Keo ki proman korte parbe - Por por 5 ti kromic sonkhar gunf

[Sudip Deb new]

Keu ki proman korte parbe - por por 5 ti kromic sonkhar gunfol 5! (factorial) diye bivajjo ?

[tarek like math]

among 1~n numbers one number is divisible by n. bcz if u divide a number suppose a by n and maximum (n-1) left as reminder then a+(n-1) is divisible by n that means every time between 1~n numbers one has n as a factor. thus every (n-1) numbers one number is divisible by (n-1). so every 1~(n-2) number has a number which divisible by (n-2). so each of n consecutive number has 1 factor between 1~n. so all 1~5 numbers are factor of multiple of 5 consecutive numbers. so 5! is a factor of (1.2...5).

[Masum]

That might be true but how do you confirm that one of the remaining $n-1$ numbers must be divisible by $n-1$, since the residues may eventually change and the fact is that we shall repeat this process.
You may try with two more processes.
First note that $5!=2^3\cdot 3\cdot5$. So prove that the product of five consecutive numbers is divisible by $8,3,5$ individually. Then,since they are co-prime, we can finish.
Alternatively, much more generally, try to prove that the product of $k$ consecutive integers is divisible by $k!$.
Hint : Think carefully about the identity,\[\binom{n}{k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\]

Also, don't use languages like bcz and etc. while posting.