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How many powers of 900 are in 50!

A) 2 B) 4 C) 6 D) 8 E) 10

Can one explain this answer clearly? This is actually a post in the GMATClub Math Tutorial (I don't know how to paste the link, sorry, am new). It says at the end that...

"We need all the prime {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!"

I did not understand this. What does "5 can provide us with only 6 pairs" mean? Is the answer only driven by that? What about 2 and 3? And if the powers had been all different for the original number say X = 2^4 3^7 5^9, then what?

Can one explain this answer clearly? This is actually a post in the GMATClub Math Tutorial (I don't know how to paste the link, sorry, am new). It says at the end that...

"We need all the prime {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!"

I did not understand this. What does "5 can provide us with only 6 pairs" mean? Is the answer only driven by that? What about 2 and 3? And if the powers had been all different for the original number say X = 2^4 3^7 5^9, then what?

If you have a problem understanding it don't worry, you won't need it for GMAT.

There is a following solution: 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

To elaborate:

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

Again don't worry about this examples too much.
_________________

Re: How many powers of 900 are in 50! [#permalink]

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08 Aug 2010, 09:17

Excellent explanation. Thanks for the detailed analysis! Also notice you moved the post, sorry about that.. Realize should have posted here to begin with.
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Re: How many powers of 900 are in 50! [#permalink]

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23 Oct 2014, 05:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: How many powers of 900 are in 50! [#permalink]

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20 Mar 2016, 06:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

My friend, please never open a brand new thread to post a problem until you already have searched extensively for the problem in question. This particular problem has been posted & discussed at least twice: how-many-powers-of-900-are-in-98781.html how-many-powers-of-900-are-in-134888.html I will ask Bunuel to merge the current post with these other posts.

Re: How many powers of 900 are in 50! [#permalink]

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21 Jan 2017, 23:44

To quickly solve such a problem, you need to factorise 900 (in this case). 900=2^2*3^2*5^2. You just need to find out the highest power of the largest prime number in 50!. As all other smaller primes will obviously be there in 50! if the largest prime is present. So 50! has 12 5's. So we have 6 pairs of 5's in 50!. Job done! 50! has 6 900s. :D
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In the pursuit of a better GMAT score. You can help me by giving me kudos if you like my post.

My friend, please never open a brand new thread to post a problem until you already have searched extensively for the problem in question. This particular problem has been posted & discussed at least twice: how-many-powers-of-900-are-in-98781.html how-many-powers-of-900-are-in-134888.html I will ask Bunuel to merge the current post with these other posts.

My friend, please never open a brand new thread to post a problem until you already have searched extensively for the problem in question. This particular problem has been posted & discussed at least twice: how-many-powers-of-900-are-in-98781.html how-many-powers-of-900-are-in-134888.html I will ask Bunuel to merge the current post with these other posts.

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

gmatclubot

Re: How many powers of 900 are in 50!
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22 Jan 2017, 10:25

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