Question Number 112613 by mnjuly1970 last updated on 08/Sep/20
$$\:\:\:\:\:….{number}\:{theory}… \\ $$$$\:\:\:\:\:\:\:{Question}\::\:\:\:\:\:\:\mathrm{I}{f}\:\:\:{a}\:,\:{b}\:,\:{c}\:\:\in\:\mathbb{N}\:\:\:;\:{then}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}!\ast{b}!\ast{c}!\mid\left({a}+{b}+{c}\right)!\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}\:.\:{july}\:\mathrm{970}# \\ $$
Commented by Aina Samuel Temidayo last updated on 08/Sep/20
$$\mathrm{Oh}!\:\mathrm{Sorry},\:\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{was}\:\mathrm{an} \\ $$$$\mathrm{equation}. \\ $$
Commented by MJS_new last updated on 08/Sep/20
$$\mathrm{then}\:\mathrm{please}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counterexample} \\ $$
Answered by MJS_new last updated on 08/Sep/20
$$\mathrm{just}\:\mathrm{a}\:\mathrm{try}… \\ $$$$\mathrm{let}\:{a}\leqslant{b}\leqslant{c};\:\mathrm{obviously}\:{c}!\mid\left({a}+{b}+{c}\right)! \\ $$$$\frac{\left({a}+{b}+{c}\right)!}{{c}!}=\left({c}+\mathrm{1}\right)\left({c}+\mathrm{2}\right)…\left({c}+{a}+{b}\right) \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{factors}\:=\:{a}+{b} \\ $$$${b}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×…×{b} \\ $$$$\mathrm{now}\:\mathrm{each}\:\mathrm{factor}\:\mathrm{of}\:{b}!\:\mathrm{is}\:\mathrm{included}\:\mathrm{at}\:\mathrm{least}\:\mathrm{once} \\ $$$$\mathrm{in}\:{b}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\left({c}+\mathrm{1}\right)\left({c}+\mathrm{2}\right)…\left({c}+{b}\right) \\ $$$$\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}: \\ $$$$\mathrm{if}\:{c}={b}+{n}\:\Rightarrow\:{b}\mid\left({c}+{b}−{n}\right);\:\mathrm{1}\leqslant{n}\leqslant{b} \\ $$$$\mathrm{the}\:\mathrm{remaining}\:\mathrm{question}\:\mathrm{is},\:\mathrm{how}\:\mathrm{to}\:\mathrm{show}\:\mathrm{the} \\ $$$$“\mathrm{remaining}''\:\mathrm{number}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:{a} \\ $$$$\mathrm{let}\:{b}+{c}={n} \\ $$$$\mathrm{then}\:\frac{\left({a}+{n}\right)!}{{n}!}=\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({a}+{n}\right) \\ $$$$\mathrm{same}\:\mathrm{argument}\:\mathrm{as}\:\mathrm{above} \\ $$
Commented by mnjuly1970 last updated on 09/Sep/20
$${thank}\:{you}\:{sir}.{excellent} \\ $$$${and}\:{admirable}.. \\ $$