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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.

Joffan
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udit ghosh
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2 Answers2

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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 $18^1=18$.

Joffan
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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.

Vidyanshu Mishra
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  • why downvote??? – Vidyanshu Mishra Dec 17 '16 at 13:46
  • 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. – Jyrki Lahtonen Dec 17 '16 at 13:46
  • @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 – Vidyanshu Mishra Dec 17 '16 at 13:50
  • 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. – Jyrki Lahtonen Dec 17 '16 at 13:52
  • 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. – Jyrki Lahtonen Dec 17 '16 at 13:54
  • @JyrkiLahtonen so I think my answer is valid only if numerator and denominator, both are greater than 18. – Vidyanshu Mishra Dec 17 '16 at 13:56
  • I doubt it. You need to calculate divisibility by powers of two and three separately, and then use uniqueness of factorization to see how high a power of 18 will appear. – Jyrki Lahtonen Dec 17 '16 at 14:00
  • @JyrkiLahtonen Okay, let me do that thing in my answer, I shall ping you whenever it is done and please reply me after that whether it is correct or not. – Vidyanshu Mishra Dec 17 '16 at 14:01
  • @JyrkiLahtonen, I identified what a little mistake I was making and now I have transformed my answer. Please take a look at it. – Vidyanshu Mishra Dec 17 '16 at 14:20
  • The principle is now ok. – Jyrki Lahtonen Dec 17 '16 at 14:40