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# Everything about Factorials on the GMAT

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Re: Everything about Factorials on the GMAT [#permalink]  17 Apr 2014, 02:09
Bunuel wrote:
agavaqif wrote:
Very nice post, indeed.is there anything about factors of factorial?

Check the links in my signature below.

Thankss. Actually after writing this post I found necessery info from links in your profile. Thanks again
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Re: Everything about Factorials on the GMAT [#permalink]  11 Jun 2014, 12:18
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.
how come we divide 32 by 5?
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Re: Everything about Factorials on the GMAT [#permalink]  11 Jun 2014, 13:13
Expert's post
sagnik2422 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. 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.
how come we divide 32 by 5?

Well, this is done according to the formula given there.
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Re: Everything about Factorials on the GMAT [#permalink]  27 Aug 2014, 05:05
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I don't know the proof but I thought this would be useful:

Where a and b are integers,

(a + b)! is divisible by a!.b!
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Re: Everything about Factorials on the GMAT [#permalink]  12 Dec 2014, 18:57
Really valuable, thanks!
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Re: Everything about Factorials on the GMAT [#permalink]  22 Apr 2015, 00:07
A more generalised method for finding powers of non prime numbers in a factorial:

Assuming we have to find the power of 12 in 30!

Now, 30 = 2 * 3 * 5.

Find what the highest power of each prime is in the given factorial (by the same method as given by Bunuel)

We have 25 + 12 + 6 + 3 + 1 = 47 2s in 50!

16 + 5 + 1 = 22 3s in 50!

10 + 2 = 12 5s in 50!

We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.

Similarly we can find powers of 4 (number of pairs of 2) etc.

Hope it helps.
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Re: Everything about Factorials on the GMAT [#permalink]  05 May 2015, 08:47
If there were option give 100 kidos at once, I would give for this post.

Thanks a hundred!!
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Re: Everything about Factorials on the GMAT [#permalink]  26 May 2015, 11:51
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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Re: Everything about Factorials on the GMAT [#permalink]  27 May 2015, 02:34
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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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Re: Everything about Factorials on the GMAT   [#permalink] 27 May 2015, 02:34

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