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.
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.
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.
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?
_________________ The sky is the limit 800 is the limit
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!?
juukkk
Re: Everything about Factorials on the GMAT [#permalink]
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"?
bb
Re: Everything about Factorials on the GMAT [#permalink]
Posted: Sat Oct 31, 2009 4:19 am
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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"?
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
Bunuel
Re: Everything about Factorials on the GMAT [#permalink]
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.
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
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Wow. I want to know how you calculated it too 
