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How many powers of 900 are in 50!

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

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New post 08 Aug 2010, 08:49
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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?

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

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New post 08 Aug 2010, 08:59
7
6
mainhoon wrote:
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?


This is from my topic: math-number-theory-88376.html or everything-about-factorials-on-the-gmat-85592.html


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

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New post 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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New post 21 Jan 2017, 18:43
1
iMyself wrote:
How many powers of 900 are in 50!
A) 2
B) 4
C) 6
D) 8
E) 10

What is the easiest way to solve this problem?

Dear iMyself,

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.

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

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New post 21 Jan 2017, 23:44
3
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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Re: How many powers of 900 are in 50!  [#permalink]

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New post 03 Aug 2018, 12:08
Bunuel wrote:
mainhoon wrote:
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?


This is from my topic: http://gmatclub.com/forum/math-number-theory-88376.html or http://gmatclub.com/forum/everything-ab ... 85592.html


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.



Bunuel this is the best thing you ever said on gmat club If you have a problem understanding it don't worry, you won't need it for GMAT. :lol:

so i dont need to understand a problem :lol:
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Re: How many powers of 900 are in 50!  [#permalink]

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New post 04 Aug 2018, 03:52
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Cool dave13

On similar lines, we learn from mistakes we make.
If we do not make mistakes, we never learn.
Does it mean that one who learns the most keep making mistakes all the time? ;)
See you more in CR forums, buddy.

Make mistakes your best friend. :-)
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Re: How many powers of 900 are in 50! &nbs [#permalink] 04 Aug 2018, 03:52
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