ganeshie8
  • ganeshie8
show that \((n-2)!\) is divisible by \(n-1\) when ever \(n \ne p+1\) where \(p\) is a prime and \(n\gt 5\)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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bubblegum.
  • bubblegum.
i took n=5 so \(\Large\frac{(5-2)!}{5-1}\) = \(\Large\frac{3 \times 2 \times 1}{4}\)=\(\Large\frac{3}{2}\) so here (n-2)! is not divisible by (n-1)
anonymous
  • anonymous
hm
ganeshie8
  • ganeshie8
corrected the question, sorry

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ganeshie8
  • ganeshie8
@bubblegum.
bubblegum.
  • bubblegum.
\(\large n \ne p+1\) \(\large (n-1) \ne p\) we can write \(\large\frac{(n-2)!}{(n-1)}\) as \(\large\frac{(n-1)!}{(n-1)^2}\) If we have any number say X lets take X=36 then the divisors are->1,2,3,4,6,9,12,18,36 we can write 36 as a product of 2 divisors like-> 1x36=36 2x18=36 3x12=36... similarly for (n-1) the divisors will obviously be smaller than (n-1) and they all will come in the factorial we know that every number X can be represented as a product of its 2 factors lets say \((n-1)=d_1 \times d_2\) where \(d_1\) and \(d_2\) are the 2 divisors of \(n-1\) multiplying them in numerator gives us (n-1) we will have \((n-1)!=(n-1)(n-2)(n-3)...d_1...d_2..2.1\) we can see we have \((n-1)^2\) term generated in the numerator and we can easily say that this numerator will be divisible by denominator which is \((n-1)^2\)
bubblegum.
  • bubblegum.
we take \(n > 5\) because we cannot represent 1,2,3,4,5 is the form of product of 2 divisors now m thing why \(n \ne p+1\)
bubblegum.
  • bubblegum.
*thinking
bubblegum.
  • bubblegum.
if \(n=1+p\) then \(n-1=p\) \((n-1)!=p!\) yeah its clear that p! will obviously not contain \(p^2\) term so clearly \((n-1)!\) will not be divisible by \((n-1)^2\)
ganeshie8
  • ganeshie8
Very nice!
bubblegum.
  • bubblegum.
thanks :)

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