How many powers of 900 are in 50!

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.