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The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\).

For example: \(4!=1*2*3*4=24\).

Properties

Factorial of a negative number is undefined.

\(0!=1\), zero factorial is defined to equal 1.

\(n!=(n-1)!*n\), valid for \(n\geq{1}\).

Trailing zeros:

Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example: 125,000 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 \(5^k\leq{n}\)

Example: How many zeros are in the end (after which no other digits follow) of 32!?

\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is \(\frac{32}{5}=6\) not 6.4). Therefore, 32! has 7 trailing zeros.

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.

Finding the powers of a prime number p, in the n!

The formula is: \(\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}\), where \(k\) must be chosen such that \(p^k\leq{n}\)

Re: Everything about Factorials on the GMAT [#permalink]
23 Oct 2009, 00:56

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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? _________________

Re: Everything about Factorials on the GMAT [#permalink]
25 Oct 2009, 21:51

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Expert's post

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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 _________________

Re: Everything about Factorials on the GMAT [#permalink]
30 Oct 2009, 15:46

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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!?

Re: Everything about Factorials on the GMAT [#permalink]
30 Oct 2009, 16:22

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Expert's post

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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. _________________

Re: Everything about Factorials on the GMAT [#permalink]
30 Oct 2009, 16:28

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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"?

Re: Everything about Factorials on the GMAT [#permalink]
31 Oct 2009, 03:19

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Expert's post

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"?

Wow. I want to know how you calculated it too _________________

Re: Everything about Factorials on the GMAT [#permalink]
03 Jan 2010, 19:16

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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 ?

Re: Everything about Factorials on the GMAT [#permalink]
03 Jan 2010, 23:34

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Expert's post

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. _________________

Re: Everything about Factorials on the GMAT [#permalink]
28 Feb 2010, 05:34

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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

Re: Everything about Factorials on the GMAT [#permalink]
06 May 2010, 01:24

1

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Hi Do we need to worry about the following concepts ? 1. last 2 digits 2. 10's digit or 100's digit 3. last non-zero digit

What about binomial theorem, fermet's little theorem and Euler's theorem? I understand that GMAT tests on understanding the concept, but these concepts can cut down the time significantly when dealing with exponent. I found myself relying on Fermet's quite often rather than cyclicity, especially when dealing with complex index. Any comment is much appreciated.

Re: Everything about Factorials on the GMAT [#permalink]
30 May 2010, 05:34

1

This post received KUDOS

Eden wrote:

Hi Do we need to worry about the following concepts ? 1. last 2 digits 2. 10's digit or 100's digit 3. last non-zero digit

What about binomial theorem, fermet's little theorem and Euler's theorem? I understand that GMAT tests on understanding the concept, but these concepts can cut down the time significantly when dealing with exponent. I found myself relying on Fermet's quite often rather than cyclicity, especially when dealing with complex index. Any comment is much appreciated.

These concepts in GMAT are as important as Newton's 2nd law of Thermodynamics

gmatclubot

Re: Everything about Factorials on the GMAT
[#permalink]
30 May 2010, 05:34

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