$17 | 29x + 18y implies 17 | 23x + 9y$
$begingroup$
So, basically the title.
Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?
Can you please explain what's the idea for doing these type of problems .. ?
elementary-number-theory divisibility
$endgroup$
add a comment |
$begingroup$
So, basically the title.
Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?
Can you please explain what's the idea for doing these type of problems .. ?
elementary-number-theory divisibility
$endgroup$
add a comment |
$begingroup$
So, basically the title.
Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?
Can you please explain what's the idea for doing these type of problems .. ?
elementary-number-theory divisibility
$endgroup$
So, basically the title.
Prove that $17 | 29x + 18y implies 17 | 23x + 9y$.
Plus, does then $17 | 11x + 8y$ ?
Can you please explain what's the idea for doing these type of problems .. ?
elementary-number-theory divisibility
elementary-number-theory divisibility
asked Dec 19 '18 at 12:13
user626177
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For the first part:
Let us first reduce it mod 17. Hence
$$29x+18y equiv 12x+y pmod{17}$$
$$23x+9y equiv 6x+9y pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
$endgroup$
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
|
show 5 more comments
$begingroup$
For the first part :
$17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$
For the second part :
$17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$
On the other hand :
$17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$
$endgroup$
add a comment |
$begingroup$
$left.begin{align}
!!bmod 17!: 0, equiv &, 18y + 29x\
equiv &, y+12x \
iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
end{align}right} $ If so then $ left{ begin{align}
&, 23x + 9,color{#c00}y\
equiv & , 6x + 9(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right} $ and $ left{ begin{align}
&, 11x + 8,color{#c00}y\
equiv &, 11x + 8(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right}$
$endgroup$
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
add a comment |
$begingroup$
Here is the ansatz:
If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.
This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.
This argument actually shows
$17 mid 29x + 18y iff 17 mid 23x + 9y$
Here is an explanation.
Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
|
show 5 more comments
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the first part:
Let us first reduce it mod 17. Hence
$$29x+18y equiv 12x+y pmod{17}$$
$$23x+9y equiv 6x+9y pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
$endgroup$
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
|
show 5 more comments
$begingroup$
For the first part:
Let us first reduce it mod 17. Hence
$$29x+18y equiv 12x+y pmod{17}$$
$$23x+9y equiv 6x+9y pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
$endgroup$
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
|
show 5 more comments
$begingroup$
For the first part:
Let us first reduce it mod 17. Hence
$$29x+18y equiv 12x+y pmod{17}$$
$$23x+9y equiv 6x+9y pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
$endgroup$
For the first part:
Let us first reduce it mod 17. Hence
$$29x+18y equiv 12x+y pmod{17}$$
$$23x+9y equiv 6x+9y pmod{17}$$
since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?
edited Dec 19 '18 at 17:17
answered Dec 19 '18 at 12:37
Maged SaeedMaged Saeed
8771417
8771417
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
|
show 5 more comments
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
1
1
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
$begingroup$
Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ ..
$endgroup$
– user626177
Dec 19 '18 at 12:43
1
1
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
$begingroup$
So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ?
$endgroup$
– user626177
Dec 19 '18 at 12:44
1
1
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
$begingroup$
@someone, it amounts to Gaussian elimination mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:45
1
1
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
$begingroup$
Got it, thought it was harder. Thanks, again :)
$endgroup$
– user626177
Dec 19 '18 at 12:52
2
2
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
$begingroup$
@someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer.
$endgroup$
– Bill Dubuque
Dec 19 '18 at 17:06
|
show 5 more comments
$begingroup$
For the first part :
$17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$
For the second part :
$17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$
On the other hand :
$17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$
$endgroup$
add a comment |
$begingroup$
For the first part :
$17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$
For the second part :
$17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$
On the other hand :
$17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$
$endgroup$
add a comment |
$begingroup$
For the first part :
$17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$
For the second part :
$17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$
On the other hand :
$17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$
$endgroup$
For the first part :
$17 mid 29x+18y Rightarrow 17 mid 17x+29x+18y Rightarrow 17 mid 46x+18y Rightarrow 17 mid 2cdot (23x+9y) Rightarrow 17 mid 23x+9y$
For the second part :
$17 mid 29x+18y Rightarrow 17 mid 17x+17y+12x+y Rightarrow 17 mid 12x+y$
On the other hand :
$17 mid 23x+9y Rightarrow 17 mid 12x+y+11x+8y Rightarrow 17 mid 11x+8y$
edited Dec 19 '18 at 13:01
answered Dec 19 '18 at 12:34
MatkoMatko
864
864
add a comment |
add a comment |
$begingroup$
$left.begin{align}
!!bmod 17!: 0, equiv &, 18y + 29x\
equiv &, y+12x \
iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
end{align}right} $ If so then $ left{ begin{align}
&, 23x + 9,color{#c00}y\
equiv & , 6x + 9(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right} $ and $ left{ begin{align}
&, 11x + 8,color{#c00}y\
equiv &, 11x + 8(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right}$
$endgroup$
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
add a comment |
$begingroup$
$left.begin{align}
!!bmod 17!: 0, equiv &, 18y + 29x\
equiv &, y+12x \
iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
end{align}right} $ If so then $ left{ begin{align}
&, 23x + 9,color{#c00}y\
equiv & , 6x + 9(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right} $ and $ left{ begin{align}
&, 11x + 8,color{#c00}y\
equiv &, 11x + 8(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right}$
$endgroup$
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
add a comment |
$begingroup$
$left.begin{align}
!!bmod 17!: 0, equiv &, 18y + 29x\
equiv &, y+12x \
iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
end{align}right} $ If so then $ left{ begin{align}
&, 23x + 9,color{#c00}y\
equiv & , 6x + 9(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right} $ and $ left{ begin{align}
&, 11x + 8,color{#c00}y\
equiv &, 11x + 8(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right}$
$endgroup$
$left.begin{align}
!!bmod 17!: 0, equiv &, 18y + 29x\
equiv &, y+12x \
iff color{#c00}y,equiv& {-}12xequivcolor{#c00}{5x}
end{align}right} $ If so then $ left{ begin{align}
&, 23x + 9,color{#c00}y\
equiv & , 6x + 9(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right} $ and $ left{ begin{align}
&, 11x + 8,color{#c00}y\
equiv &, 11x + 8(color{#c00}{5x})\
equiv &, 51x,equiv, 0
end{align}right}$
answered Dec 19 '18 at 17:01
Bill DubuqueBill Dubuque
212k29195648
212k29195648
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
add a comment |
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
2
2
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
$begingroup$
Clear and simple!
$endgroup$
– user626177
Dec 19 '18 at 17:08
add a comment |
$begingroup$
Here is the ansatz:
If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.
This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.
This argument actually shows
$17 mid 29x + 18y iff 17 mid 23x + 9y$
Here is an explanation.
Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
|
show 5 more comments
$begingroup$
Here is the ansatz:
If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.
This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.
This argument actually shows
$17 mid 29x + 18y iff 17 mid 23x + 9y$
Here is an explanation.
Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.
$endgroup$
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
|
show 5 more comments
$begingroup$
Here is the ansatz:
If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.
This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.
This argument actually shows
$17 mid 29x + 18y iff 17 mid 23x + 9y$
Here is an explanation.
Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.
$endgroup$
Here is the ansatz:
If $a(29x + 18y)equiv 23x + 9y bmod 17$ for some $a in mathbb Z$, then the result follows easily.
This has to work for all $x,y$ and so $29a equiv 23 bmod 17$, which gives $a equiv 9 bmod 17$. This $a$ also works for $18aequiv 9 bmod 17$.
This argument actually shows
$17 mid 29x + 18y iff 17 mid 23x + 9y$
Here is an explanation.
Consider the matrix $pmatrix{ 29 & 18 \ 23 & 9}$.
Its determinant is $-153$, which is $0$ mod $17$.
Therefore, its two rows are linearly dependent mod $17$.
In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)
For the second part, consider the matrix $pmatrix{ 29 & 18 \ 11 & 8}$.
edited Dec 19 '18 at 12:40
answered Dec 19 '18 at 12:18
lhflhf
166k10171396
166k10171396
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
|
show 5 more comments
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
$begingroup$
Adapted from math.stackexchange.com/a/2947271/589
$endgroup$
– lhf
Dec 19 '18 at 12:18
1
1
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) ..
$endgroup$
– user626177
Dec 19 '18 at 12:23
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
@someone, just compute $9(29x + 18y)$ mod $17$.
$endgroup$
– lhf
Dec 19 '18 at 12:25
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
Oh, but whats the trick with 9 ?
$endgroup$
– user626177
Dec 19 '18 at 12:26
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
$begingroup$
@someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$.
$endgroup$
– Shaun
Dec 19 '18 at 12:29
|
show 5 more comments
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