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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?
[QuestionfromFrixsir]…IonceplayedthegameTetris.Yougotpointsforeachfilledline,dependingonhowmanylinesyoufilledatonce:1line=1point2lines=3points3lines=6points4lines=10pointsIfyoumade100points,whichpossiblecombinationsoflinesdidyoufillwhenyoufilled31lines?
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′ssaytheplayerfilledatimes1lineatonce,btimes2linesatonce,ctimes3linesatonceanddtimes4linesatonce.hefilledtotally31linesandgottotally100points.itistofindthenonnegativeintergersolutionsoffollowingequations:a+2b+3c+4d=31a+3b+6c+10d=100(ii)−(ii):b+3(c+2d)=69letc+2d=ub+3u=69itsgeneralsolutionis⇒b=3k⇒u=−k+23generalsolutionforc+2d=1isc=2h−1d=−h+1generalsolutionforc+2d=uisthenc=(2h−1)(−k+23)=2h(−k+23)+k−23d=(−h+1)(−k+23)=−h(−k+23)−k+23leth(−k+23)=m⇒c=2m+k−23⇒d=−m−k+23a=31−2b−3c−4d=31−2(3k)−3(2m+k−23)−4(−m−k+23)⇒a=−5k−2m+8summary:{a=−5k−2m+8b=3kc=k+2m−23d=−k−m+23a=−5k−2m+8⩾0⇒5k+2m⩽8b=3k⩾0⇒k⩾0c=k+2m−23⩾0⇒k+2m⩾23d=−k−m+23⩾0⇒k+m⩽23therearenosuchkandmwhichfulfillallconditions,seediagram.thatmeansthereisnosolution.i.e.theplayercan′tfill31linesandgetatthesametime100points.
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?
dididosomethingwrongandthereexistsatleastavalidsolution?
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.
Youareright.AsyousurelyhadnoticedIwrotethisquestionasanexamplewhichcannotbesolvedbysimplysolvingasystemoflinearequations.

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