62The 2nd question can be solved by realizing that for n≥4, the lst digit will be 3 so no perfect square.
30Yeah.. I noticed that too now. :) Sir but for n=4, we aren't getting a perfect square..
Sir can we prove the first one from induction?
30
Ashish Kothari
·2011-09-12 23:02:05
@pidohbuhs - I got absolutely nothing out of your post. [2]
211)
We will use two proposition:
1)x\geq 0,y\geq 0, \lfloor{xy}\rfloor\geq \lfloor{x}\rfloor \lfloor{y}\rfloor
2)The exponent of any prime p in n! is given by
ord_p(n!)= \sum_{k=1}^{\infty}\left \lfloor{\frac{n}{p^k}} \rfloor
We will be able to prove the fact that (n!)! \mid (n!)^{(n-1)!} if we can prove that
ord_p((n!)!)\geq ord_p((n!)^{(n-1)!})
It's a bit obvious that ord_p((n!)^{(n-1)!})= (n-1)!ord_p(n!)
So we have to prove that \sum_{k=1}^{\infty}\lfloor{\frac{n!}{p^k}}\rfloor\geq (n-1)!\sum_{k=1}^{\infty}\lfloor{\frac{n}{p^k}}\rfloor
Note that \sum_{k=1}^{\infty}\lfloor{\frac{n!}{p^k}}\rfloor= \sum_{k=1}^{\infty}\lfloor{\frac{(n-1)!n}{p^k}}\rfloor \geq (n-1)!\sum_{k=1}^{\infty}\lfloor\frac{n}{p^k}\rfloor by proposition 1.
So we have proved the statement.
21
Shubhodip
·2011-09-12 23:12:55
actually its a very common idea in proving this kind of problems.
like m! n! (m+n)! | (2m)! (2n)!
m! (n!)m| (mn)! etc
341I like to use the result that the product of n consecutive numbers is divisible by n!
So since (n!)! is composed of (n-1)! sets of n consecutive numbers, the whole product is divisible by (n!)(n-1)!
211.A solution is given in Pre College Maths book.
Consider a situation where n! objects are there. Out of those they are organised in (n-1)! groups each having n objects. All the n objects in each group is identical.
So, no. of permutations of n! objects is \frac{(n!)!}{n!n!...n!}=\frac{(n!)!}{n!^{(n-1)!}}
So, n!(n-1)!| (n!)!
