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# How many zeros does 100! end with? (anybody know a nifty way

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How many zeros does 100! end with? (anybody know a nifty way [#permalink]

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25 May 2008, 11:56
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How many zeros does 100! end with?

(anybody know a nifty way to do this, what if was how many zeros does 4500! end with?)
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25 May 2008, 12:07
one zero corresponds to 2*5. As we have 2 more often in n! than 5. So, we have to count 5.

4500!

(4500-5)/5+1=900
(4500-25)/25+1=180
(4500-125)/125+1=36
(4375-625)/625+1=7
(3125-3125)/3125+1=1

So 4500! has 900+180+36+7+1=1124 zeros
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25 May 2008, 14:21
We'll know the number of ending zeros of 100! by finding the largest power of 10 contained in 100!
Why? Because for example 1000= 10^3;so 1000 has 3 ending zeros.
900= 9*10^2; so 900 has 2 ending zeros.
Now to solve the problem, we first need to break down 10 into prime numbers. 10 = 2*5.
We then need to find how many 2s (call it x) and how many 5s (call it y) go in 100!
Once we do that, we'll take the lower number between x and y.
Clearly, there will be more 2s than 5s in 100!; so let's save time by focusing on y (the number of 5).
Let's do it:
Divide 100 by 5 and the resulting quotient by 5 repeatedly until the quotient of the division is less than 5, which is the divisor, and you stop.
We only write out and take the quotients in the divisions.
100/5 = 20; 20/5= 4. Stop! since 4 is less than 5.
Let's now add all the quotients: y= 20 + 4 = 24.
24 is the largest power of 5 in 100!
24 is also the largest power of 10 contained in 100!
100! has 24 ending zeros!
That's all folks!
Asan Azu, The GMAT Doctor.
Asan is a very experienced GMAT teacher and tutor that can be reached at http://www.GMATLounge.com

For 4500, the same methodology applies. Remember that we only take the quotients in the successive divisions by 5.
4500/5 = 900; 900/5 = 180; 180/5 = 36; 36/5 = 7; 7/5 = 1; Stop! since 1 is less than 5.
Now add the quotients : y= 900+180+36+7+7 = 1124.
1124 is the largest power of 10 contained in 4500!
4500! has 1124 ending zeros.
Done.

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Last edited by TheGMATDoctor on 13 Jul 2010, 13:12, edited 3 times in total.
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25 May 2008, 21:37
You may ask why I use (4500-5)/5+1=900 rather than 4500/5
I use this formula as an universal and reliable formula. For instance,

How many 5 between 6 and 99, and between 7 and 100.

1) Use simple approach leads to (99-6)/5=18.6 --> 18 and (100-7)/5=18.6 --> 18
2) Use reliable approach leads to (95-5)/5 = 18 and (100-5)/5 = 19

Are you see the difference? By the way, I saw a few GMAT problems that test this trap....
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26 May 2008, 00:35
walker wrote:
You may ask why I use (4500-5)/5+1=900 rather than 4500/5
I use this formula as an universal and reliable formula. For instance,

How many 5 between 6 and 99, and between 7 and 100.

1) Use simple approach leads to (99-6)/5=18.6 --> 18 and (100-7)/5=18.6 --> 18
2) Use reliable approach leads to (95-5)/5 = 18 and (100-5)/5 = 19

Are you see the difference? By the way, I saw a few GMAT problems that test this trap....

Walker; What about the "+1" in 2) ?
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26 May 2008, 01:24
farend wrote:
Walker; What about the "+1" in 2) ?

Of course, You are right! (95-5)/5+1 = 19 and (100-5)/5 +1 = 20

P.S. Sometimes it is better to drink a cup of coffee just after I waked up than to start with GMATclub
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26 May 2008, 19:50
irishspring wrote:
How many zeros does 100! end with?

(anybody know a nifty way to do this, what if was how many zeros does 4500! end with?)

my old is old approach:

1. How many zeros does 100! end with?

100/5 = 20
100/25=4
so 24

2. How many zeros does 4500! end with?
4500/5 = 900
4500/25=180
4500/125 = 36
4500/625=7
4500/3125=1
total = 900+180+36+7+1 = 1124
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29 May 2008, 11:41
The DOCTOR is the real deal!
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29 May 2008, 18:51
Doctor - help us on the verbal forums too
Re: Challenge Problem   [#permalink] 29 May 2008, 18:51
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# How many zeros does 100! end with? (anybody know a nifty way

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