Question-207864

Question Number 207864 by efronzo1 last updated on 29/May/24

$$\:\:\:\:\:\underbrace{\:} \\ $$

Answered by Rasheed.Sindhi last updated on 29/May/24

2^(5m) ∙5^(2n) ∙k=2020^(2020) =(2^2 .5.101)^(2020) 2^(5m) =2^(4040) ∧ 5^(2n) =5^(2020) ∧k=101^(2020) 5m≤4040 ∧ 2n≤2020 m_(max) =((4040)/5)=808 ∧ n_(max) =((2020)/2)=1010 max(n+2m)=n_(max) +2m_(max) =1010+2(808)=2626

$$\mathrm{2}^{\mathrm{5}{m}} \centerdot\mathrm{5}^{\mathrm{2}{n}} \centerdot{k}=\mathrm{2020}^{\mathrm{2020}} =\left(\mathrm{2}^{\mathrm{2}} .\mathrm{5}.\mathrm{101}\right)^{\mathrm{2020}} \\ $$$$\mathrm{2}^{\mathrm{5}{m}} =\mathrm{2}^{\mathrm{4040}} \wedge\:\mathrm{5}^{\mathrm{2}{n}} =\mathrm{5}^{\mathrm{2020}} \wedge{k}=\mathrm{101}^{\mathrm{2020}} \\ $$$$\mathrm{5}{m}\leqslant\mathrm{4040}\:\wedge\:\mathrm{2}{n}\leqslant\mathrm{2020} \\ $$$${m}_{{max}} =\frac{\mathrm{4040}}{\mathrm{5}}=\mathrm{808}\:\wedge\:{n}_{{max}} =\frac{\mathrm{2020}}{\mathrm{2}}=\mathrm{1010} \\ $$$${max}\left({n}+\mathrm{2}{m}\right)={n}_{{max}} +\mathrm{2}{m}_{{max}} \\ $$$$=\mathrm{1010}+\mathrm{2}\left(\mathrm{808}\right)=\mathrm{2626} \\ $$

Facebook Tweet Pin

Question-207864

Leave a Reply Cancel reply