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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]  08 Aug 2010, 07:49
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]  08 Aug 2010, 07:59
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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 $$18!=12^8*a$$ --> $$x=8$$.

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Re: How many powers of 900 are in 50! [#permalink]  08 Aug 2010, 08:17
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]  23 Oct 2014, 04:29
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Re: How many powers of 900 are in 50!   [#permalink] 23 Oct 2014, 04:29
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