1.

P=n(n^(2)-1)(n^(2)-4)(n^(2)-9)…(n^(2)-100) is always divisible by, (n in I)

Answer»

`2!3!4!5!6!`
`(5!)^(4)`
`(10!)^(2)`
`10!11!`

SOLUTION :`(a,b,C,d)`
We know that product of `r` consecutive integer is divisible by `r!`
We have `P=(N-10)(n-9)(n-8)…(n+10)` is PRODUT of `21` consecutive integers
Which is divisible by `21!`
`[(n-10)(n-9)(n-8)...n][(n+1)(n+2)(n+3)...(n+10)]`
Which is divisible by `11! 10!`
`[(n-10)...(n-6)]xx[(n-5)...(n-1)]xx[n(n+1)...(n+4)]xx[(n+5)...(n+9)]xx(n+10)`
which is divisible by `(5!)^(4)`.
Also it can be shown that `P` is divisible by `2!3!4!5!6!`


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