Question Number 13389 by Tinkutara last updated on 19/May/17
![Prove that [x] + [2x] + [4x] + [8x] + [16x] + [32x] = 12345 has no solution.](https://www.tinkutara.com/question/Q13389.png)
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{x}\right]\:+\:\left[\mathrm{2}{x}\right]\:+\:\left[\mathrm{4}{x}\right]\:+\:\left[\mathrm{8}{x}\right]\:+\:\left[\mathrm{16}{x}\right]\:+\:\left[\mathrm{32}{x}\right]\:=\:\mathrm{12345} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$
Commented by prakash jain last updated on 20/May/17
![x=n+y , n∈Z^+ , 0≤y<1 [x] + [2x] + [4x] + [8x] + [16x] + [32x] =n+[y]+2n+[2y]+..+32n+[32y] =63n+[y]+[2y]+[4y]+[8y]+[16y]+[32y] =12345 ⇒n≤195 x=195+y 0≤y<1 [y]+[2y]+[4y]+[8y]+[16y]+[32y]=60 max([y])=0 max([2y])=1 max([4y])=3 max([8y])=7 max([16y)=15 max([32y])=31 max([y]+[2y]+[4y]+[8y]+[16y]+[32y])=57 no value of y is possible which will give [y]+[2y]+[4y]+[8y]+[16y]+[32y]=60 hence no solution](https://www.tinkutara.com/question/Q13423.png)
$${x}={n}+{y}\:,\:{n}\in\mathbb{Z}^{+} ,\:\mathrm{0}\leqslant{y}<\mathrm{1} \\ $$$$\left[{x}\right]\:+\:\left[\mathrm{2}{x}\right]\:+\:\left[\mathrm{4}{x}\right]\:+\:\left[\mathrm{8}{x}\right]\:+\:\left[\mathrm{16}{x}\right]\:+\:\left[\mathrm{32}{x}\right] \\ $$$$={n}+\left[{y}\right]+\mathrm{2}{n}+\left[\mathrm{2}{y}\right]+..+\mathrm{32}{n}+\left[\mathrm{32}{y}\right] \\ $$$$=\mathrm{63}{n}+\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right] \\ $$$$=\mathrm{12345} \\ $$$$\Rightarrow{n}\leqslant\mathrm{195} \\ $$$${x}=\mathrm{195}+{y} \\ $$$$\mathrm{0}\leqslant{y}<\mathrm{1} \\ $$$$\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right]=\mathrm{60} \\ $$$${max}\left(\left[{y}\right]\right)=\mathrm{0} \\ $$$${max}\left(\left[\mathrm{2}{y}\right]\right)=\mathrm{1} \\ $$$${max}\left(\left[\mathrm{4}{y}\right]\right)=\mathrm{3} \\ $$$${max}\left(\left[\mathrm{8}{y}\right]\right)=\mathrm{7} \\ $$$${max}\left(\left[\mathrm{16}{y}\right)=\mathrm{15}\right. \\ $$$${max}\left(\left[\mathrm{32}{y}\right]\right)=\mathrm{31} \\ $$$${max}\left(\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right]\right)=\mathrm{57} \\ $$$${no}\:{value}\:{of}\:{y}\:{is}\:{possible}\:{which} \\ $$$${will}\:{give}\: \\ $$$$\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right]=\mathrm{60} \\ $$$${hence}\:{no}\:{solution} \\ $$
Commented by mrW1 last updated on 20/May/17
$${very}\:{nice}! \\ $$
Commented by mrW1 last updated on 20/May/17
![but, can we find a solution for [y]+[2y]+[4y]+[8y]+[16y]+[32y]≤57? or how can we find?](https://www.tinkutara.com/question/Q13452.png)
$${but},\:{can}\:{we}\:{find}\:{a}\:{solution}\:{for} \\ $$$$\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right]\leqslant\mathrm{57}? \\ $$$${or}\:{how}\:{can}\:{we}\:{find}? \\ $$
Commented by prakash jain last updated on 20/May/17
![y=.99 or .999 will give [y]+[2y]+[4y]+[8y]+[16y]+[32y]=57 For other value, say 50 we can try various option 0+1+3+6+13+27=50 so we can pick y=.85 [y]=0 ]2y]=1 [4y]=3 [8y]=6 [16y]=13 [32y]=27](https://www.tinkutara.com/question/Q13456.png)
$${y}=.\mathrm{99}\:\mathrm{or}\:.\mathrm{999} \\ $$$${will}\:{give} \\ $$$$\left[{y}\right]+\left[\mathrm{2}{y}\right]+\left[\mathrm{4}{y}\right]+\left[\mathrm{8}{y}\right]+\left[\mathrm{16}{y}\right]+\left[\mathrm{32}{y}\right]=\mathrm{57} \\ $$$$\mathrm{For}\:\mathrm{other}\:\mathrm{value},\:{say}\:\mathrm{50} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{try}\:\mathrm{various}\:\mathrm{option} \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{3}+\mathrm{6}+\mathrm{13}+\mathrm{27}=\mathrm{50} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{pick}\:{y}=.\mathrm{85} \\ $$$$\left[{y}\right]=\mathrm{0} \\ $$$$\left.\right]\left.\mathrm{2}{y}\right]=\mathrm{1} \\ $$$$\left[\mathrm{4}{y}\right]=\mathrm{3} \\ $$$$\left[\mathrm{8}{y}\right]=\mathrm{6} \\ $$$$\left[\mathrm{16}{y}\right]=\mathrm{13} \\ $$$$\left[\mathrm{32}{y}\right]=\mathrm{27} \\ $$
Commented by mrW1 last updated on 20/May/17
$${Thanks}! \\ $$