Question

What is the highest power of $18$ contained in $50C25$?

Comments

As we find no of zeros (or highest power of $10$) in $n!$ by calculating no of pairs of $(2,5)$ in $n!$, same way we can find highest power of $18$ by counting no of pairs of $(2,3^2)$ in given term.

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do you mean like this -

min(50C25/2 + 50C25/2² + 50C25/2³ +........., 50C25/9 + 50C25/9² +....)

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@Akriti

No, Do seperately for both $50$! and $25$! each for $2$ and $3$ and then substract denominator from numerator .

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okay..thanks kapil..:)

Answer 1

We can rewrite it as 50!/(25!*25!)

18= 3*3*2

So we must find number of 2's and 3's in 50! & 25!

2's in 50!

$\frac{50}{2}+\frac{50}{4}+\frac{50}{8}+\frac{50}{16}+\frac{50}{32}$

25+12+6+3+1=47

3's in 50!

$\frac{50}{3}+\frac{50}{9}+\frac{50}{27}$

16+5+1= 22

Similarly we find no of 2's and 3's in 25! which is 22 & 10 respectively

50!/25!*25! no of 2's= 47-44=3 no of 3's=22-20=3

so only 18¹ is possible 

Hence highest power is 1