of 8, 6 factors of 16, 3 factors of 32, and 1 factor of 65. So 100! has:

2^50 * 2^25 * 2*12 * 2^6 * 2^3 * 2^1 = 2^97 as the largest power of 2 as a factor

It also has 3^33 * 3^11 * 3^3 * 3^1 = 3^48 as the largest power of 3 as a factor

the answer is 2^97 * 3^48.

Hope this helped!

P.S. You can also write this as 2 * 12^48.

They are not fractions and do not have denominators

Unless of course, you choose to write them as 12^50/1 and 100/1, in which case the GCD is 1