Question

show that \((n-2)!\) is divisible by \(n-1\) when ever \(n \ne p+1\)

where \(p\) is a prime and \(n\gt 5\)

Answer 1

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)

Answer 2

hm

Answer 3

corrected the question, sorry

Answer 4

@bubblegum.

Answer 5

\(\large n \ne p+1\)
\(\large (n-1) \ne p\)

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 \((n-1)^2\)

Answer 6

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 \(n \ne p+1\)

Answer 7

*thinking

Answer 8

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 \((n-1)^2\)

Answer 9

Very nice!

Answer 10

thanks :)