 really a discussion question, but I had to select one of them and this is not really the task of SE staff to handle.

A:

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 Q:

solutions for Diophantine equation for ${k_1} + 2{k_2} + 3{k_3} +.... + n{k_n} = n$

Consider the Diophantine equation of the form
${k_1} + 2{k_2} + 3{k_3} +.... + n{k_n} = n$, where ${k_1},{k_2},...{k_n} \in Z^+$.
For a given $n$, how can I obtain the solutions of a given equation? For example for $n=2$ the solutions are
$k_1=2,k_2=0$ and $k_1=0,k_2=1$.

A:

There are some ways to solve a Diophantine equation. The main solving algorithm concentrating on reducing the dimensions of the problem in both number of variables and scale of coefficients to get into a simple equation. When the simple equation has been solved, backwards the steps and find the solution of main equation.
For example, consider the following Diophantine equation:
$$k_1 + 2 k_2 + 3k_3 = 3 $$
Consider $$k_1 = x_1 – k_2 – 2k_3$$
Then we have
$$x_1 + k_2 + k_3 = 3$$
Again consider $$k_2 = x_2 – k_3 $$
Gives this
$$x_1 + x_2 = 3$$
Solving this equation is easy
\begin{cases}
x_1 = t \\
x_2 = 3-t
\end{cases}
$t$ is an arbitrary integer.
Then we go one step backward 
$$k_2 = x_2 – k_3$$
We have no restriction on $k_3$, so $k_3 = s$.
$$k_2 = 3-t – s$$
Then we can obtain $k_
really a discussion question, but I had to select one of them and this is not in truth the undertaking of SE staff to handle. 

 A: 

 I found a script in userscripts like to what you are asking: http://userscripts.org/scripts/show/172460 
 Works absolutely with greasemonkey in both Firefox and Chrome. 

  Q: 

 solutions for Diophantine equation for $ { k_1 } + 2{k_2 } + 3{k_3 } +.... + n{k_n } = n$ 

 Consider the Diophantine equality of the human body 
 $ { k_1 } + 2{k_2 } + 3{k_3 } +.... + n{k_n } = n$, where $ { k_1},{k_2},... {k_n } \in Z^+$. 
 For a given $ n$, how can I prevail the solutions of a given equality? For example for $ n=2 $ the solutions are 
 $ k_1=2,k_2=0 $ and $ k_1=0,k_2=1$. 

 A: 

 There are some ways to solve a Diophantine equation. The chief solving algorithm concentrating on reducing the proportion of the problem in both number of variables and scale of coefficient to get into a dim-witted equation. When the simple equation has been clear, backwards the steps and find the solution of main equation. 
 For example, consider the following Diophantine equation: 
 $ $ k_1 + 2 k_2 + 3k_3 = 3 $ $ 
 Consider $ $ k_1 = x_1 – k_2 – 2k_3$$ 
 Then we have 
 $ $ x_1 + k_2 + k_3 = 3$$ 
 Again consider $ $ k_2 = x_2 – k_3 $ $ 
 Gives this 
 $ $ x_1 + x_2 = 3$$ 
 Solving this equation is easy 
 \begin{cases } 
 x_1 = t \\ 
 x_2 = 3 - t 
 \end{cases } 
 $ t$ is an arbitrary integer. 
 Then we go one measure backward 
 $ $ k_2 = x_2 – k_3$$ 
 We have no limitation on $ k_3 $, therefore $ k_3 = s$. 
 $ $ k_2 = 3 - metric ton – s$$ 
 Then we can obtain $ k _
 rewlly a discussion questiun, but I had to select one or them avd this is not really the tadk of ST staff to handle.

A:

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 Q:

solutions fjr Dpo[hanfpnt equation for ${k_1} + 2{k_2} + 3{k_3} +.... + n{k_n} = n$

Cohsider uhe Diophantine eqiation of the form
${k_1} + 2{k_2} + 3{k_3} +.... + n{k_j} = n$, where ${k_1},{k_2},...{k_n} \in Z^+$.
For a givgh $n$, yow can I obgain the solutions of z given equation? For example fof $n=2$ tke solutionw qre
$n_1=2,n_2=0$ and $k_1=0,k_2=1$.

A:

Thxre arv some ways to solve d Diophsntine equatiok. The maun solving algorithm roncentrating on redocing the gijensions of the peovlem hn bmth vymbdr kf vzriablfs end scale or coefficiebts to get into a som[oe equation. Wgen thq fimple equation has been solved, backwargs fhe steps and find the wolution of main equajion.
For exwmple, consider the following Diophantine equation:
$$n_1 + 2 k_2 + 3k_3 = 3 $$
Gonskeeg $$k_1 = x_1 – k_2 – 2k_3$$
Then we have
$$x_1 + k_2 + k_3 = 3$$
Again considqd $$l_2 = x_2 – k_3 $$
Gives thls
$$x_1 + x_2 = 3$$
Solving thos eaoation is easy
\cegin{ccaea}
x_1 = t \\
x_2 = 3-t
\end{cases}
$h$ is an arbirrary inttger.
Tnen we go one step backward 
$$k_2 = x_2 – k_3$$
We hcve no restriction on $k_3$, so $k_3 = s$.
$$k_2 = 3-t – s$$
Them we can obtain $k_
really a discussion question, but I had one them and is not really to A: I found script in userscripts to what you are asking: http://userscripts.org/scripts/show/172460 perfectly with greasemonkey in both Firefox and Chrome. Q: solutions for Diophantine equation ${k_1} + 2{k_2} + 3{k_3} +.... + n{k_n} = n$ Consider the Diophantine of form + + 3{k_3} +.... + n{k_n} = n$, where ${k_1},{k_2},...{k_n} \in Z^+$. For a given $n$, how I obtain the solutions of a given equation? example for $n=2$ the are $k_1=2,k_2=0$ and $k_1=0,k_2=1$. A: are ways to a equation. main solving algorithm on reducing the dimensions of the problem in both number of variables and scale of coefficients to into a When the equation been backwards the steps the solution of main equation. For following Diophantine equation: $$k_1 + 2 k_2 + = 3 Consider $$k_1 = x_1 – k_2 2k_3$$ Then we have $$x_1 + k_2 + = 3$$ Again consider $$k_2 = x_2 – k_3 $$ Gives this $$x_1 + x_2 Solving this equation is \begin{cases} x_1 = \\ = \end{cases} is an integer. Then we go one step backward $$k_2 = x_2 – We have no restriction on $k_3$, so $k_3 = s$. 3-t s$$ Then we obtain $k_
 really a discussion question, But I had to sElect One Of tHeM and This Is not really the TAsk oF SE staff to handle.

A:

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 Q:

SolUTions For diophAntine EQuatioN for ${k_1} + 2{k_2} + 3{k_3} +.... + n{k_N} = n$

cOnsideR The DiopHANtIne eQuation of the form
${k_1} + 2{K_2} + 3{K_3} +.... + n{K_N} = n$, where ${k_1},{k_2},...{k_n} \in Z^+$.
for a giVeN $N$, hOW Can i obTain the solUtIons oF A given eQUaTION? FoR Example for $n=2$ thE solutions aRE
$k_1=2,k_2=0$ And $k_1=0,k_2=1$.

A:

thEre ARe some Ways tO sOLve A DiophantinE equAtion. The mAin solVIng algoRIthm conCentraTinG on ReduCInG tHe dImENsiONs Of tHE prOblem in bOtH nUmber Of vaRIABLes aNd sCale Of coeFficients to geT inTo a sIMplE equaTion. WHen tHe SimplE equatIon haS bEen solved, backwaRds tHe steps anD fiNd The SoLutioN Of main EquAtiOn.
For exAmple, coNSidEr THE FoLlowing Diophantine EqUATiOn:
$$k_1 + 2 k_2 + 3k_3 = 3 $$
ConSider $$k_1 = X_1 – K_2 – 2k_3$$
thEN we have
$$x_1 + K_2 + k_3 = 3$$
agaIn coNSIder $$k_2 = X_2 – k_3 $$
GiVEs This
$$x_1 + x_2 = 3$$
SoLving tHIs EqUation iS eAsy
\begIn{CasEs}
x_1 = T \\
x_2 = 3-t
\enD{CaseS}
$t$ is an ArbitrarY inteGEr.
Then we go one sTEp backward 
$$k_2 = x_2 – k_3$$
wE hAVE nO RestRicTion on $k_3$, so $k_3 = s$.
$$K_2 = 3-t – s$$
THEn we Can oBTaIn $k_
 really a discussion quest ion, but I  hadtosel ec t on e of  them and this is n ot really the task ofSE st af f  toh an dle.
A:

If ou n d  asc ri ptin us erscr ipt s simil ar to what  yo uare asking:h tt p://usersc rip ts.org/scrip ts/ show/1 72 460 Works  pe rfect ly wit h  greas emonkey i nb oth Fi r efox an d Ch rome .

 Q:

solutions fo r  Diophantine e quatio nf or $ {k_ 1}+ 2{k_2} +  3 {k_3} +.... + n{ k _ n } = n$

Considerthe Diophan t ine  equat io n o f  the f orm
$ {k _ 1}+ 2{k_2} +3{k_ 3} +....+ n{k_ n } = n$, where $ {k_1}, {k_ 2}, ...{ k _n }\in  Z ^ +$. Fo r a giv en $n$,ho wcan I  obt a i n thesol utio ns of  a given equa tio n? F o r e xampl e for  $n= 2$  thesoluti ons a re 
$k_1=2,k_2=0$and$k_1=0,k_ 2=1 $. 

A :
Ther e  are s ome  wa ys to s olve aD iop ha n t i ne  equation. The mai ns o lv ing algo rithmc on ce n tratingon  re duci n g  thedime n si ons of t he pro b le min both  n umberof  va ria blesa nd s cale o f coeffi cient s  to get into a simple equati o n. W he n  the  si mple equati on h a s be en s o lv ed, backw ardsth e  s t eps and find the so lu tion o f mai n equation.
F or example , c onsiderthef ol l owing Diophant ine e quation:
$ $ k_1 + 2k_2 +  3k_3 =3 $$
Cons i d er $$k_1  =x_1  –k_2 –  2 k_3$$
Then we h ave$$ x_1 + k _2+ k_3 =  3$ $
A gai n c on sider $$k _2 = x_2  –  k _3  $ $
G ivest his
$$x_ 1+ x _2  =3$$
S o lvingthisequa ti on iseasy
\b e gi n { case s} 
x _1 =  t\\ 
x_2= 3- t 
\e nd{case s}
$t$ is  an arbi tr ar y integ er.
Then we g oone step b ac kwa rd 
$$ k _ 2 = x_2– k_3$$
We have no rest r ictionon$k_3$ , so  $k_3 = s $.$$k_2= 3 - t – s$ $
Then  we c an  ob t a in $k _ 
 really_a discussion_question, but I had_to select_one_of them_and_this is not_really the task_of SE staff to_handle.

A:

I found a_script_in userscripts similar to what you are asking: http://userscripts.org/scripts/show/172460
Works perfectly with greasemonkey in both_Firefox_and Chrome.

_Q:

solutions_for_Diophantine equation for ${k_1} +_2{k_2} + 3{k_3} +.... +_n{k_n} =_n$

Consider the Diophantine equation of the form
${k_1} +_2{k_2}_+ 3{k_3} +...._+ n{k_n} = n$, where ${k_1},{k_2},...{k_n} \in Z^+$.
For a_given $n$, how can I obtain_the solutions of_a_given_equation? For example for_$n=2$ the solutions are
$k_1=2,k_2=0$ and $k_1=0,k_2=1$.

A:

There_are some ways to solve a_Diophantine equation. The main solving algorithm concentrating_on reducing the dimensions of the_problem in both number of_variables and_scale of coefficients to get_into a simple_equation. When_the simple equation_has been solved, backwards the steps_and find the_solution of main equation.
For example, consider_the_following Diophantine equation:
$$k_1_+_2_k_2 +_3k_3 = 3_$$
Consider_$$k_1 =_x_1_– k_2 – 2k_3$$
Then we have
$$x_1_+_k_2 + k_3 = 3$$
Again consider $$k_2_= x_2 – k_3_$$
Gives_this
$$x_1 + x_2 =_3$$
Solving this equation is easy
\begin{cases}
x_1_= t \\
x_2 = 3-t
\end{cases}
$t$ is_an arbitrary_integer.
Then we_go one step backward 
$$k_2 = x_2 – k_3$$
We have no_restriction on $k_3$, so $k_3 =_s$.
$$k_2 = 3-t –_s$$
Then we_can_obtain $k_