enigma123 wrote:

How many terminating zeroes does 200! have?

(A) 40

(B) 48

(C) 49

(D) 55

(E) 64

Factors of 10 create the zeros. In order to count how many tens and thus how many zeros, use the method described by

hyena1986 - with one minor adjustment.

There will be many more powers of 2 than there are powers of 5. Every 5 will have a matching 2 (to get 5*2=10), but not every 2 will have a matching 5.

So consider

only powers of 5. Divide the factorial number (w/o !) by separate and increasing powers of 5.

Don't worry about remainders (we just need how many powers of \(5^{n}\) exist in the factorial, not how many "fit" perfectly).

\(\frac{200}{5^{1}}\) = 40

\(\frac{200}{5^{2}}\) = \(\frac{200}{25}\) = 8

\(\frac{200}{5^{3}}\) = \(\frac{200}{125}\) = 1

\(5^4\) is 600. Too large.

Add the results: 49 + 8 + 1 = 49.

Answer C

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