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# Everything about Factorials on the GMAT

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Joined: 27 May 2012
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19 Jun 2012, 07:25
3
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.

bunuel

How did we decide to use 5 in point no . 1, I mean what is the complete question to the example

Is it no. of trailing zero's when 32! is divided by 5 , In the second case its apparent that we are dividing by the prime factors .

But in order to find the no. of trailing zero's of n! what should we divide by ?
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19 Jun 2012, 07:53
1
stne wrote:
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.

bunuel

How did we decide to use 5 in point no . 1, I mean what is the complete question to the example

Is it no. of trailing zero's when 32! is divided by 5 , In the second case its apparent that we are dividing by the prime factors .

But in order to find the no. of trailing zero's of n! what should we divide by ?

Not sure understand your question. Anyway:

1st example above asks: "How many zeros are in the end (after which no other digits follow) of 32!?" Here the answer is 32/5+32/5^2=6+1=7. Notice that you take only the quotient into account and that the last denominator (5^2) must be less than numerator.

2nd example above asks: "What is the highest power of 2 in 25!?"

I think it would be easier to understand if you see the questions below:
p-and-q-are-integers-if-p-is-divisible-by-10-q-and-cannot-109038.html
how-many-zeros-does-100-end-with-100599.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html
find-the-number-of-trailing-zeros-in-the-product-of-108248.html
find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html
if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html
if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html
if-10-2-5-2-is-divisible-by-10-n-what-is-the-greatest-106060.html

Hope it helps.
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23 Sep 2013, 08:29
Can someone please explain this part of the above topic??? thanks
"The formula actually counts the number of factors 5 in , 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."
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24 Sep 2013, 00:35
skamran wrote:
Can someone please explain this part of the above topic??? thanks
"The formula actually counts the number of factors 5 in , 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."

Consider 12!. Now, if we prime factorize 12! we'll get that the power of 2 in 12! will naturally be higher than the power of 5. To get one trailing zeros we need one 2 and one 5. Thus if we know what is the power of 5 in 12! we'll know how many trailing zeros 12! has.

Does this make sense?
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26 May 2015, 12:51
Bunuel wrote:
noboru wrote:

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

If I go to excel and calculate factorial 50! and then divide by 900^6, i still have room for 900^7 which then still yields a positive result. Could you explain that further?

3.04141E+64 =50!
5.31441E+17 =900^6
5.722948210942210000000000E+46 =Quotient (50!/900^6)

Thank you very much
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27 May 2015, 03:34
1
reto wrote:
Bunuel wrote:
noboru wrote:

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

If I go to excel and calculate factorial 50! and then divide by 900^6, i still have room for 900^7 which then still yields a positive result. Could you explain that further?

3.04141E+64 =50!
5.31441E+17 =900^6
5.722948210942210000000000E+46 =Quotient (50!/900^6)

Thank you very much

Use better calculator: http://www.wolframalpha.com/input/?i=50%21%2F900%5E6
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19 Dec 2017, 03:24
Bunuel wrote:
FACTORIALS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, Bunuel

--------------------------------------------------------

Definition

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 $$p^k\leq{n}$$

Example:
What is the power of 2 in 25!?

$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$.

_________________________________________________________________________________________________
Questions to practice:
http://gmatclub.com/forum/if-60-is-writ ... 01752.html
http://gmatclub.com/forum/how-many-zero ... 00599.html
http://gmatclub.com/forum/find-the-numb ... 08249.html
http://gmatclub.com/forum/find-the-numb ... 08248.html
http://gmatclub.com/forum/if-n-is-the-p ... 01187.html
http://gmatclub.com/forum/if-m-is-the-p ... 08971.html
http://gmatclub.com/forum/if-p-is-a-nat ... 08251.html
http://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
http://gmatclub.com/forum/p-and-q-are-i ... 09038.html
http://gmatclub.com/forum/question-abou ... 08086.html
http://gmatclub.com/forum/if-n-is-the-p ... 06289.html
http://gmatclub.com/forum/what-is-the-g ... 05746.html
http://gmatclub.com/forum/if-d-is-a-pos ... 26692.html
http://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
http://gmatclub.com/forum/how-many-zero ... 42479.html

Bunuel
could you please explain what this mean ? >> The formula actually counts the number of factors 5 in n!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.
Thank you!
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19 Dec 2017, 03:26
1
1
gracecud46 wrote:
Bunuel wrote:
FACTORIALS

This post is a part of [GMAT MATH BOOK]

created by: Bunuel
edited by: bb, Bunuel

--------------------------------------------------------

Definition

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 $$p^k\leq{n}$$

Example:
What is the power of 2 in 25!?

$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$.

_________________________________________________________________________________________________
Questions to practice:
http://gmatclub.com/forum/if-60-is-writ ... 01752.html
http://gmatclub.com/forum/how-many-zero ... 00599.html
http://gmatclub.com/forum/find-the-numb ... 08249.html
http://gmatclub.com/forum/find-the-numb ... 08248.html
http://gmatclub.com/forum/if-n-is-the-p ... 01187.html
http://gmatclub.com/forum/if-m-is-the-p ... 08971.html
http://gmatclub.com/forum/if-p-is-a-nat ... 08251.html
http://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
http://gmatclub.com/forum/p-and-q-are-i ... 09038.html
http://gmatclub.com/forum/question-abou ... 08086.html
http://gmatclub.com/forum/if-n-is-the-p ... 06289.html
http://gmatclub.com/forum/what-is-the-g ... 05746.html
http://gmatclub.com/forum/if-d-is-a-pos ... 26692.html
http://gmatclub.com/forum/if-10-2-5-2-i ... 06060.html
http://gmatclub.com/forum/how-many-zero ... 42479.html

Bunuel
Could you please explain what >> The formula actually counts the number of factors 5 in n!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
means from above? Thank you!

We are getting 0 by multiplying 5 by 2. n! will have more 2's than 5's (in some cases equal number but you cannot have more 5's than 2's). So, if we count the number of 5's we;d be sure that there will be enough 2's for these 5's to make 0. So, the formula counts this number, the number of 5's in n!.

Hope it's clear.
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15 May 2018, 03:24
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

This is a type of question you see on GMAT at times framed as follows:-
Q. What is the greatest integer k such that 2^k is a factor of 25!?
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31 Jul 2018, 06:26
Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichC

What is the rationale behind choosing 5 and its powers in the denominator?
for e.g. in terminating decimals I need to know whether the denominator contains either of 2 or 5 or its powers, else the long division may go indefinitely.

If I do not recall this formula in exam under timed condition, is there any alternate approach to this?
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01 Aug 2018, 05:22
Bunuel chetan2u KarishmaB niks18 GMATPrepNow EMPOWERgmatRichC

What is the rationale behind choosing 5 and its powers in the denominator?
for e.g. in terminating decimals I need to know whether the denominator contains either of 2 or 5 or its powers, else the long division may go indefinitely.

If I do not recall this formula in exam under timed condition, is there any alternate approach to this?

I have explained why you need 2 and/or 5 in the denominator for a terminating decimal in this post:
https://www.veritasprep.com/blog/2013/1 ... -the-gmat/

In number properties questions, if you are unable to recall the property, look for a pattern by using some simple numbers.
I am not sure to which question you are referring but if you point it out, I can tell you how to solve it without "knowing" the property.
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01 Aug 2018, 06:59
Thank you KarishmaB for your two cents.

Below is a relevant quote from your blog:

Quote:
Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s.You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?

Why do we not need to need to know whether b is greater than (or not) 1? I could not understand highlighed text in context of long division, Can you elaborate?

I would be grateful if you could explore an additional approach for factorial problem. TIA.
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02 Aug 2018, 03:31
1
Thank you KarishmaB for your two cents.

Below is a relevant quote from your blog:

Quote:
Say, a and b are two integers.

a/b = a * 1/b

For a/b to be terminating, 1/b must be a terminating decimal. What happens when you start dividing 1 by b? You add a decimal point and start adding 0s.You will get 1 followed by as many 0s as you require in the numerator. 10/100/1000/10000 etc have only two prime divisors: 2 and 5. If the denominator has 2s or 5s or both, we will be able to terminate the decimal by choosing the required multiple of 10. If there are any other primes, we will never be able to divide a multiple of 10 completely and hence the decimal will not terminate. It is obvious, isn’t it?

Why do we not need to need to know whether b is greater than (or not) 1? I could not understand highlighed text in context of long division, Can you elaborate?

I would be grateful if you could explore an additional approach for factorial problem. TIA.

If b = 1, a/b is an integer and the question of terminating/non-terminating decimal doesn't arise. b cannot be 0 since then a/b is not defined for b = 0.
b would take values out of 2/3/4/5... etc

When you divide 1 by 3 by long division, how do you do it?
Attachment:

decimals13-05.gif [ 7.14 KiB | Viewed 2806 times ]

You will keep getting 0s after the decimal but it will never be fully divisible.
Only if you divide by 2 and/or 5 or some combination of them will it be fully divisible. 10/100/1000/10000 etc have only 2 and 5 as factors.
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29 Nov 2019, 13:23
Bunuel wrote:
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

Hi Bunuel,

I get everything apart from the part in bold. Can you explain please?
Re: Everything about Factorials on the GMAT   [#permalink] 29 Nov 2019, 13:23

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