Question

Beautiful problem with a beautiful solution (if you see it!)

Answer 1

I don't see anything yet

Answer 2

\[D = \left| \begin{matrix}n!&(n+1)! &(n+2)! \\ (n+1)!&(n+2)!&(n+3)!\\(n+2)!&(n+3)!&(n+4)!\end{matrix}\right|\]Show that \(n\) divides \(\dfrac{D}{(n!)^3} - 4\).

Answer 3

factor out n! from each row maybe

Answer 4

Yup, and...?

Answer 5

or even better factor out n! from first row, (n+1)! from second row and (n+2)! from third row

Answer 6

\[D = n!(n+1)!(n+2)!\left| \begin{matrix}1&n+1 &(n+1)(n+2) \\ 1&n+2 &(n+2)(n+3)\\1&n+3&(n+3)(n+4)\end{matrix}\right|\]

Answer 7

Next we can reduce it easily or we can split that into product of two determinants i guess...

Answer 8

isnt it (n+2)(n+1) ? @ganeshie8

Answer 9

This is one of the problems where you have to do \({exactly}\) what the problem is trying to convey to you. \(D/(n!)^3\) is given as one of the terms? Just calculate it using properties of determinants. \(n\) divides a certain polynomial? Apply the Factor Theorem.

Answer 10

@divu.mkr you can factor n! out of the first row, (n+1)! out of the 2nd row , and (n+2)! out of the third row

Answer 11

$$
D = n!(n+1)!(n+2)!\left| \begin{matrix}1&n+1 &(n+1)(n+2) \\ 1&n+2 &(n+2)(n+3)\\1&n+3&(n+3)(n+4)\end{matrix}\right| = n!(n+1)!(n+2)!~\cdot 2
$$

Answer 12

now the problem simplifies to

Answer 13

\[\begin{align}D &= n!(n+1)!(n+2)!\left| \begin{matrix}1&n+1 &(n+1)(n+2) \\ 1&n+2 &(n+2)(n+3)\\1&n+3&(n+3)(n+4)\end{matrix}\right|\\~\\
&=n!(n+1)!(n+2)!\left| \begin{matrix}1&n+1 &(n+1)(n+2) \\ 0&1 &2(n+2)\\0&1 &2(n+3)\end{matrix}\right|\\~\\
&=n!(n+1)!(n+2)! [2(n+3) - 2(n+2)]\\~\\
& = n!^3(n+1)^2(n+2)[2]
\end{align}\]

\(\implies \dfrac{D}{n!^3} = 2(n+1)^2(n+2) \equiv 4 \pmod{n}\)

Answer 14

Ofcourse we may also start wid the expression D/n!^3, and divide each row by n! etc...

Answer 15

$$
D = n!(n+1)!(n+2)!\left| \begin{matrix}1&n+1 &(n+1)(n+2) \\ 1&n+2 &(n+2)(n+3)\\1&n+3&(n+3)(n+4)\end{matrix}\right| = n!(n+1)!(n+2)!~\cdot 2 \\
\therefore\\

\dfrac{D}{(n!)^3} - 4 = \frac{n!(n+1)!(n+2)!~\cdot 2}{n!^2}-4\\
= 2(n+1)^2(n+2)-4

$$

Answer 16

One may, of course, notice the fact that\[\frac{D}{(n!)^3} = D'(n) =\left| \begin{matrix}1 & n +1&(n+1)(n+2) \\ n+1&(n+1)(n+2)&(n+1)(n+2)(n+3) \\ (n+1)(n+2) & (n+1)(n+2)(n+3) &(n+1)(n+2)(n+3)(n+4)\end{matrix} \right|\]And hence, show that \(D(0) = 4\).

Answer 17

Nice :) \(D(0) = 4 \implies (n-4)\) is a factor of \(D(n)\)

Answer 18

The actual polynomial is \(p(n) = D(n) - 4\) so we're effectively showing \(p(\color{red}0) = 0 = \color{green}{4} - 4\) or that \(n - \color{red}0\) is a factor of \(p(n)\) by showing that \(\color{green}{D(0) = 4}\).

Answer 19

Clever indeed !

Answer 20

Now I see everything

Answer 21

http://en.wikipedia.org/wiki/Wilson%27s_theorem