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Re: Everything about Factorials on the GMAT [#permalink]
18 Jan 2012, 03:37

Expert's post

theptrk wrote:

Im getting a problem trying to use the formula. When looking for power of 3 in 35! i do 35/3 + 35/9 + 35/27 = 11+3+1 = 15

But I've multiplied out the factorial 32! and get 18 3's including the square for 9 and the cubed for 27. Am i doing something wrong?

Yes, as formula is correct then it must be that you have miscalculated.

By the way here is complete factorization of 35!: 35!=2^{32}*3^{15}*5^8*7^5*11^3*13^2*17^2*19*23*29*31, so you can see that the power of 3 is indeed 15. _________________

Re: Everything about Factorials on the GMAT [#permalink]
15 Mar 2012, 01:09

Expert's post

Impenetrable wrote:

Bunuel wrote:

If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

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

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

How come I get completely different results when I use my calculator to get 25! and 2^22?

Did I understand it wrong?

25! is a huge number, not many calculators can handle it. Check whether it gives you the following result: 25!=15,511,210,043,330,985,984,000,000. _________________

Re: Everything about Factorials on the GMAT [#permalink]
19 Jun 2012, 06:25

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 ? _________________

Re: Everything about Factorials on the GMAT [#permalink]
19 Jun 2012, 06:53

Expert's post

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!?"

GMAT Club Math Book (Question about Factorials) [#permalink]
24 Jun 2012, 14:42

Hello Community,

I needed some help understanding something in the GMAT Club Math book. In the Factorial section, it gives an example of how to find power of non-prime as follows: How many powers of 900 are in 50!? I understand that we must figure the prime factorization of 900 and the powers of those prime numbers 50!. I'm confused by the following: "We need all prime 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!." Can someone clarify this for me? Thanks

Re: GMAT Club Math Book (Question about Factorials) [#permalink]
24 Jun 2012, 19:37

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Alterego wrote:

Hello Community,

I needed some help understanding something in the GMAT Club Math book. In the Factorial section, it gives an example of how to find power of non-prime as follows: How many powers of 900 are in 50!? I understand that we must figure the prime factorization of 900 and the powers of those prime numbers 50!. I'm confused by the following: "We need all prime 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!." Can someone clarify this for me? Thanks

Hi,

To find the power of 900 (=2^2*3^2*5^2), you should be able to find the pairs of 2's, 3's & 5's. Now, if you check 50!, it is 1*2*3*4.....*50 Multiples of 2 are 2, 4, 6, 8... Multiples of 3 are 3, 6, 9, ... Multiples of 5 are 5, 10, 15, .... So, 2 occurs in every second number, 3 occurs in every third number, 5 occurs in every fifth number, So, you have many 2's & 3's available for each 5. Thus, if we can find the pairs of 5's we can find the power of 900.

To find the number of 5's 50/5 = 10 (numbers in 50! that will have a 5) 50/2 = 2 (numbers in 50! that will provide two 5's) Total power of 5 = 12. For one 900, it requires two power of 5's. Since, 6 pairs of 5 are available, power of 900 would be 6.

Re: Everything about Factorials on the GMAT [#permalink]
27 Dec 2012, 09:22

Expert's post

This is just awesome Bunuel...Kudos !

Now apart from your combination problem collections,we are really lucky enough to get the required theory for GMAT on the same.Thanks a ton..

Few quick clarifications required from you my friend: 1.On this thread you've mentioned that this part is included in GMAT Math book however on the GMAT Math book download page it's mentioned that this post is not included in the same.Is the GMAT Math book not the updated one ? Actually I'm facing issue while downloading the GMAT Math book,the pdf file is getting corrupted somehow in my system after the download completes.So,I'm unable to verify it. I'm still trying to troubleshoot the issue.

Re: Everything about Factorials on the GMAT [#permalink]
23 Sep 2013, 07:29

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."

Re: Everything about Factorials on the GMAT [#permalink]
23 Sep 2013, 23:35

Expert's post

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.

Re: GMAT Club Math Book (Question about Factorials) [#permalink]
01 Oct 2013, 08:59

cyberjadugar wrote:

So, you have many 2's & 3's available for each 5. Thus, if we can find the pairs of 5's we can find the power of 900.

To find the number of 5's 50/5 = 10 (numbers in 50! that will have a 5) 50/2 = 2 (numbers in 50! that will provide two 5's) Total power of 5 = 12. For one 900, it requires two power of 5's. Since, 6 pairs of 5 are available, power of 900 would be 6.

Let me know, if the concept is clear to you now.

Regards,

Having problem understanding this part.. Total power of 5 = 12. For one 900, it requires two power of 5's. Since, 6 pairs of 5 are available, power of 900 would be 6.

So i get the total power of 5 = 12 and from that for one 900, it requires two power of 5's (as done in the prime factorization)

But I'm not seeing the 6 pairs of 5.. any further clarification?

Re: GMAT Club Math Book (Question about Factorials) [#permalink]
02 Oct 2013, 03:04

Expert's post

kcx214 wrote:

cyberjadugar wrote:

So, you have many 2's & 3's available for each 5. Thus, if we can find the pairs of 5's we can find the power of 900.

To find the number of 5's 50/5 = 10 (numbers in 50! that will have a 5) 50/2 = 2 (numbers in 50! that will provide two 5's) Total power of 5 = 12. For one 900, it requires two power of 5's. Since, 6 pairs of 5 are available, power of 900 would be 6.

Let me know, if the concept is clear to you now.

Regards,

Having problem understanding this part.. Total power of 5 = 12. For one 900, it requires two power of 5's. Since, 6 pairs of 5 are available, power of 900 would be 6.

So i get the total power of 5 = 12 and from that for one 900, it requires two power of 5's (as done in the prime factorization)

But I'm not seeing the 6 pairs of 5.. any further clarification?