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Question Number 163313 by SLVR last updated on 06/Jan/22
Find the non negative integer  solutions of 2x+3y+5z=60
$${Find}\:{the}\:{non}\:{negative}\:{integer} \\ $$$${solutions}\:{of}\:\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{5}{z}=\mathrm{60} \\ $$
Answered by mr W last updated on 06/Jan/22
let 2x+3y=5u  5u+5z=60  u+z=12  ⇒u=n with n ∈ Z  ⇒z=12−n    2x+3y=1  x=3k−1 with k ∈ Z  y=−2k+1  2x+3y=5u=5n  x=5n(3k−1)=3(5nk)−5n=3m−5n  y=5n(−2k+1)=−2(5nk)+5n=−2m+5n    general solution:   { ((x=3m−5n)),((y=−2m+5n)),((z=12−n)) :}     (with m,n∈Z)  non−negative solutions:  z=12−n≥0 ⇒n≤12  x=3m−5n≥0  y=−2m+5n≥0  ⇒0≤n≤12  ⇒((5n)/3)≤m≤((5n)/2)  totally there are 71 solutions:  n=0: m=0  n=1: m=2  n=2: m=3, 4  n=3: m=5, 6, 7  ....  n=12: m=20, 21, ..., 30
$${let}\:\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{5}{u} \\ $$$$\mathrm{5}{u}+\mathrm{5}{z}=\mathrm{60} \\ $$$${u}+{z}=\mathrm{12} \\ $$$$\Rightarrow{u}={n}\:{with}\:{n}\:\in\:\mathbb{Z} \\ $$$$\Rightarrow{z}=\mathrm{12}−{n} \\ $$$$ \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{1} \\ $$$${x}=\mathrm{3}{k}−\mathrm{1}\:{with}\:{k}\:\in\:\mathbb{Z} \\ $$$${y}=−\mathrm{2}{k}+\mathrm{1} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{5}{u}=\mathrm{5}{n} \\ $$$${x}=\mathrm{5}{n}\left(\mathrm{3}{k}−\mathrm{1}\right)=\mathrm{3}\left(\mathrm{5}{nk}\right)−\mathrm{5}{n}=\mathrm{3}{m}−\mathrm{5}{n} \\ $$$${y}=\mathrm{5}{n}\left(−\mathrm{2}{k}+\mathrm{1}\right)=−\mathrm{2}\left(\mathrm{5}{nk}\right)+\mathrm{5}{n}=−\mathrm{2}{m}+\mathrm{5}{n} \\ $$$$ \\ $$$${general}\:{solution}: \\ $$$$\begin{cases}{{x}=\mathrm{3}{m}−\mathrm{5}{n}}\\{{y}=−\mathrm{2}{m}+\mathrm{5}{n}}\\{{z}=\mathrm{12}−{n}}\end{cases}\:\:\:\:\:\left({with}\:{m},{n}\in\mathbb{Z}\right) \\ $$$${non}−{negative}\:{solutions}: \\ $$$${z}=\mathrm{12}−{n}\geqslant\mathrm{0}\:\Rightarrow{n}\leqslant\mathrm{12} \\ $$$${x}=\mathrm{3}{m}−\mathrm{5}{n}\geqslant\mathrm{0} \\ $$$${y}=−\mathrm{2}{m}+\mathrm{5}{n}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{n}\leqslant\mathrm{12} \\ $$$$\Rightarrow\frac{\mathrm{5}{n}}{\mathrm{3}}\leqslant{m}\leqslant\frac{\mathrm{5}{n}}{\mathrm{2}} \\ $$$${totally}\:{there}\:{are}\:\mathrm{71}\:{solutions}: \\ $$$${n}=\mathrm{0}:\:{m}=\mathrm{0} \\ $$$${n}=\mathrm{1}:\:{m}=\mathrm{2} \\ $$$${n}=\mathrm{2}:\:{m}=\mathrm{3},\:\mathrm{4} \\ $$$${n}=\mathrm{3}:\:{m}=\mathrm{5},\:\mathrm{6},\:\mathrm{7} \\ $$$$…. \\ $$$${n}=\mathrm{12}:\:{m}=\mathrm{20},\:\mathrm{21},\:…,\:\mathrm{30} \\ $$
Commented by SLVR last updated on 06/Jan/22
Wow...really great enough..we  are blessed...with your service
$${Wow}…{really}\:{great}\:{enough}..{we} \\ $$$${are}\:{blessed}…{with}\:{your}\:{service} \\ $$
Commented by mr W last updated on 06/Jan/22
Commented by mr W last updated on 06/Jan/22
note:  number of non negative solutions of  2x+3y+5z=60 is the coef. of term x^(60)   in the expansion of   (1+x^2 +x^4 +...)(1+x^3 +x^6 +...)(1+x^5 +x^(10) +...)  =(1/((1−x^2 )(1−x^3 )(1−x^5 ))). that is 71.
$$\underline{{note}:} \\ $$$${number}\:{of}\:{non}\:{negative}\:{solutions}\:{of} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{5}{z}=\mathrm{60}\:{is}\:{the}\:{coef}.\:{of}\:{term}\:{x}^{\mathrm{60}} \\ $$$${in}\:{the}\:{expansion}\:{of}\: \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…\right)\left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +…\right)\left(\mathrm{1}+{x}^{\mathrm{5}} +{x}^{\mathrm{10}} +…\right) \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{x}^{\mathrm{5}} \right)}.\:{that}\:{is}\:\mathrm{71}. \\ $$
Commented by mr W last updated on 06/Jan/22
Commented by Rasheed.Sindhi last updated on 06/Jan/22
M RE _(THAN_(P∈RF∈⊂T  !) )   Multimedia(graph)  & number of solutions    are  in addition!  ThanX sir!
$$\underset{\underset{\mathcal{P}\in\mathcal{RF}\in\subset\mathcal{T}\:\:!} {\mathrm{THAN}}} {\mathcal{M} \mathcal{RE}\:} \\ $$$$\mathbb{M}\mathrm{ultimedia}\left(\mathrm{graph}\right) \\ $$$$\&\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\: \\ $$$$\:\mathrm{are}\:\:\mathrm{in}\:\mathrm{addition}! \\ $$$$\mathcal{T}{han}\mathcal{X}\:\boldsymbol{{sir}}! \\ $$
Commented by mr W last updated on 06/Jan/22
thanks sirs!  graph helps very much to find out if   errors are made. therefore i like to   work with graph.
$${thanks}\:{sirs}! \\ $$$${graph}\:{helps}\:{very}\:{much}\:{to}\:{find}\:{out}\:{if}\: \\ $$$${errors}\:{are}\:{made}.\:{therefore}\:{i}\:{like}\:{to}\: \\ $$$${work}\:{with}\:{graph}. \\ $$
Commented by Tawa11 last updated on 06/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Jan/22
SHARING: https://youtu.be/fw1kRz83Fj0
Commented by SLVR last updated on 31/Jan/22
Respected prof.W...evaluation  of x^(60)  with out original mul  tlication all terms complnent  wise..kindly  with general  term of expansion.please
$${Respected}\:{prof}.{W}…{evaluation} \\ $$$${of}\:{x}^{\mathrm{60}} \:{with}\:{out}\:{original}\:{mul} \\ $$$${tlication}\:{all}\:{terms}\:{complnent} \\ $$$${wise}..{kindly}\:\:{with}\:{general} \\ $$$${term}\:{of}\:{expansion}.{please} \\ $$
Commented by mr W last updated on 31/Jan/22
please express clearly what you want  to have!
$${please}\:{express}\:{clearly}\:{what}\:{you}\:{want} \\ $$$${to}\:{have}! \\ $$
Commented by SLVR last updated on 02/Feb/22
Sir...I mean to say ..  coeffitient of x^(60)  in (1−x^2 )(1−x^3 )(1−x^5 )  not by multiplying all terms  but possibility of general term of(1−x)^(−1)   C_r ^(n+r−1)  ... Or getting the coeffitient  of x^(60)   as 71 in a easy way???  Please..sir...
$${Sir}…{I}\:{mean}\:{to}\:{say}\:.. \\ $$$${coeffitient}\:{of}\:{x}^{\mathrm{60}} \:{in}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)\left(\mathrm{1}−{x}^{\mathrm{5}} \right) \\ $$$${not}\:{by}\:{multiplying}\:{all}\:{terms} \\ $$$${but}\:{possibility}\:{of}\:{general}\:{term}\:{of}\left(\mathrm{1}−{x}\right)^{−\mathrm{1}} \\ $$$${C}_{{r}} ^{{n}+{r}−\mathrm{1}} \:…\:{Or}\:{getting}\:{the}\:{coeffitient} \\ $$$${of}\:{x}^{\mathrm{60}} \:\:{as}\:\mathrm{71}\:{in}\:{a}\:{easy}\:{way}??? \\ $$$${Please}..{sir}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 02/Feb/22
in fact there is no easy way to   determine the coefficient of a   general term.
$${in}\:{fact}\:{there}\:{is}\:{no}\:{easy}\:{way}\:{to}\: \\ $$$${determine}\:{the}\:{coefficient}\:{of}\:{a}\: \\ $$$${general}\:{term}. \\ $$

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