Trailiiiiiing zer0s

Hi Mathsbuddy,

In these types of questions, when the answers are so "close" together, you really have to be thorough with your thinking and your work. You named all of the multiples of 5 and 10 that exist in 60!, but is that REALLY what the question is asking for?

Take the number 50 and multiply it by ANY even number. How many zeroes will you end up with (hint: it's NOT 1).

Now take the number 25. If you multiply it by 2, then you get one 0. What happens when you multiply it by 4 though?

The reason why those extra zeros exist is because both 25 and 50 include two 5s within their respective prime factorizations.

GMAT assassins aren't born, they're made,
Rich

Ramit88 wrote:how many trailing zeroes does 60! have?

a 12
b 9
c 15
d 14
e 13

To determine the number of trailing zeros in a number, 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 60!, 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 in 60!, we can use the following shortcut in which we divide 60 by 5 and then divide the quotient of 60/5 by 5 and continue this process until we no longer get a nonzero quotient.

60/5 = 12

12/5 = 2 (we can ignore the remainder)

Since 2/5 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 60!.

Thus, there are 12 + 2 = 14 factors of 5 within 60!, and thus there will be fourteen 5-and-2 pairs, indicating that there are 14 trailing zeros.

Answer: D