Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 176847 by Ar Brandon last updated on 27/Sep/22

If x, y ∈Z 23!=2^x 5^y k and k is an even number, then what is max(x+y) ?

$$\mathrm{If}\:{x},\:{y}\:\in\mathbb{Z} \\ $$$$\mathrm{23}!=\mathrm{2}^{{x}} \mathrm{5}^{{y}} {k} \\ $$$$\mathrm{and}\:{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{number},\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\: \\ $$$$\mathrm{max}\left({x}+{y}\right)\:? \\ $$

Answered by BaliramKumar last updated on 27/Sep/22

determinant ((2,(23)),(2,(11)),(2,5),(2,2),(2,1),(,0)) & determinant ((5,(23)),(5,4),(,0)) determinant ((),()) x = 11+5+2+1+0 = 19 y = 4+0 = 4 23! = 2^(19) ×5^4 ×x [where ′x′ will be odd] 23! = 2^(18) ×5^4 ×(2x) 23! = 2^(18) ×5^4 ×k [2x = k] max(x+y) = 18+4 = determinant (((22)))

$$ \\ $$$$\begin{array}{|c|c|c|c|c|c|}{\mathrm{2}}&\hline{\mathrm{23}}\\{\mathrm{2}}&\hline{\mathrm{11}}\\{\mathrm{2}}&\hline{\mathrm{5}}\\{\mathrm{2}}&\hline{\mathrm{2}}\\{\mathrm{2}}&\hline{\mathrm{1}}\\{}&\hline{\mathrm{0}}\\\hline\end{array}\:\:\:\:\:\:\&\:\:\:\:\:\:\begin{array}{|c|c|c|}{\mathrm{5}}&\hline{\mathrm{23}}\\{\mathrm{5}}&\hline{\mathrm{4}}\\{}&\hline{\mathrm{0}}\\\hline\end{array}\begin{array}{|c|c|}\\\\\hline\end{array} \\ $$$${x}\:=\:\mathrm{11}+\mathrm{5}+\mathrm{2}+\mathrm{1}+\mathrm{0}\:=\:\mathrm{19} \\ $$$${y}\:=\:\mathrm{4}+\mathrm{0}\:=\:\mathrm{4} \\ $$$$\mathrm{23}!\:=\:\mathrm{2}^{\mathrm{19}} ×\mathrm{5}^{\mathrm{4}} ×{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{where}\:'{x}'\:{will}\:{be}\:{odd}\right] \\ $$$$\mathrm{23}!\:=\:\mathrm{2}^{\mathrm{18}} ×\mathrm{5}^{\mathrm{4}} ×\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{23}!\:=\:\mathrm{2}^{\mathrm{18}} ×\mathrm{5}^{\mathrm{4}} ×{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{2}{x}\:=\:{k}\right] \\ $$$${max}\left({x}+{y}\right)\:=\:\mathrm{18}+\mathrm{4}\:=\:\begin{array}{|c|}{\mathrm{22}}\\\hline\end{array} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 28/Sep/22

Thanks

Answered by mr W last updated on 27/Sep/22

⌊((23)/2)⌋+⌊((23)/2^2 )⌋+⌊((23)/2^3 )⌋+⌊((23)/2^4 )⌋=11+5+2+1=19 ⌊((23)/5)⌋=4 ⇒23!=2^(19) ×5^4 ×... such that (...) is even, it must contain at least one factor 2. ⇒x_(max) =18, y_(max) =4 ⇒(x+y)_(max) =18+4=22

$$\lfloor\frac{\mathrm{23}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{23}}{\mathrm{2}^{\mathrm{2}} }\rfloor+\lfloor\frac{\mathrm{23}}{\mathrm{2}^{\mathrm{3}} }\rfloor+\lfloor\frac{\mathrm{23}}{\mathrm{2}^{\mathrm{4}} }\rfloor=\mathrm{11}+\mathrm{5}+\mathrm{2}+\mathrm{1}=\mathrm{19} \\ $$$$\lfloor\frac{\mathrm{23}}{\mathrm{5}}\rfloor=\mathrm{4} \\ $$$$\Rightarrow\mathrm{23}!=\mathrm{2}^{\mathrm{19}} ×\mathrm{5}^{\mathrm{4}} ×... \\ $$$${such}\:{that}\:\left(...\right)\:{is}\:{even},\:{it}\:{must}\:{contain} \\ $$$${at}\:{least}\:{one}\:{factor}\:\mathrm{2}. \\ $$$$\Rightarrow{x}_{{max}} =\mathrm{18},\:{y}_{{max}} =\mathrm{4} \\ $$$$\Rightarrow\left({x}+{y}\right)_{{max}} =\mathrm{18}+\mathrm{4}=\mathrm{22} \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thank you ��

Answered by Rasheed.Sindhi last updated on 27/Sep/22

2 4 6 8 10 12 14 16 18 20 22......11 4 8 12 16 20..........5 8 16..................2 16.....................1 11+5+2+1=19 2′s k ieven maximum x is 18 5 10 15 20...........4 maximum y 4 max(x+y)=18+4=22

$$\mathrm{2}\:\mathrm{4}\:\mathrm{6}\:\mathrm{8}\:\mathrm{10}\:\mathrm{12}\:\mathrm{14}\:\mathrm{16}\:\mathrm{18}\:\mathrm{20}\:\mathrm{22}......\mathrm{11} \\ $$$$\mathrm{4}\:\mathrm{8}\:\mathrm{12}\:\mathrm{16}\:\mathrm{20}..........\mathrm{5} \\ $$$$\mathrm{8}\:\mathrm{16}..................\mathrm{2} \\ $$$$\mathrm{16}.....................\mathrm{1} \\ $$$$\mathrm{11}+\mathrm{5}+\mathrm{2}+\mathrm{1}=\mathrm{19}\:\mathrm{2}'{s} \\ $$$${k}\:{ieven}\:{maximum}\:{x}\:{is}\:\mathrm{18} \\ $$$$\mathrm{5}\:\mathrm{10}\:\mathrm{15}\:\mathrm{20}...........\mathrm{4}\: \\ $$$${maximum}\:{y}\:\:\mathrm{4} \\ $$$${max}\left({x}+{y}\right)=\mathrm{18}+\mathrm{4}=\mathrm{22} \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thanks ��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com