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

bruel i just dont understand that part. CAn you explain it plz
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fatihaysu wrote:
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

bruel i just dont understand that part. CAn you explain it plz


\(50!=900^xa=(2^2*3^2*5^2)^x*a\), where \(x\) is the highest possible value of 900 and \(a\) is the product of other multiples of \(50!\).

\(50!=2^{47}*3^{22}*5^{12}*b=(2^2*3^2*5^2)^6*(2^{35}*3^{10})*b=900^{6}*(2^{35}*3^{10})*b\), where \(b\) is the product of other multiples of \(50!\). So \(x=6\).

Below is another example:

Suppose we have the number \(18!\) and we are asked to to determine the power of \(12\) in this number. Which means to determine the highest value of \(x\) in \(18!=12^x*a\), where \(a\) is the product of other multiples of \(18!\).

\(12=2^2*3\), so we should calculate how many 2-s and 3-s are in \(18!\).

Calculating 2-s: \(\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16\). So the power of \(2\) (the highest power) in prime factorization of \(18!\) is \(16\).

Calculating 3-s: \(\frac{18}{3}+\frac{18}{3^2}=6+2=8\). So the power of \(3\) (the highest power) in prime factorization of \(18!\) is \(8\).

Now as \(12=2^2*3\) we need twice as many 2-s as 3-s. \(18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a\). So \(18!=12^8*a\) --> \(x=8\).

Hope it's clear.
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Re: Everything about Factorials on the GMAT [#permalink]
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.
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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: Everything about Factorials on the GMAT [#permalink]
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?
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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.
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Re: Everything about Factorials on the GMAT [#permalink]
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?
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Re: Everything about Factorials on the GMAT [#permalink]
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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.
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