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
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
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$begingroup$
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
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
$endgroup$
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
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
$endgroup$
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
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
edited Nov 1 '14 at 9:07
200_success
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
edited Nov 1 '14 at 9:07
200_success
666515
edited Nov 1 '14 at 9:07
200_success
666515
edited Nov 1 '14 at 9:07
200_success
666515
666515
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
asked Nov 1 '14 at 3:26
LearnmoreLearnmore
17.8k325101
17.8k325101
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
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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
454k36430817
$endgroup$
$begingroup$
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 $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
Taking the equation mod $3,5$ and $7$, we get: $x=5p, y=3q, z=7r+1$ where $p,q,rinmathbb Z$. Putting these back into the original equation, we get $3p+2q+r=19$ which is much easier to solve. $(0,0,134),(30,0,8),(0,27,8)$ etc. are all solutions.
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
370415
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Solving the equation certainly gives an affirmative answer.
$$
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,
$$
hence
the solution in integers is $x=5n$, $y=3m$, $z=-21n-14m+134$, where $n$, $minmathbb{Z}$.
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
edited Nov 1 '14 at 3:56
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
answered Nov 1 '14 at 3:50
Przemysław ScherwentkePrzemysław Scherwentke
11.9k52751
11.9k52751
add a comment |
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
$endgroup$
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$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
454k36430817
$endgroup$
$begingroup$
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 $$b$ divides $c$. This can be extended by induction to several variables.
$endgroup$
– André Nicolas
Nov 1 '14 at 3:36
add a comment |
$begingroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
$endgroup$
The gcd of the coefficients is $1$. In particular, it divides $2010$, so there is an integer solution.
If we want to look more closely, we can see that for example we can take $x=0$. The equation $70y+15z=2010$ has a solution, since the gcd of $70$ and $15$ divides $2010$.
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
answered Nov 1 '14 at 3:28
André NicolasAndré Nicolas
454k36430817
454k36430817
$begingroup$
How did you get such a condition.Can you share any articles in this regard or any books.Looking forward to you
$endgroup$
– Learnmore
Nov 1 '14 at 3:33
1
$begingroup$
It is a standard early result in number theory (almost any book) that the congruence $ax+by=c$ has an integer olution if and only if the gcd of $a$ and $$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
370415
$endgroup$
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
370415
$endgroup$
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
370415
$endgroup$
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
370415
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
370415
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
370415
answered Dec 7 '18 at 9:28
RhaldrynRhaldryn
370415
370415
add a comment |
add a comment |
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