How many terminating zeroes does 200! have?

This is how I'd solve it: $$200!=200\cdot199\cdot198\cdot\cdots\cdot3\cdot2\cdot1$$ Now, to get a terminating zero, I'll pick the numbers that are multiple of 5:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50,..., 180, 185, 190, 195, 200.

I have 40 numbers.

Now, the numbers that are also multiples of 25, add a zero. For example, 4*25=100, 2*50=100, 75*12=900, . . . A total of 8 numbers.

And 125 add two zeros, 125*8=1000.

Hence, we have 40 terminating zeros + 8 terminating zeros + 1 terminating zero = 49 terminating zeros.

And hence I picked the option C

Roland2rule wrote:How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64

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. Note that each 5-and-2 pair creates a “10,” which contributes one trailing zero.

Since we know there are fewer 5s in 200! than there are 2s, 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!. This means that there are 49 5-and-2 pairs in 200!; therefore, there are 49 terminating zeros in 200!.

Answer: C