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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. |
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