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.htmlIf 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*5^2Find 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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