manhattan challenge problem -- terminating zeros

This type of problem has been discussed before, but here goes:

The number of terminating zeroes in 200! is actually the number of 10s you have in this product (sine when you multiply say 3 by 10, it adds a zero to the end of it = 30). Now, the prime factorization for 10 = 2*5. There are plenty of 2s in 200! (at least one every other number, in even numbers), so the number of 10s in 200! will be equal to the number of 5s in 200!. There is a formula for determining this and your answer will be:
200/5 + 200/(5^2) + 200/(5^3) = 40 + 8 + 1 = 49.

The formula you need to determine the number of 5s in any given permutation (i.e. n!) will be the sum of the rapports of n and positive powers of 5 smaller than n:

n/(5^1) + n/(5^2) + n/(5^3) + ...

I stopped at 5^3 because 5^3 = 125 < 200, but 5^4 = 625 > 200.

To better understand this type of problem, here's another example: how many 5s do you have in 60! ?
The answer will be 60/5 + 60/25 = 12 + 2 = 14. We stop at 25 = 5^2 since 5^3 = 125 is greater than 60.

Hope this helps.

Phew!!!! Man this is new.. I dint know this stuff.. thanks buddy..

nh8404052006 wrote:How many terminating zeroes does 200! have?

OA: 49
Please explain, thank

Source: Manhattan challenge problem
https://gmat-maths.blocked/search?u ... results=20

Recall that each 5-and-2 pair (which makes 10 when multiplied) in the factorization of a number results in one trailing zero. Thus, we need to determine the number of 5-and-2 pairs within the prime factorization of that number.

Since we know there are fewer 5s than 2s in 200!, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 200!, we can use the following shortcut in which we divide 200 by 5, then divide the quotient of 200/5 by 5 and continue this process until we no longer get a nonzero quotient.

200/5 = 40

40/5 = 8

8/5 = 1 (we can ignore the remainder)

Since 1/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 200!.

Thus, there are 40 + 8 + 1 = 49 factors of 5 within 200!

Since there are 49 factors of 5 within 200!, there are 49 5-and-2 pairs and thus
49 trailing zeros.