the number of zeroes will be determined by the number of 5s and 2s. since 5*2 gives us one zero, the number of zeroes will be the number of (5*2) that we get. So in first case it is 19 and in second case it is 16. Hence A is greater than B.

m=2^{16}3^{17}4^{18}5^{19}

n=2^{19}3^{18}4^{17}5^{16}
We can rewrite it as following:

m= 2^{16}..2^{36}...5^{19} = 2^{52}..5^{19} = 19 zeroes
n= 2^{19}..2^{34}...5^{16} = 2^{53}..5^{16} = 16 zeroes

So A is greater.

A

Can someone explain this in detail?

Answer: A
The parameters that can produce 0 are 2 and 5.
n= 2^19 * 3^18 * 4^17 * 5^16 = 2^16 * 5^16 * 3^18 * 4^17 (there are 16 2s and 16 5s, we didn’t consider 4 because there are only 16 5s, if there were more than 19 5s we used 4s after 2s. So there are 16 zeros in the end of n)
m= 2^16 * 3^17 * 4 ^18 * 5^19= 5^19 * 2^52 * 3^17 = 5^19 * 2^19 * 2^33 * 3^17 (there are 19 multiplication of 2 with 5. So there are 19 zeros in the end of m)
So the number of zeros in m are more than n. And the answer is A.

I think the answer should have been C..a pair of 5 and 2 creates a zero..and among these integers whichever is raised to a lower power, you take that number i.e. 2^19 5^16 would make 16 zeros. Can someone please verify?

The easiest way to solve this problem is to look at all of the factors when combined result in a product that ends in zero which is 2,4, and 5. It would be easier to take out the 4 and 5s first to combine to make 20, then combine the 2s to get 40. What makes the top one different is when you combine the numbers there is an extra 5 left over that can be added to a get an extra zero:

m= 2^16 3^17 4^18 5^19 => combine the 4s and 5s => (20)^18 2^16 3^17 (5) => combine the 2s and 20s => (40)^16 (20)^2 3^17 (5)= combine the remaining 5 with either 40 or 20=> (200) 40^15 20^2 3^17 count the zeroes based on the exponents for each number with zeroes and you should get 19 for m. Repeat the same process for n and you should get 16 zeroes in total. I hope this helps.