Existence of integer solution to 63x+70y+15z=2010
$begingroup$
I have an equation $63x+70y+15z=2010$. The question asks me to conclude whether it has an integral solution or not? Any help on how to proceed?
number-theory diophantine-equations
edited Nov 1 '14 at 9:07
200_success
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
$endgroup$
add a comment |
$begingroup$
I have an equation $63x+70y+15z=2010$. The question asks me to conclude whether it has an integral solution or not? Any help on how to proceed?
number-theory diophantine-equations
edited Nov 1 '14 at 9:07
200_success
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
$endgroup$
add a comment |
$begingroup$
I have an equation $63x+70y+15z=2010$. The question asks me to conclude whether it has an integral solution or not? Any help on how to proceed?
number-theory diophantine-equations
edited Nov 1 '14 at 9:07
200_success
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
$endgroup$
I have an equation $63x+70y+15z=2010$. The question asks me to conclude whether it has an integral solution or not? Any help on how to proceed?
number-theory diophantine-equations
number-theory diophantine-equations
edited Nov 1 '14 at 9:07
200_success
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
edited Nov 1 '14 at 9:07
200_success
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
edited Nov 1 '14 at 9:07
200_success
669515
edited Nov 1 '14 at 9:07
200_success
669515
edited Nov 1 '14 at 9:07
200_success
669515
669515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.7k32495
17.7k32495
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Solving the equation certainly gives an affirmative answer.
$$
z=frac{-21x}{5}+frac{-14y}{3}+134,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
$endgroup$
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1000790%2fexistence-of-integer-solution-to-63x70y15z-2010%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solving the equation certainly gives an affirmative answer.
$$
z=frac{-21x}{5}+frac{-14y}{3}+134,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
$endgroup$
add a comment |
$begingroup$
Solving the equation certainly gives an affirmative answer.
$$
z=frac{-21x}{5}+frac{-14y}{3}+134,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
$endgroup$
add a comment |
$begingroup$
Solving the equation certainly gives an affirmative answer.
$$
z=frac{-21x}{5}+frac{-14y}{3}+134,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
$endgroup$
Solving the equation certainly gives an affirmative answer.
$$
z=frac{-21x}{5}+frac{-14y}{3}+134,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
edited Nov 1 '14 at 3:56
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
11.9k52751
add a comment |
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
$endgroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
452k36423808
452k36423808
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
$begingroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
$endgroup$
add a comment |
$begingroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
$endgroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
319414
319414
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1000790%2fexistence-of-integer-solution-to-63x70y15z-2010%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
