I'd use the exact same logic Jay employed here.

Here's a neat shortcut you can use, once you know you're looking for the number of 5's in the prime factorization of 53! * 5 * 11:

Number of terms in 53! with at least one five in its prime factorization:: 53/5 = 10 + 3/5, so there are 10 terms with at least one five.

Number of terms in 53! with at least two fives in its prime factorization:: 53/(5^2) = 2 + 3/25, so there are 2 terms with at least two fives.

(Because 53/(5^3) < 1, there are no terms that have three fives.)

Number of 5's in 53!: 10 + 2 = 12

If there are twelve 5's in 53! and one more five in the produce 53! * 5 * 11, there will be a total of 12 + 1 = 13 fives.

VJesus12 wrote:How many trailing Zeroes does 53! + 54! have?

(A) 12
(B) 13
(C) 14
(D) 15
(E) 16

To determine the number of trailing zeros, we need to determine the number of 5-and-2 pairs within the prime factorization of 53! + 54!. Let's start by simplifying 53! + 54!.

53! + 54! = 53!(1 + 54) = 53!(55)

Since we know there are fewer 5s than 2s in 53!(55), 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 53!, we can use the following shortcut in which we divide 53 by 5, then divide the quotient of 53/5 by 5 and continue this process until we no longer get a nonzero quotient.

53/5 = 10 (we can ignore the remainder)

10/5 = 2

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 53!.

Thus, there are 10 + 2 = 12 factors of 5 within 53!.

Finally, we see that there is one factor of 5 within 55.

Since there are 13 factors of 5 within 53!(55), there are thirteen 5-and-2 pairs and thus 13 trailing zeros.

Answer: B