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  Everything about Factorials on the GMAT [#permalink]
New postPosted: Mon Oct 05, 2009 5:02 am 
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If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35! ?

The formula is:
\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3} ... till n>k^x

What is the power of 2 in 25!
\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sun Oct 25, 2009 10:51 pm 
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noboru wrote:
Point 1 is just point 2 for k=5, isnt it?


Yes, it is. I've separated them as for GMAT generally we need only trailing zeros and almost never other prime's power.

shalva wrote:
Can you please post this one too? It's still interesting, though may not be usable for GMAT.


It's better to illustrate it on the example:
How many powers of 900 are in 50!
900=2^2*3^2*5^2


Find the power of 2:
\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}=25+12+6+3+1=47

= 2^{47}


Find the power of 3:
\frac{50}{3}+\frac{50}{9}+\frac{50}{27}=16+5+1=22

=3^{22}


Find the power of 5:
\frac{50}{5}+\frac{50}{25}=10+2=12

=5^{12}

We need all of them (2,3,5) to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!
900^6

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Fri Oct 30, 2009 5:22 pm 
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juukkk wrote:
Bunuel wrote:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!




If you actually go and check 32! in Excel the result will be 263130836933694000000000000000000000

So more like 21 zeros... I really hope Excel is making a mistake because of how neat is your formula but someone please explain!?


32! = 263130836933693530167218012160000000 This is what 32! really equals to.
32!= 263130836933694000000000000000000000 Accoroding to Excell. Don't worry it's just rounded, so formula is correct.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Tue Jul 13, 2010 8:29 am 
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yasar434 wrote:
can anyone tell me how to calculate the no.of factors for 20!. well this was a small number if i would have to answer a bigger number like no.of factors for 720! what will be the solution.


You won't need this for GMAT but still let's see if we can do it:

First we should make prime factorization of 20!. 20! will have all primes from 0 to 20, so we should find the powers of these primes in 20!.

Powers of 2 --> \frac{20}{2}+\frac{20}{4}+\frac{20}{8}+\frac{20}{16}=10+5+2+1=18;
Powers of 3 --> \frac{20}{3}+\frac{20}{9}=6+2=8;
Powers of 5 --> \frac{20}{5}=4;
Powers of 7 --> \frac{20}{7}=2;
Powers of 11 --> \frac{20}{11}=1;
Powers of 13 --> \frac{20}{13}=1;
Powers of 17 --> \frac{20}{17}=1;
Powers of 19 --> \frac{20}{19}=1.

So 20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1.

Next: How to Find the Number of Factors of an Integer.

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.


So, the # of positive factors of 20!=2^{18}*3^8*5^4*7^2*11^1*13^1*17^1*19^1 will be (18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1)=19*9*5*3*2*2*2*2=41040.

The same way we can find for 720!, but we'll need much more time.

Hope it helps.

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  Re: Number Properties from GMATPrep [#permalink]
New postPosted: Wed Oct 21, 2009 1:20 pm 
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This post has been split off the original discussion and cleaned up for reference.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sat Oct 31, 2009 10:07 pm 
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Wiki has the the answer to 32! on trailing zeros article.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sat Feb 27, 2010 11:18 pm 
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arturocb86 wrote:
Could you point some Gmat style problems that test the second property? :)

Thank you and by the way... great post!


Hi, welcome to the Gmat Club. Below are the links to the problems about this concept:

m12-q4-72884.html#p669768
unit-s-digit-of-a-86818.html#p659138
gmatprep-ds-product-of-first-30-integers-87454.html#p657434

Hope it helps.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Mon Jun 21, 2010 4:22 am 
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Pipp wrote:
What about finding the power a factorial in n!? ex: the power of 16! in 50!


I think best way to solve the above question is to determine the prime number just below 16 which is 13

13*3 = 39 => (16!)^3 will be the power in 50!

as for (16!)^4 we do not have 13^4 in 50!.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Fri Oct 23, 2009 1:56 am 
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Bunuel wrote:
If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35! ?

The formula is:
\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3} ... till n>k^x

What is the power of 2 in 25!
\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.


Point 1 is just point 2 for k=5, isnt it?

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Fri Oct 23, 2009 5:57 am 
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Bunuel

Valuable post! +1

Quote:
There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.


Can you please post this one too? It's still interesting, though may not be usable for GMAT


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  Re: Everything about Factorials on the GMAT [#permalink]
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Bunuel wrote:

How many zeros are in the end (after which no other digits follow) of 32!?
\frac{32}{5}+\frac{32}{5^2}=6+1=7 (denominator must be less than 32, 5^2=25 is less)

So there are 7 zeros in the end of 32!




If you actually go and check 32! in Excel the result will be 263130836933694000000000000000000000

So more like 21 zeros... I really hope Excel is making a mistake because of how neat is your formula but someone please explain!?


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Fri Oct 30, 2009 5:28 pm 
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Bunuel wrote:

32! = 263130836933693530167218012160000000 This is what 32! really equals to.
32!= 263130836933694000000000000000000000 Accoroding to Excell. Don't worry it's just rounded, so formula is correct.


Kudos given :) and formula memorized already. From where did you got the what "32! really equals to"?


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sat Oct 31, 2009 4:19 am 
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juukkk wrote:
Bunuel wrote:

32! = 263130836933693530167218012160000000 This is what 32! really equals to.
32!= 263130836933694000000000000000000000 Accoroding to Excell. Don't worry it's just rounded, so formula is correct.


Kudos given :) and formula memorized already. From where did you got the what "32! really equals to"?


:shock: Wow. I want to know how you calculated it too

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sun Nov 01, 2009 8:05 pm 
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Nice post, thanks! I loathe factorials :x


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sun Nov 01, 2009 8:12 pm 
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Also, my favorite calculator program (Speedcrunch: won't let me post URLs, so google it I guess!) does factorials quite easily; just type 32!


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  Re: Everything about Factorials on the GMAT [#permalink]
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Hi,

I just went through this thread. I understand point# 1 about trailing zeroes. However, for the life of me, I cant understand Point#2.
Can anyone please explain me what are we really trying to solve in "2. Finding the number of powers of a prime number k, in the n!."

For the question "What is the power of 2 in 25!", the answer is given as 22. What does it mean ?

Thanks


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Mon Jan 04, 2010 12:34 am 
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aim-high wrote:
Hi,

I just went through this thread. I understand point# 1 about trailing zeroes. However, for the life of me, I cant understand Point#2.
Can anyone please explain me what are we really trying to solve in "2. Finding the number of powers of a prime number k, in the n!."

For the question "What is the power of 2 in 25!", the answer is given as 22. What does it mean ?

Thanks


25! is some number, let's say x. Power of 2 (highest power, 2 will have) in 25!, means the power of 2 in prime factorization of x.

For example: 5!=120=2^3*3*5, so the power of 2 in 5! is 3.

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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Mon Feb 15, 2010 6:51 pm 
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excellent!
thank you


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sat Feb 27, 2010 10:07 am 
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Could you point some Gmat style problems that test the second property? :)

Thank you and by the way... great post!


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  Re: Everything about Factorials on the GMAT [#permalink]
New postPosted: Sun Feb 28, 2010 6:34 am 
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Great post thanks. I have 1 question and 1 suggestion.
- How often do these questions come up on the GMAT? Of all the practice questions/tests I've done, I haven't seen it once. Additionally, the Kaplan maths refresher didn't mention anything about this formula.

Suggestion:
I think if you changed the wording of the first formula to:
n/5 + n/5^2 + .... + n/5^k while 5^k < n

and the wording of the second formula to:
n/k + n/k^2 + .... + n/k^x while k^x < n

it would make it clearer.


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