Question Number 21404 by Tinkutara last updated on 23/Sep/17
$$\mathrm{Prove}\:\mathrm{that}\:\left(\mathrm{6}{n}\right)!\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}^{\mathrm{2}{n}} .\mathrm{3}^{{n}} . \\ $$
Answered by dioph last updated on 23/Sep/17
![if n > 1, there are at least 3n even numbers and 2n multiples of 3 in [1, 6n], so that (6n)! is divisible by 2^(3n) .3^(2n) and, in particular, by 2^(2n) .3^n](https://www.tinkutara.com/question/Q21416.png)
$$\mathrm{if}\:{n}\:>\:\mathrm{1},\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{3}{n}\:\mathrm{even} \\ $$$$\mathrm{numbers}\:\mathrm{and}\:\mathrm{2}{n}\:\mathrm{multiples}\:\mathrm{of}\:\mathrm{3}\:\mathrm{in} \\ $$$$\left[\mathrm{1},\:\mathrm{6}{n}\right],\:\mathrm{so}\:\mathrm{that}\:\left(\mathrm{6}{n}\right)!\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by} \\ $$$$\mathrm{2}^{\mathrm{3}{n}} .\mathrm{3}^{\mathrm{2}{n}} \:\mathrm{and},\:\mathrm{in}\:\mathrm{particular},\:\mathrm{by}\:\mathrm{2}^{\mathrm{2}{n}} .\mathrm{3}^{{n}} \\ $$
Commented by Tinkutara last updated on 23/Sep/17
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{lines}? \\ $$
Commented by dioph last updated on 23/Sep/17
$$\mathrm{from}\:\mathrm{every}\:\mathrm{2}\:\mathrm{consecutive}\:\mathrm{numbers} \\ $$$$\mathrm{1}\:\mathrm{is}\:\mathrm{even}\:\mathrm{and}\:\mathrm{for}\:\mathrm{every}\:\mathrm{3}\:\mathrm{consecutive} \\ $$$$\mathrm{numbers}\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$
Commented by Tinkutara last updated on 23/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$