Menu Close

n-n-20-n-




Question Number 187215 by Humble last updated on 14/Feb/23
((√(n!))/n)=(√(20)) ; n=?
$$\frac{\sqrt{{n}!}}{{n}}=\sqrt{\mathrm{20}}\:;\:{n}=? \\ $$
Answered by ajfour last updated on 14/Feb/23
((n!)/n^2 )=20  (n−1)!=20(n−1)+20  (n−2)!=20+((20)/(n−1))  n−1=1,2,4,5  (eligible candidates)  lets check which  qualifies ( .  .)  but also  (n−2)!>20>3!  ⇒  n>5     n−1=5    4!=20+4  ✓  n=6
$$\frac{{n}!}{{n}^{\mathrm{2}} }=\mathrm{20} \\ $$$$\left({n}−\mathrm{1}\right)!=\mathrm{20}\left({n}−\mathrm{1}\right)+\mathrm{20} \\ $$$$\left({n}−\mathrm{2}\right)!=\mathrm{20}+\frac{\mathrm{20}}{{n}−\mathrm{1}} \\ $$$${n}−\mathrm{1}=\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{5}\:\:\left({eligible}\:{candidates}\right) \\ $$$${lets}\:{check}\:{which}\:\:{qualifies}\:\left(\:.\:\:.\right) \\ $$$${but}\:{also}\:\:\left({n}−\mathrm{2}\right)!>\mathrm{20}>\mathrm{3}! \\ $$$$\Rightarrow\:\:{n}>\mathrm{5}\:\:\: \\ $$$${n}−\mathrm{1}=\mathrm{5}\:\: \\ $$$$\mathrm{4}!=\mathrm{20}+\mathrm{4}\:\:\checkmark \\ $$$${n}=\mathrm{6} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *