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Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a...
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I was checking the following number theory excercise:
Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.
I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.
Any help will be really appreciated.
number-theory modular-arithmetic
asked Sep 26 '18 at 5:10
mrazmraz
39118
$endgroup$
add a comment |
$begingroup$
I was checking the following number theory excercise:
Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.
I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.
Any help will be really appreciated.
number-theory modular-arithmetic
asked Sep 26 '18 at 5:10
mrazmraz
39118
$endgroup$
add a comment |
$begingroup$
I was checking the following number theory excercise:
Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.
I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.
Any help will be really appreciated.
number-theory modular-arithmetic
asked Sep 26 '18 at 5:10
mrazmraz
39118
$endgroup$
I was checking the following number theory excercise:
Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.
I made the $gcd$ between $12$ and $36$ and my conclusion was that the numbers are $12$ and $24$ but I think is not correct.
Any help will be really appreciated.
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Sep 26 '18 at 5:10
mrazmraz
39118
asked Sep 26 '18 at 5:10
mrazmraz
39118
edited Jan 21 at 5:15
mraz
asked Sep 26 '18 at 5:10
mrazmraz
39118
asked Sep 26 '18 at 5:10
mrazmraz
39118
asked Sep 26 '18 at 5:10
mrazmraz
39118
39118
add a comment |
add a comment |
2 Answers
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$begingroup$
Take a look at the expression:
$c equiv a pmod b$
If $gcd(a,b)=d rightarrow gcd(c,d) = d$
Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$
Then $gcd(24, 36) = 2^2cdot 3 = 12$.
So the remainder has the form $12y$. How many $y's$ are there until $35$?
Right: $12cdot 0, 12cdot1$ and $12cdot2=24$
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
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$begingroup$
$n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
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2 Answers
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2 Answers
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$begingroup$
Take a look at the expression:
$c equiv a pmod b$
If $gcd(a,b)=d rightarrow gcd(c,d) = d$
Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$
Then $gcd(24, 36) = 2^2cdot 3 = 12$.
So the remainder has the form $12y$. How many $y's$ are there until $35$?
Right: $12cdot 0, 12cdot1$ and $12cdot2=24$
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
$endgroup$
add a comment |
$begingroup$
Take a look at the expression:
$c equiv a pmod b$
If $gcd(a,b)=d rightarrow gcd(c,d) = d$
Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$
Then $gcd(24, 36) = 2^2cdot 3 = 12$.
So the remainder has the form $12y$. How many $y's$ are there until $35$?
Right: $12cdot 0, 12cdot1$ and $12cdot2=24$
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
$endgroup$
add a comment |
$begingroup$
Take a look at the expression:
$c equiv a pmod b$
If $gcd(a,b)=d rightarrow gcd(c,d) = d$
Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$
Then $gcd(24, 36) = 2^2cdot 3 = 12$.
So the remainder has the form $12y$. How many $y's$ are there until $35$?
Right: $12cdot 0, 12cdot1$ and $12cdot2=24$
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
$endgroup$
Take a look at the expression:
$c equiv a pmod b$
If $gcd(a,b)=d rightarrow gcd(c,d) = d$
Since $24 = 2^3cdot 3$ and $36 = 2^2cdot 3^2$
Then $gcd(24, 36) = 2^2cdot 3 = 12$.
So the remainder has the form $12y$. How many $y's$ are there until $35$?
Right: $12cdot 0, 12cdot1$ and $12cdot2=24$
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
answered Sep 26 '18 at 6:05
kub0xkub0x
1,7411921
1,7411921
add a comment |
add a comment |
$begingroup$
$n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
$n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
$endgroup$
add a comment |
$begingroup$
$n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
$endgroup$
$n$ has to be a multiple of $operatorname{gcd}(24,36)=12$. So you left out $0$.
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
answered Sep 26 '18 at 6:14
Chris CusterChris Custer
14.1k3827
14.1k3827
add a comment |
add a comment |
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