What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$?
How will I be able to find the answer to such questions? Is there any special technique to find the answer to such problems? Thank you.
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3$\begingroup$ this question is relevant. $\endgroup$– luluDec 17, 2016 at 13:39
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2$\begingroup$ This formula may be even more relevant :-) See also here. $\endgroup$– Jyrki LahtonenDec 17, 2016 at 13:56
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$\begingroup$ @Rohan I don't think it is correct to tamper with original question, even if it is obvious. $\endgroup$– user312097Dec 17, 2016 at 14:05
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2 Answers
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$18 = 2\cdot 3^2$. We can find the power of a small prime in a large factorial by successive division to find base divisibility, then divisibility by squares, etc. So the multiplicity of powers of $2$ in $50!$, $v_2(50!),$ is $$ v_2(50!) = \left\lfloor\frac{50}{2}\right\rfloor + \left\lfloor\frac{50}{4}\right\rfloor + \left\lfloor\frac{50}{8}\right\rfloor + \cdots = 25+12+6+3+1 = 47$$
and similarly $v_2(25!)=22$, $v_3(50!)=16+5+1 = 22$ and $v_3(25!)=8+2 =10$, so
$$v_2\left(\frac{50!}{25!25!}\right) = 47-2\cdot22=3 \\ v_3\left(\frac{50!}{25!25!}\right) = 22-2\cdot 10=2 $$
and only $2$ available powers of $3$ means that $v_{18}\left(\frac{50!}{25!25!}\right)=1$ - the highest power of $18$ dividing the given expression is $18^1=18$.
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HINT:Use the de polignac formula. As written first, answer was a little vague with many errors (I forgot that initially de polignac formula is used to find highest power of primes and 18 is not a prime). I m presenting a new one now.
Since $18=2\times 3\times3$ (Prime factorisation). So, a better idea is just to find the highest power of $2,3,3$ in the numerator and denominator. Then just subtract the number of powers of $2,3,3$ in denominator from the number of powers of $2,3,3$ in numerator. Arrange the remaining powers in such a manner so that they form highest possible power of $18$. That will be the answer.
answered Dec 17, 2016 at 13:42
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$\begingroup$ Hmm. If we had $\dfrac{36}{6\cdot6}$ instead, then you seem to suggest that because $36$ is divisible by $18$, but neither of the sixes is, the fraction would be divisible by $18$ also? I didn't downvote, but this is incomplete, and not very helpful in that something tells me not all readers will know of that divisibility formula at all, and even those who do, won't recognize it by that name. Particularly because Lulu already gave a link to an on-site description of the formula. $\endgroup$ Dec 17, 2016 at 13:46
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$\begingroup$ @JyrkiLahtonen I thought de polignac formula work for factorials only. So there must be one of two things happening. 1.you missed the factorial sign 2.formula is wrong. ???? Which is the case?? Regards $\endgroup$ Dec 17, 2016 at 13:50
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$\begingroup$ The formula works for factorials only. But I thought you would see the error in your thinking if I used simpler numbers in that example. HINT: 18 is not a prime. $\endgroup$ Dec 17, 2016 at 13:52
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$\begingroup$ But, since you insist, consider $$\frac{6!}{3!3!}.$$ The numerator is a multiple of $18$, neither of the factorials in the denominator is. Yet the fraction is not divisible by $18$ because it is not divisible by three. $\endgroup$ Dec 17, 2016 at 13:54