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Find-all-integer-solutions-of-the-system-35x-63y-45z-1-x-lt-9-y-lt-5-z-lt-7-

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Question Number 19198 by Tinkutara last updated on 07/Aug/17
Find all integer solutions of the system: 35x + 63y + 45z = 1, ∣x∣ < 9, ∣y∣ < 5, ∣z∣ < 7.
Findallintegersolutionsofthesystem:35x+63y+45z=1,∣x∣<9,∣y∣<5,∣z∣<7.
Commented by mrW1 last updated on 09/Aug/17
(−10,2,5) (−1,−3,5) (8,−3,−2) (−1,2,−2) is this correct?
(−10,2,5)(−1,−3,5)(8,−3,−2)(−1,2,−2)isthiscorrect?
Commented by mrW1 last updated on 13/Aug/17
what is kvs jmo?
whatiskvsjmo?
Commented by Tinkutara last updated on 13/Aug/17
It is a Mathematics Olympiad in India.
ItisaMathematicsOlympiadinIndia.
Commented by mrW1 last updated on 13/Aug/17
Once you mentioned a nice book with many interesting mathematical questions. Can you tell me the book title and its author please.
Onceyoumentionedanicebookwithmanyinterestingmathematicalquestions.Canyoutellmethebooktitleanditsauthorplease.
Commented by Tinkutara last updated on 13/Aug/17
bdmath.com/books Mathematical Olympiad Treasures Search for this book on the given site.
bdmath.com/booksMathematicalOlympiadTreasuresSearchforthisbookonthegivensite.
Commented by mrW1 last updated on 13/Aug/17
wow, so many books even as pdf! thank you sir!
wow,somanybooksevenaspdf!thankyousir!
Commented by Tawa1 last updated on 04/Oct/18
Sir, i don′t understand how to search the site. i typed bdmath.com/books. it is not showing anything sir
Sir,idon′tunderstandhowtosearchthesite.itypedbdmath.com/books.itisnotshowinganythingsir
Answered by mrW1 last updated on 13/Aug/17
let′s find at first the general solution of 35x + 63y + 45z = 1 ⇒7(5x+9y)+45z=1 ⇒5x+9y=u ...(i) ⇒7u+45z=1 ...(ii) a particular solution of (ii) is: u_1 =−32 and z_1 =5 the general solution of (ii) is: u=−32+45m z=5−7m (with m=any integer) a particular solution of 5x+9y=1 is x_1 =2 and y_1 =−1 the general solution of 5x+9y=1 is x=2+9k y=−1−5k (with k=any integer) the general solution of (i) is then x=(2+9k)u=2×(−32+45m)+9uk=−64+90m+9n y=(−1−5k)u=−1×(−32+45m)−5uk=32−45m−5n with n=uk=any integer the requested general solution for 35x + 63y + 45z = 1 is therefore: { ((x=−64+90m+9n)),((y=32−45m−5n)),((z=5−7m)) :} with m, n=any integer for ∣z∣<7 −7<5−7m<7 −12<−7m<2 12/7>m>−2/7 ⇒m=0, 1 with m=0 x=−64+9n y=32−5n for ∣x∣<9 −9<−64+9n<9 55<9n<73 55/9<n<73/9 ⇒n=7,8 (∗) for ∣y∣<5 −5<32−5n<5 −37<−5n<−27 37/5>n>27/5 ⇒n=6,7 (∗) ⇒m=0, n=7 with m=1 x=−64+90+9n=26+9n y=32−45−5n=−13−5n for ∣x∣<9 −9<26+9n<9 −35<9n<−17 −39/9<n<−17/9 ⇒n=−4,−3,−2 (∗) for ∣y∣<5 −5<−13−5n<5 8<−5n<18 −8/5>n>−18/5 ⇒n=−3,−2 (∗) ⇒m=1, n=−3 or −2 now we have totally 3 solutions (m,n)=(0,7), (1,−3),(1,−2) or (x,y,z)=(−1,−3,5) (x,y,z)=(−1,2,−2) (x,y,z)=(−8,−3,−2)
let′sfindatfirstthegeneralsolutionof35x+63y+45z=1⇒7(5x+9y)+45z=1⇒5x+9y=u…(i)⇒7u+45z=1…(ii)aparticularsolutionof(ii)is:u1=−32andz1=5thegeneralsolutionof(ii)is:u=−32+45mz=5−7m(withm=anyinteger)aparticularsolutionof5x+9y=1isx1=2andy1=−1thegeneralsolutionof5x+9y=1isx=2+9ky=−1−5k(withk=anyinteger)thegeneralsolutionof(i)isthenx=(2+9k)u=2×(−32+45m)+9uk=−64+90m+9ny=(−1−5k)u=−1×(−32+45m)−5uk=32−45m−5nwithn=uk=anyintegertherequestedgeneralsolutionfor35x+63y+45z=1istherefore:{x=−64+90m+9ny=32−45m−5nz=5−7mwithm,n=anyintegerfor∣z∣<7−7<5−7m<7−12<−7m<212/7>m>−2/7⇒m=0,1withm=0x=−64+9ny=32−5nfor∣x∣<9−9<−64+9n<955<9n<7355/9<n<73/9⇒n=7,8(∗)for∣y∣<5−5<32−5n<5−37<−5n<−2737/5>n>27/5⇒n=6,7(∗)⇒m=0,n=7withm=1x=−64+90+9n=26+9ny=32−45−5n=−13−5nfor∣x∣<9−9<26+9n<9−35<9n<−17−39/9<n<−17/9⇒n=−4,−3,−2(∗)for∣y∣<5−5<−13−5n<58<−5n<18−8/5>n>−18/5⇒n=−3,−2(∗)⇒m=1,n=−3or−2nowwehavetotally3solutions(m,n)=(0,7),(1,−3),(1,−2)or(x,y,z)=(−1,−3,5)(x,y,z)=(−1,2,−2)(x,y,z)=(−8,−3,−2)
Commented by Tinkutara last updated on 13/Aug/17
If particular solution is u_1 = −32, how is general solution u_1 = −32 + 45m?
Ifparticularsolutionisu1=−32,howisgeneralsolutionu1=−32+45m?
Commented by mrW1 last updated on 13/Aug/17
This is the basic to solve such questions. let′s have a look at ax+by=c. let′s say the greates common divisor of a and b is δ, i.e. δ=gcd(a,b). we see that a solution exists if and only if δ is a divisor of c. then: ax+by=c⇔(a/δ)x+(b/δ)y=(c/δ) or we write it as a′x+b′y=c′ with a′=(a/δ), etc. a′ and b′ are relatively prime and a′x+b′y=c′ has always solution. if a′x+b′y=c′ has a particular solution x_1 and y_1 , i.e. a′x_1 +b′y_1 =c′, then its general solution can be expressed as: x=x_1 +b′n y=y_1 −a′n (with n=any integer) this can be easily checked: a′(x_1 +b′n)+b′(y_1 −a′n) =a′x_1 +a′b′n+b′y_1 −b′a′n =a′x_1 +b′y_1 =c′ certainly you can also write the general solution as x=x_1 −b′n y=y_1 +a′n but this makes no difference since n can be any integer value. now we go back to your question. 7u+45z=1 δ=gcd(a,b)=gcd(7,45)=1 ⇒a′=7, b′=45, c′=1 7u+45z=1 has a particular solution: u_1 =−32 and z_1 =5 according to above its general solution is u=u_1 +b′m=−32+45m (or −32−45m) z=z_1 −a′m=5−7m (or 5+7m) I hope I could explain clearly enough.
Thisisthebasictosolvesuchquestions.let′shavealookatax+by=c.let′ssaythegreatescommondivisorofaandbisδ,i.e.δ=gcd(a,b).weseethatasolutionexistsifandonlyifδisadivisorofc.then:ax+by=c⇔aδx+bδy=cδorwewriteitasa′x+b′y=c′witha′=aδ,etc.a′andb′arerelativelyprimeanda′x+b′y=c′hasalwayssolution.ifa′x+b′y=c′hasaparticularsolutionx1andy1,i.e.a′x1+b′y1=c′,thenitsgeneralsolutioncanbeexpressedas:x=x1+b′ny=y1−a′n(withn=anyinteger)thiscanbeeasilychecked:a′(x1+b′n)+b′(y1−a′n)=a′x1+a′b′n+b′y1−b′a′n=a′x1+b′y1=c′certainlyyoucanalsowritethegeneralsolutionasx=x1−b′ny=y1+a′nbutthismakesnodifferencesincencanbeanyintegervalue.nowwegobacktoyourquestion.7u+45z=1δ=gcd(a,b)=gcd(7,45)=1⇒a′=7,b′=45,c′=17u+45z=1hasaparticularsolution:u1=−32andz1=5accordingtoaboveitsgeneralsolutionisu=u1+b′m=−32+45m(or−32−45m)z=z1−a′m=5−7m(or5+7m)IhopeIcouldexplainclearlyenough.
Commented by Tinkutara last updated on 13/Aug/17
Thank you very much Sir! This explanation is very useful.
ThankyouverymuchSir!Thisexplanationisveryuseful.

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