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Question Number 181007 by mr W last updated on 20/Nov/22
[Question from Frix sir] ... I once played the game Tetris. You got points for each filled line, depending on how many lines you filled at once: 1 line = 1 point 2 lines = 3 points 3 lines = 6 points 4 lines = 10 points If you made 100 points, which possible combinations of lines did you fill when you filled 31 lines?
$$\left[{Question}\:{from}\:{Frix}\:{sir}\right] \\ $$$$…\:\mathrm{I}\:\mathrm{once}\:\mathrm{played}\:\mathrm{the}\:\mathrm{game}\:\mathrm{Tetris}.\: \\ $$$$\mathrm{You}\:\mathrm{got}\:\mathrm{points}\:\mathrm{for}\:\mathrm{each}\:\mathrm{filled}\:\mathrm{line},\: \\ $$$$\mathrm{depending}\:\mathrm{on}\:\mathrm{how}\:\mathrm{many}\:\mathrm{lines}\:\mathrm{you}\: \\ $$$$\mathrm{filled}\:\mathrm{at}\:\mathrm{once}: \\ $$$$\mathrm{1}\:\mathrm{line}\:=\:\mathrm{1}\:\mathrm{point} \\ $$$$\mathrm{2}\:\mathrm{lines}\:=\:\mathrm{3}\:\mathrm{points} \\ $$$$\mathrm{3}\:\mathrm{lines}\:=\:\mathrm{6}\:\mathrm{points} \\ $$$$\mathrm{4}\:\mathrm{lines}\:=\:\mathrm{10}\:\mathrm{points} \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{made}\:\mathrm{100}\:\mathrm{points},\:\mathrm{which}\:\mathrm{possible} \\ $$$$\mathrm{combinations}\:\mathrm{of}\:\mathrm{lines}\:\mathrm{did}\:\mathrm{you}\:\mathrm{fill}\:\mathrm{when} \\ $$$$\mathrm{you}\:\mathrm{filled}\:\mathrm{31}\:\mathrm{lines}? \\ $$
Answered by mr W last updated on 20/Nov/22
let′s say the player filled a times 1 line at once, b times 2 lines at once, c times 3 lines at once and d times 4 lines at once. he filled totally 31 lines and got totally 100 points. it is to find the non negative interger solutions of following equations: a+2b+3c+4d=31 a+3b+6c+10d=100 (ii)−(ii): b+3(c+2d)=69 let c+2d=u b+3u=69 its general solution is ⇒b=3k ⇒u=−k+23 general solution for c+2d=1 is c=2h−1 d=−h+1 general solution for c+2d=u is then c=(2h−1)(−k+23)=2h(−k+23)+k−23 d=(−h+1)(−k+23)=−h(−k+23)−k+23 let h(−k+23)=m ⇒c=2m+k−23 ⇒d=−m−k+23 a=31−2b−3c−4d=31−2(3k)−3(2m+k−23)−4(−m−k+23) ⇒a=−5k−2m+8 summary: { ((a=−5k−2m+8)),((b=3k)),((c=k+2m−23)),((d=−k−m+23)) :} a=−5k−2m+8≥0 ⇒5k+2m≤8 b=3k≥0 ⇒k≥0 c=k+2m−23≥0 ⇒k+2m≥23 d=−k−m+23≥0 ⇒k+m≤23 there are no such k and m which fulfill all conditions, see diagram. that means there is no solution. i.e. the player can′t fill 31 lines and get at the same time 100 points.
$${let}'{s}\:{say}\:{the}\:{player}\:{filled} \\ $$$${a}\:{times}\:\mathrm{1}\:{line}\:{at}\:{once}, \\ $$$${b}\:{times}\:\mathrm{2}\:{lines}\:{at}\:{once}, \\ $$$${c}\:{times}\:\mathrm{3}\:{lines}\:{at}\:{once}\:{and} \\ $$$${d}\:{times}\:\mathrm{4}\:{lines}\:{at}\:{once}. \\ $$$$ \\ $$$${he}\:{filled}\:{totally}\:\mathrm{31}\:{lines}\:{and}\:{got} \\ $$$${totally}\:\mathrm{100}\:{points}. \\ $$$${it}\:{is}\:{to}\:{find}\:{the}\:{non}\:{negative}\:{interger} \\ $$$${solutions}\:{of}\:{following}\:{equations}: \\ $$$${a}+\mathrm{2}{b}+\mathrm{3}{c}+\mathrm{4}{d}=\mathrm{31} \\ $$$${a}+\mathrm{3}{b}+\mathrm{6}{c}+\mathrm{10}{d}=\mathrm{100} \\ $$$$\left({ii}\right)−\left({ii}\right): \\ $$$${b}+\mathrm{3}\left({c}+\mathrm{2}{d}\right)=\mathrm{69} \\ $$$${let}\:{c}+\mathrm{2}{d}={u} \\ $$$${b}+\mathrm{3}{u}=\mathrm{69} \\ $$$${its}\:{general}\:{solution}\:{is} \\ $$$$\Rightarrow{b}=\mathrm{3}{k} \\ $$$$\Rightarrow{u}=−{k}+\mathrm{23} \\ $$$$ \\ $$$${general}\:{solution}\:{for}\:{c}+\mathrm{2}{d}=\mathrm{1}\:{is} \\ $$$${c}=\mathrm{2}{h}−\mathrm{1} \\ $$$${d}=−{h}+\mathrm{1} \\ $$$${general}\:{solution}\:{for}\:{c}+\mathrm{2}{d}={u}\:{is}\:{then} \\ $$$${c}=\left(\mathrm{2}{h}−\mathrm{1}\right)\left(−{k}+\mathrm{23}\right)=\mathrm{2}{h}\left(−{k}+\mathrm{23}\right)+{k}−\mathrm{23} \\ $$$${d}=\left(−{h}+\mathrm{1}\right)\left(−{k}+\mathrm{23}\right)=−{h}\left(−{k}+\mathrm{23}\right)−{k}+\mathrm{23} \\ $$$${let}\:{h}\left(−{k}+\mathrm{23}\right)={m} \\ $$$$\Rightarrow{c}=\mathrm{2}{m}+{k}−\mathrm{23} \\ $$$$\Rightarrow{d}=−{m}−{k}+\mathrm{23} \\ $$$$ \\ $$$${a}=\mathrm{31}−\mathrm{2}{b}−\mathrm{3}{c}−\mathrm{4}{d}=\mathrm{31}−\mathrm{2}\left(\mathrm{3}{k}\right)−\mathrm{3}\left(\mathrm{2}{m}+{k}−\mathrm{23}\right)−\mathrm{4}\left(−{m}−{k}+\mathrm{23}\right) \\ $$$$\Rightarrow{a}=−\mathrm{5}{k}−\mathrm{2}{m}+\mathrm{8} \\ $$$$ \\ $$$${summary}: \\ $$$$\begin{cases}{{a}=−\mathrm{5}{k}−\mathrm{2}{m}+\mathrm{8}}\\{{b}=\mathrm{3}{k}}\\{{c}={k}+\mathrm{2}{m}−\mathrm{23}}\\{{d}=−{k}−{m}+\mathrm{23}}\end{cases} \\ $$$${a}=−\mathrm{5}{k}−\mathrm{2}{m}+\mathrm{8}\geqslant\mathrm{0}\:\Rightarrow\mathrm{5}{k}+\mathrm{2}{m}\leqslant\mathrm{8} \\ $$$${b}=\mathrm{3}{k}\geqslant\mathrm{0}\:\Rightarrow{k}\geqslant\mathrm{0} \\ $$$${c}={k}+\mathrm{2}{m}−\mathrm{23}\geqslant\mathrm{0}\:\Rightarrow{k}+\mathrm{2}{m}\geqslant\mathrm{23} \\ $$$${d}=−{k}−{m}+\mathrm{23}\geqslant\mathrm{0}\:\Rightarrow{k}+{m}\leqslant\mathrm{23} \\ $$$${there}\:{are}\:{no}\:{such}\:{k}\:{and}\:{m}\:{which}\: \\ $$$${fulfill}\:{all}\:{conditions},\:{see}\:{diagram}. \\ $$$$ \\ $$$${that}\:{means}\:{there}\:{is}\:{no}\:{solution}.\:{i}.{e}. \\ $$$${the}\:{player}\:{can}'{t}\:{fill}\:\mathrm{31}\:{lines}\:{and}\:{get}\: \\ $$$${at}\:{the}\:{same}\:{time}\:\mathrm{100}\:{points}. \\ $$
Commented by mr W last updated on 20/Nov/22
Commented by mr W last updated on 20/Nov/22
did i do something wrong and there exists at least a valid solution?
$${did}\:{i}\:{do}\:{something}\:{wrong}\:{and}\:{there} \\ $$$${exists}\:{at}\:{least}\:{a}\:{valid}\:{solution}? \\ $$
Commented by Frix last updated on 20/Nov/22
You are right. As you surely had noticed I wrote this question as an example which cannot be solved by simply solving a system of linear equations.
$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{As}\:\mathrm{you}\:\mathrm{surely}\:\mathrm{had}\:\mathrm{noticed} \\ $$$$\mathrm{I}\:\mathrm{wrote}\:\mathrm{this}\:\mathrm{question}\:\mathrm{as}\:\mathrm{an}\:\mathrm{example}\:\mathrm{which} \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{simply}\:\mathrm{solving}\:\mathrm{a} \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{linear}\:\mathrm{equations}. \\ $$

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