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How many terminating zeroes does 200! have?

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How many terminating zeroes does 200! have?  [#permalink]

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New post 26 Dec 2011, 11:31
4
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A
B
C
D
E

Difficulty:

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Question Stats:

71% (00:55) correct 29% (01:46) wrong based on 128 sessions

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How many terminating zeroes does 200! have?

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

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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 26 Dec 2011, 21:45
2
1
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.
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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 26 Dec 2011, 23:03
hyena1986 wrote:
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.


nice explained .
could you please clarify whether zeros coming from 10,20......200 also included in this calculation as you have given above?
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How many terminating zeroes does 200! have?  [#permalink]

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New post 10 Aug 2017, 07:41
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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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 10 Aug 2017, 07:50
Counting the number of 5's gives us the number of trailing zeroes

There are 40 multiples of 5 within 1 to 200

Every multiple of 25 contributes, an additional 5. Since there are 8 multiples of 25, we have 8 additional 5's

Also 125, a multiple of 25, contains one more 5.

So 40+8+1 = 49
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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 10 Aug 2017, 07:52
hyena1986 wrote:
number property states

n!/p + n!/p^2 + n!/p^3 +......

where p is the prime number

to find restricting zeros we need to find number of 5's or 2's in 200! as (10=5*2)

200!/2=100 2's
200/2^2=50 2's
200/2^3=25 2's
200!/2^4 =12 2's
200/2^5=6 2's
200/2^6=3 2's
200/2^7=1 2's
total there are

100+50+25+12+6+3+1= 197 2's

number of 5's in 200! are
40+8+1=49

5 being less in number becomes the restriction for number of 10's in the equation.
hence the answer will be C.


In a given range, the number of 5's is always less than the number of 2's. So its always sufficient to count the number of 5's in these kind of problems.
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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 15 Aug 2017, 10:49
enigma123 wrote:
How many terminating zeroes does 200! have?

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


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.

Answer: C
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Re: How many terminating zeroes does 200! have?  [#permalink]

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New post 04 Oct 2017, 05:56
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enigma123 wrote:
How many terminating zeroes does 200! have?

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


No. of zeroes at the end of 200! = [200/5]+[200/25]+[200/125] = 40 + 8 + 1 = 49
where [x] denotes greatest integer function less than equal to x

Answer C
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Re: How many terminating zeroes does 200! have? &nbs [#permalink] 04 Oct 2017, 05:56
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