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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

\(50!=900^xa=(2^2*3^2*5^2)^x*a\), 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 \(18!=12^8*a\) --> \(x=8\).

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

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13 Jul 2010, 05:21

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.

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.

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

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02 Aug 2010, 02:49

Bunuel,

Do you have some thoughts over Factorials and Remainders? Like what will be the remainder when 20! is divided by x, where x will be a number greater than 20(may be a non even/prime for which the answer isnt pretty obvious)? Just a random thought, not sure whether such questions appear in GMAT.

Btw, Kudos for your post. I had learnt these formulas long back. This thread helped in refreshing them.

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

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02 Aug 2010, 12:12

Everyone Where can I find a compilation such information for Quant? The OG is not elaborative on such properties. Is there a single place where all such information resides - perhaps it is here somewhere? Can you please point me to the same? Thanks mainhoon
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Re: Everything about Factorials on the GMAT [#permalink]

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08 Feb 2011, 00:19

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.

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

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16 Jul 2011, 06:29

Dear Bunuel,

I did not get the below part .pls explain briefly . "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"
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