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mgmat 700 level zeros sum

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mgmat 700 level zeros sum

by vkb16 » Tue Aug 11, 2009 4:26 am
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

6
12
14
42
56

OA is [spoiler]C but I think is B - my method is--
from 60 to 51, there are two 10s (60, and 65x62)
from 50-41 there are two 10s (50, and 55x52)
similarly, there are in total 6x2=12 zeroes.
I didnt quite get mgmat's explanation[/spoiler]

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by DanaJ » Tue Aug 11, 2009 5:29 am
You see this type of question a lot. You're looking for the number of trailing zeroes in 60!, which means that you need the number of 10s in 60!. Think of it this way: 7000 has three trailing zeroes, since 7000 = 10*10*10*7 (three 10s).

The number of 10s is actually provided by the number of 5s in 60!, since 10 = 2*5. There are plenty of 2s in 60! (at least one every other number, in even numbers), so 5s are more rare. This is why you're trying to count those.

There's this really short formula for finding out the number of 5s in 60! (a formula discovered on this forum):

60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

What you have to do is divide the factorial by the powers of 5 up to the greatest power smaller than 60 (5^2 = 25 - fits; 5^3 = 125 - doesn't fit).

Another example: The number of trailing zeroes in 140:

140/5 + 140/(5^2) + 140/(5^3) = 28 + 5 + 1 = 34.


LATER EDIT: the reason you got only 12 zeroes is because you didn't count the fact that 25 has two 5s and so does 50. While you did count them once, they should be counted twice!
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by ket » Fri Nov 20, 2009 4:23 am
I am confused about this problem. I am ok with explanation of C it seems logical. I just wanted to check and calculated factorial of 60 in EXCEl with formula FACT(60), and the result is not 14 zeros :O ... I must be doing something wrong but the excel answer is this (23 zeros!)

8,320,987,112,741,390,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

would be nice to hear smbdy's explanation...

Cheers :)
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by Ian Stewart » Fri Nov 20, 2009 9:29 am
When Excel produces that value, it's rounding it off.
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by ssuarezo » Fri Dec 04, 2009 12:24 pm
DanaJ wrote: There's this really short formula for finding out the number of 5s in 60! (a formula discovered on this forum):

60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

Another example: The number of trailing zeroes in 140:

140/5 + 140/(5^2) + 140/(5^3) = 28 + 5 + 1 = 34.
Hi Dana:

Why in your first example you went upto 5^2: 60/5 + 60/(5^2) .. but in the second example you went upto 5^3: 140/5 + 140/(5^2) + 140/(5^3) .. how do I know what's the limit??? Since it's a common question, I need to understand it.

Thanks
Silvia.
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by Stuart@KaplanGMAT » Fri Dec 04, 2009 12:43 pm
ssuarezo wrote:
DanaJ wrote: There's this really short formula for finding out the number of 5s in 60! (a formula discovered on this forum):

60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

Another example: The number of trailing zeroes in 140:

140/5 + 140/(5^2) + 140/(5^3) = 28 + 5 + 1 = 34.
Hi Dana:

Why in your first example you went upto 5^2: 60/5 + 60/(5^2) .. but in the second example you went upto 5^3: 140/5 + 140/(5^2) + 140/(5^3) .. how do I know what's the limit??? Since it's a common question, I need to understand it.

Thanks
Silvia.
Good question!

We divide by the greatest power of 5 that's less than or equal to our number.

Since:

5^1 =5
5^2 = 25
5^3 = 125
5^4 = 625,

for any number from 5 to 20, we just divide by 5;

for any number from 25 to 120, we divide by 5 and 5^2;

for any number from 125 to 620, we divide by 5, 5^2 and 5^3; and

so on, and so on and so on...
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by barira1 » Fri Dec 04, 2009 9:47 pm
Dear Ian ,

Can you please explain this question again .
really finding it difficult


regards,
Barira
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by Stuart@KaplanGMAT » Fri Dec 04, 2009 10:25 pm
barira1 wrote:Dear Ian ,

Can you please explain this question again .
really finding it difficult


regards,
Barira
Let's go back to Dana's explanation, but ignore the formula for now.

As Dana noted, one can determine how many 0s are at the end of a number by counting the number of factors of 10 that the number has, since one 10 = one 0 at the end.

Now 10 = 2*5; so for every factor pair 2/5 that a number has, it will have a factor of 10.

60! has WAY more 2s in it than 5s (since 2 is smaller, it's a much more frequent factor... hmm.. sounds like a good name for a math reward program... "frequent factor points"), so we don't need to count the 2s, we can just count the 5s and be sure that there will be enough 2s to pair off.

Since 60! = 1*2*3*... *58*59*60, it shouldn't take too long to count the factors of 5.

First, we have 5, 10, 15, 20, ..., 50, 55, 60. We could just count them all and see there are 12 or we could do 60/5 = 12 to see that there are going to be 12 factors of 5.

HOWEVER, we're not done, since some of those numbers have two factors of 5.

25=5*5, so that's an extra 5; 50=5*5*2, so that's also an extra 5.

Therefore, we have 12 + 1 + 1 = 14 factors of 5, which will give us 14 factor pairs of 2/5, which will give us 14 factors of 10, which will give us 14 0s at the end of our number.
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by barira1 » Tue Dec 08, 2009 3:22 am
Dear Stauart,

Thank u so much ,
Clear now :)
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by AegisAshore » Fri Aug 06, 2010 6:27 pm
Stuart Kovinsky wrote:
barira1 wrote:Dear Ian ,

Can you please explain this question again .
really finding it difficult


regards,
Barira
Let's go back to Dana's explanation, but ignore the formula for now.

As Dana noted, one can determine how many 0s are at the end of a number by counting the number of factors of 10 that the number has, since one 10 = one 0 at the end.

Now 10 = 2*5; so for every factor pair 2/5 that a number has, it will have a factor of 10.

60! has WAY more 2s in it than 5s (since 2 is smaller, it's a much more frequent factor... hmm.. sounds like a good name for a math reward program... "frequent factor points"), so we don't need to count the 2s, we can just count the 5s and be sure that there will be enough 2s to pair off.

Since 60! = 1*2*3*... *58*59*60, it shouldn't take too long to count the factors of 5.

First, we have 5, 10, 15, 20, ..., 50, 55, 60. We could just count them all and see there are 12 or we could do 60/5 = 12 to see that there are going to be 12 factors of 5.

HOWEVER, we're not done, since some of those numbers have two factors of 5.

25=5*5, so that's an extra 5; 50=5*5*2, so that's also an extra 5.

Therefore, we have 12 + 1 + 1 = 14 factors of 5, which will give us 14 factor pairs of 2/5, which will give us 14 factors of 10, which will give us 14 0s at the end of our number.
Stuart,

Thank you for the clear explanation. However, while doing this problem, I had a relavent question come to mind. What if they just asked for the number of significant digits that 60! was equal to? For example, 10!=3628800 = 7 significant digits. Thank you for your help.
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by Stuart@KaplanGMAT » Fri Aug 06, 2010 6:40 pm
AegisAshore wrote:
Stuart,

Thank you for the clear explanation. However, while doing this problem, I had a relavent question come to mind. What if they just asked for the number of significant digits that 60! was equal to? For example, 10!=3628800 = 7 significant digits. Thank you for your help.
If they ask for the relevant digits, then you're not writing the GMAT - that's WAY beyond the scope of the test.
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by AegisAshore » Fri Aug 06, 2010 6:46 pm
Stuart Kovinsky wrote:
AegisAshore wrote:
Stuart,

Thank you for the clear explanation. However, while doing this problem, I had a relavent question come to mind. What if they just asked for the number of significant digits that 60! was equal to? For example, 10!=3628800 = 7 significant digits. Thank you for your help.
If they ask for the relevant digits, then you're not writing the GMAT - that's WAY beyond the scope of the test.
I like that answer, thank you. I won't worry about it then.
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by agarwalmanoj2000 » Wed Jul 13, 2011 12:53 pm
60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

What you have to do is divide the factorial by the powers of 5 up to the greatest power smaller than 60 (5^2 = 25 - fits; 5^3 = 125 - doesn't fit).
10 factorial should have 2 zeroes at end because it has 2 multiples of 5 i.e. 5 and 10 but, on applying the above formula the result is 10/5 + 10/(5^1) = 2 + 2= 4 zeroes at end.

Please advise.
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by krishnasty » Wed Jul 13, 2011 8:47 pm
agarwalmanoj2000 wrote:
60/5 + 60/(5^2) = 12 + 2 = 14 (we only take the quotient)

What you have to do is divide the factorial by the powers of 5 up to the greatest power smaller than 60 (5^2 = 25 - fits; 5^3 = 125 - doesn't fit).
10 factorial should have 2 zeroes at end because it has 2 multiples of 5 i.e. 5 and 10 but, on applying the above formula the result is 10/5 + 10/(5^1) = 2 + 2= 4 zeroes at end.

Please advise.
agarwalmanoj2000, i think you missed out on the correct formula.
the formula says , if we are asked the number of trailing zeroes in x!, the formula is:

x/(5^1) + x/(5^2) + x/(5^3).....till demoninator is greater than the numerator.
now, if you neeed to find out number of trailing zeroes in 10!, then it would be :

10/(5^1) only...coz 5^2 = 25 and 25>10
10/5 = 2 and thats your ans.

what you did is to take 10/5 and 10/(5^1) as the formula, where as both of them means the same thing..

i hope its much clear now.
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by amit2k9 » Wed Jul 13, 2011 10:11 pm
12+2 after dividing by 5 and 25 each.
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