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# How many terminating zeroes does 200! have? (A) 40 (B)

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Director
Joined: 26 Feb 2006
Posts: 865
How many terminating zeroes does 200! have? (A) 40 (B)  [#permalink]

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05 Jul 2007, 19:00
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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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Manager
Joined: 12 Apr 2007
Posts: 159

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05 Jul 2007, 19:11
C?

There's 40 multiples of 5, then count 200, 175, 150, 100, 75, 50, and 25 twice, so that's 47...then we have to count 125 three times, so that's 49.

i THINK thats what the question is asking...
Senior Manager
Joined: 04 Jun 2007
Posts: 341

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06 Jul 2007, 14:36
1
200! = 200 * 199* 198* 197*.............* 4* 3 * 2 * 1

there will be one zero for every factor of 10 in 200!

20 factors end in 0 (200, 190, ....20, 10)

also 10= 5*2

20 factors end in 5 ( 195, 175,...15)
20 factors end in 2 (192, 172, ....12)

5*2= 10, so combining the above factors to end in zero's, we get 20 more factors ending in zero's

thus, 200! has 20 + 20 = 40 terminating zeros.

Manager
Joined: 25 Jul 2006
Posts: 97

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06 Jul 2007, 14:41
r019h wrote:
200! = 200 * 199* 198* 197*.............* 4* 3 * 2 * 1

there will be one zero for every factor of 10 in 200!

20 factors end in 0 (200, 190, ....20, 10)

also 10= 5*2

20 factors end in 5 ( 195, 175,...15)
20 factors end in 2 (192, 172, ....12)

5*2= 10, so combining the above factors to end in zero's, we get 20 more factors ending in zero's

thus, 200! has 20 + 20 = 40 terminating zeros.

not limited by factors of 2, since 200! has very many factors of 2 (every even number has one or more factors of 2)
only limited by factors of 5, which is correctly calculated by the previous post = 49
Director
Joined: 26 Feb 2006
Posts: 865
Re: PS: terminating zeros  [#permalink]

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06 Jul 2007, 22:54
Himalayan wrote:
How many terminating zeroes does 200! have?

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

OA is C.

200/5 = 40
200/25 = 8
200/125 = 1
total = 49
VP
Joined: 08 Jun 2005
Posts: 1136

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07 Jul 2007, 06:36
jimmyjamesdonkey wrote:
What is a terminating zero?

let's take the number 30,000 for example.

It has four (i.e. 0000) terminating zeros. You can understand that 200! has a lot of terminating zeros in it since 200! = 200*199*198*197*196 and so on.

lets work with our example (the number 30,000).

factoring 30,000 to its prime numbers:

30,000 | 2
15,000 | 3
5,000 | 5
1,000 | 2
500 | 5
100 | 5
20 | 5
4 | 2
2

30,000 = 2^4*5^4*3.

As you can see, the number of terminating zeros is determined by the number of 10 (or 5*2) in the multiplication.

since we have four times (5*2) then the number of terminating zeros will be 4.

VP
Joined: 08 Jun 2005
Posts: 1136

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07 Jul 2007, 07:17
jimmyjamesdonkey wrote:
So how did you figure out 200!?

You only need to know how many times five is repeating in 200! = 200*199*198*197 and so on ... you need to understand that the number of fives will determine the terminating zeros. I know i wrote that the number of 2*5 will determine the number of terminating zeros, but in 200! for every 5 you have at least ! one 2 so you can count fives and not 5*2.

a safe way is to sit and count all of them one by one:

starting with 1*2*3*4*5*6*7... and 8*9*10 this will take you forever to do.

a better (and quicker) way is to think logically about the number 200!

Since you are digging for fives, you can just ignore numbers that don't have five as their factor (i.e 199,198,197) and just focus on numbers that do have five as their factor (i.e 5,10,200)

200 = 5*40
195 = 5*39
190 = 5*38

So the answer looks like an easy 40 (40 fives in 200!) but that's wrong since you have more fives hiding, 200 = 5*5*8 , and you miscounted nine more fives:

5*5*8 = 200 (you only counted one five here)
5*5*7 = 175
5*5*6 = 150
5*5*5 = 125 (here you missed two fives ! you only counted one)
5*5*4 = 100
5*5*3 = 75
5*5*2 = 50
5*5*1 = 25

so lets sum things up:

40 + 8 + 1 = 49

the answer is 49

CIO
Joined: 09 Mar 2003
Posts: 459

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07 Jul 2007, 08:58
KillerSquirrel wrote:
jimmyjamesdonkey wrote:
What is a terminating zero?

let's take the number 30,000 for example.

It has four (i.e. 0000) terminating zeros. You can understand that 200! has a lot of terminating zeros in it since 200! = 200*199*198*197*196 and so on.

lets work with our example (the number 30,000).

factoring 30,000 to its prime numbers:

30,000 | 2
15,000 | 3
5,000 | 5
1,000 | 2
500 | 5
100 | 5
20 | 5
4 | 2
2

30,000 = 2^4*5^4*3.

As you can see, the number of terminating zeros is determined by the number of 10 (or 5*2) in the multiplication.

since we have four times (5*2) then the number of terminating zeros will be 4.

That'a a nice explanation.

By the way, your picture's a little freaky
Director
Joined: 26 Feb 2006
Posts: 865

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07 Jul 2007, 12:33
ian7777 wrote:
That'a a nice explanation.

By the way, your picture's a little freaky

ian7777,

His picture is to scare the beast (GMAT) not us. imo, he/she would like to kill the difficult gmatp roblems and is doing the same.

I am also considering changing my avatar to scare the difficult gmat questions.

07/07/07 (day once in 1000 years)
Director
Joined: 01 May 2007
Posts: 772

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07 Jul 2007, 13:25
Understanding everthing but the COUNTING the 5s.

I understood the counting the 2*5's for finding the # of terminating zeros in 30,000...Can you give a more detailed explanation as to why we are counting just the 5s and not the 10s to find the terminating zeros of 200!

Thanks.
VP
Joined: 08 Jun 2005
Posts: 1136

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07 Jul 2007, 13:41
jimmyjamesdonkey wrote:
Understanding everthing but the COUNTING the 5s.

I understood the counting the 2*5's for finding the # of terminating zeros in 30,000...Can you give a more detailed explanation as to why we are counting just the 5s and not the 10s to find the terminating zeros of 200!

Thanks.

There are more twos then fives in 200! , for every even number has two in it and every 2^n = (8 - 16 - 32 - 64) has more then one "twos" in it. So you can assume that you have more then 49 twos (for every five there is at least one two).

10! = 10*9*8*7*6*5*4*3*2*1 = (5*2)*(3*3)*(2*2*2)*7*(2*3)*5*(2*2)*3*2*1

8 twos = 2^8 (more twos then fives)
2 fives = 5^2
4 threes = 3^4
1 seven = 7

2^8*5^2*3^4*7 = 10! = 3,628,800 (two terminating zeros !)

If you want you can count tens - in that case 10! = 10*10*2^6*3^4*7. counting fives is easier.

Manager
Joined: 11 Jun 2006
Posts: 247

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07 Jul 2007, 14:05
This is a good technique to learn. Just so I follow, let's say I want to know how many 3's there are in 200!, does this work?

200/3 = 66
200/9 = 22
200/27 = 7
200/81 = 1

66+22+7+1 = 97

There are 97, 3's in 200!

I think there was a question posted last week that used this technique as well. Does someone have a link? Himalayan I think you posted it.
VP
Joined: 08 Jun 2005
Posts: 1136

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07 Jul 2007, 14:16
baer wrote:
This is a good technique to learn. Just so I follow, let's say I want to know how many 3's there are in 200!, does this work?

200/3 = 66
200/9 = 22
200/27 = 7
200/81 = 1

66+22+7+1 = 97

There are 97, 3's in 200!

I think there was a question posted last week that used this technique as well. Does someone have a link? Himalayan I think you posted it.

I think you have a misprint here since 200/81 = 2.46 , but thats a very good way - If anyone can think about a shortcut like this it must be Himalayan

198 = 3*66
195 = 3*65

so 66 threes, and:

198 = 3*3*22
189 = 3*3*21

22 more threes, and

189 = 3*3*3*7
162 = 3*3*3*6

seven more threes

162 = 3*3*3*3*2
81 = 3*3*3*3

last two

66+22+7+2 = 97 threes

Senior Manager
Joined: 04 Jun 2007
Posts: 341
Re: PS: terminating zeros  [#permalink]

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08 Jul 2007, 10:18
Himalayan wrote:
Himalayan wrote:
How many terminating zeroes does 200! have?

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

OA is C.

200/5 = 40
200/25 = 8
200/125 = 1
total = 49

I like this method. Will this work for other similar questions? Also, why are we dividing 200 by 5, 25, and 125 only? Why not other factors of 5? Apparently, I am not getting this question.
VP
Joined: 08 Jun 2005
Posts: 1136
Re: PS: terminating zeros  [#permalink]

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08 Jul 2007, 12:07
r019h wrote:
Himalayan wrote:
Himalayan wrote:
How many terminating zeroes does 200! have?

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

OA is C.

200/5 = 40
200/25 = 8
200/125 = 1
total = 49

I like this method. Will this work for other similar questions? Also, why are we dividing 200 by 5, 25, and 125 only? Why not other factors of 5? Apparently, I am not getting this question.

5^1 = 5
5^2 = 25
5^3 = 125 (under 200)
5^4 = 625 (too big)

Senior Manager
Joined: 04 Jun 2007
Posts: 341
Re: PS: terminating zeros  [#permalink]

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09 Jul 2007, 12:14
KillerSquirrel wrote:
r019h wrote:
Himalayan wrote:
Himalayan wrote:
How many terminating zeroes does 200! have?

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

OA is C.

200/5 = 40
200/25 = 8
200/125 = 1
total = 49

I like this method. Will this work for other similar questions? Also, why are we dividing 200 by 5, 25, and 125 only? Why not other factors of 5? Apparently, I am not getting this question.

5^1 = 5
5^2 = 25
5^3 = 125 (under 200)
5^4 = 625 (too big)

yah I was wondering if it was because 625 was over 200, but couldn't be certain. thanks for the clarification Killer!

now can i use the same method for the following-

How many terminating zeroes does 87! have?
so 87/5 =17.4 = 17
and 87/25 = 3.48 = 3 or 4?

then the answer will be 20 or 21

using baer's example of how many three's there are in 200!, say I want to find out how many 3's there are in 18!

18/3 = 6
18/9 = 2
so 6 + 2= 8 3's in 18!
that's not correct since 18! = 6,402,373,705,728,000 (hence only two 3's)

guess this method only works for terminating zeroes?
Manager
Joined: 11 Jun 2006
Posts: 247
Re: PS: terminating zeros  [#permalink]

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09 Jul 2007, 12:43
r019h wrote:
using baer's example of how many three's there are in 200!, say I want to find out how many 3's there are in 18!

18/3 = 6
18/9 = 2
so 6 + 2= 8 3's in 18!
that's not correct since 18! = 6,402,373,705,728,000 (hence only two 3's)

guess this method only works for terminating zeroes?

When we say we want the "number of 3's," we are talking about the number of 3's if you prime factor the number, not the actual number of digits.

However, the number of 10's in a number, will give you the number of zeros.
Senior Manager
Joined: 04 Jun 2007
Posts: 341
Re: PS: terminating zeros  [#permalink]

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09 Jul 2007, 19:21
baer wrote:
r019h wrote:
using baer's example of how many three's there are in 200!, say I want to find out how many 3's there are in 18!

18/3 = 6
18/9 = 2
so 6 + 2= 8 3's in 18!
that's not correct since 18! = 6,402,373,705,728,000 (hence only two 3's)

guess this method only works for terminating zeroes?

When we say we want the "number of 3's," we are talking about the number of 3's if you prime factor the number, not the actual number of digits.

However, the number of 10's in a number, will give you the number of zeros.

thanks, makes a lot more sense
Current Student
Joined: 28 Dec 2004
Posts: 3250
Location: New York City
Schools: Wharton'11 HBS'12

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10 Jul 2007, 14:09
yup C for me too

killersquirell. haha like the sn..
Director
Joined: 26 Feb 2006
Posts: 865

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12 Jul 2007, 15:24
well discussed.

thanx everybody!!!!!!!!!!!!

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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&nbs [#permalink] 12 Jul 2007, 15:24

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