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n-m-Z-n-n-n-1-n-2-if-n-m-where-we-know-the-value-m-but-not-n-can-we-solve-for-n-




Question Number 4234 by Filup last updated on 04/Jan/16
n, m∈Z  n!=n(n−1)(n−2)...    if n!=m where we know the value  m but not n, can we solve for n?
$${n},\:{m}\in\mathbb{Z} \\ $$$${n}!={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)… \\ $$$$ \\ $$$$\mathrm{if}\:{n}!={m}\:\mathrm{where}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{value} \\ $$$${m}\:\mathrm{but}\:\mathrm{not}\:{n},\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{for}\:{n}? \\ $$
Commented by prakash jain last updated on 04/Jan/16
One option:  n=m/2/3/4/5/6..  until no more divisions  are possible.  m=720  720/2=360,360/3=120,120/4=30,30/5=6
$$\mathrm{One}\:\mathrm{option}: \\ $$$${n}={m}/\mathrm{2}/\mathrm{3}/\mathrm{4}/\mathrm{5}/\mathrm{6}..\:\:\mathrm{until}\:\mathrm{no}\:\mathrm{more}\:\mathrm{divisions} \\ $$$$\mathrm{are}\:\mathrm{possible}. \\ $$$${m}=\mathrm{720} \\ $$$$\mathrm{720}/\mathrm{2}=\mathrm{360},\mathrm{360}/\mathrm{3}=\mathrm{120},\mathrm{120}/\mathrm{4}=\mathrm{30},\mathrm{30}/\mathrm{5}=\mathrm{6} \\ $$

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