solving $151x − 294 equiv 44 pmod7$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I've solved normal congruence equations like $ax equiv b pmodm$ but now I am trying to solve $ 1 5 1 x − 294 equiv44pmod7$. How do I solve this one? Can I just add $294$ to both sides and solve as normal using Euclidean algorithm?
I read the answer by quanta in this question but it is still not clear to me.
Can anyone please elaborate more on this?
Thanks for any help.
modular-arithmetic
edited Aug 26 at 20:28
Davide Morgante
2,523623
asked Aug 26 at 19:28
Wyuw
1388
add a comment |Â
up vote
0
down vote
favorite
I've solved normal congruence equations like $ax equiv b pmodm$ but now I am trying to solve $ 1 5 1 x − 294 equiv44pmod7$. How do I solve this one? Can I just add $294$ to both sides and solve as normal using Euclidean algorithm?
I read the answer by quanta in this question but it is still not clear to me.
Can anyone please elaborate more on this?
Thanks for any help.
modular-arithmetic
edited Aug 26 at 20:28
Davide Morgante
2,523623
asked Aug 26 at 19:28
Wyuw
1388
2
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've solved normal congruence equations like $ax equiv b pmodm$ but now I am trying to solve $ 1 5 1 x − 294 equiv44pmod7$. How do I solve this one? Can I just add $294$ to both sides and solve as normal using Euclidean algorithm?
I read the answer by quanta in this question but it is still not clear to me.
Can anyone please elaborate more on this?
Thanks for any help.
modular-arithmetic
edited Aug 26 at 20:28
Davide Morgante
2,523623
asked Aug 26 at 19:28
Wyuw
1388
I've solved normal congruence equations like $ax equiv b pmodm$ but now I am trying to solve $ 1 5 1 x − 294 equiv44pmod7$. How do I solve this one? Can I just add $294$ to both sides and solve as normal using Euclidean algorithm?
I read the answer by quanta in this question but it is still not clear to me.
Can anyone please elaborate more on this?
Thanks for any help.
modular-arithmetic
edited Aug 26 at 20:28
Davide Morgante
2,523623
asked Aug 26 at 19:28
Wyuw
1388
edited Aug 26 at 20:28
Davide Morgante
2,523623
edited Aug 26 at 20:28
Davide Morgante
2,523623
edited Aug 26 at 20:28
Davide Morgante
2,523623
2,523623
asked Aug 26 at 19:28
Wyuw
1388
asked Aug 26 at 19:28
Wyuw
1388
asked Aug 26 at 19:28
Wyuw
1388
1388
2
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33
add a comment |Â
2
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33
2
2
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We have that
$$151x − 294 equiv 44 pmod7 iff 4x-0equiv 2 pmod7 iff 2xequiv 1 pmod7$$
and by Euclidean algorithm we can find
$$4cdot 2-1cdot 7=1$$
therefore $4$ is the inverse of $2 pmod 7$ and we find
$$4cdot 2xequiv 4cdot 1 pmod7 implies xequiv 4 pmod7$$
answered Aug 26 at 19:31
gimusi
70.3k73786
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
add a comment |Â
up vote
2
down vote
The first right thing to do is to simplify all coefficients and write them in the range of $0,1,2,3,4,5,6$, using Euclidian division.
$$151=21times7 +4 quad;quad 294 = 42times 7 quad;quad 44=7times 6 + 2$$
So the equation is the same as
$$4xequiv 2 pmod7$$
All we need to do is find the inverse of $4pmod7$. After a few tries, we find that it is $2$ since $2times 4 equiv 8 equiv 1 pmod7$. Hence, we get
$$x equiv 2times 2 equiv 4pmod7$$
as the answer of your equation.
answered Aug 26 at 19:33
Suzet
2,421527
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We have that
$$151x − 294 equiv 44 pmod7 iff 4x-0equiv 2 pmod7 iff 2xequiv 1 pmod7$$
and by Euclidean algorithm we can find
$$4cdot 2-1cdot 7=1$$
therefore $4$ is the inverse of $2 pmod 7$ and we find
$$4cdot 2xequiv 4cdot 1 pmod7 implies xequiv 4 pmod7$$
answered Aug 26 at 19:31
gimusi
70.3k73786
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
add a comment |Â
up vote
2
down vote
accepted
We have that
$$151x − 294 equiv 44 pmod7 iff 4x-0equiv 2 pmod7 iff 2xequiv 1 pmod7$$
and by Euclidean algorithm we can find
$$4cdot 2-1cdot 7=1$$
therefore $4$ is the inverse of $2 pmod 7$ and we find
$$4cdot 2xequiv 4cdot 1 pmod7 implies xequiv 4 pmod7$$
answered Aug 26 at 19:31
gimusi
70.3k73786
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We have that
$$151x − 294 equiv 44 pmod7 iff 4x-0equiv 2 pmod7 iff 2xequiv 1 pmod7$$
and by Euclidean algorithm we can find
$$4cdot 2-1cdot 7=1$$
therefore $4$ is the inverse of $2 pmod 7$ and we find
$$4cdot 2xequiv 4cdot 1 pmod7 implies xequiv 4 pmod7$$
answered Aug 26 at 19:31
gimusi
70.3k73786
We have that
$$151x − 294 equiv 44 pmod7 iff 4x-0equiv 2 pmod7 iff 2xequiv 1 pmod7$$
and by Euclidean algorithm we can find
$$4cdot 2-1cdot 7=1$$
therefore $4$ is the inverse of $2 pmod 7$ and we find
$$4cdot 2xequiv 4cdot 1 pmod7 implies xequiv 4 pmod7$$
answered Aug 26 at 19:31
gimusi
70.3k73786
edited Aug 26 at 19:36
answered Aug 26 at 19:31
gimusi
70.3k73786
answered Aug 26 at 19:31
gimusi
70.3k73786
answered Aug 26 at 19:31
gimusi
70.3k73786
70.3k73786
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
add a comment |Â
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning?
– Wyuw
Aug 26 at 19:35
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
When we work $mod m$ we can reduce all the terms to the remainder $0le r<m$.
– gimusi
Aug 26 at 19:37
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
Could we also say that $x = -3$?
– Wyuw
Aug 27 at 0:03
1
1
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 pmod 7$ but also $-3 pmod 7$ is a correct result.
– gimusi
Aug 27 at 0:50
add a comment |Â
up vote
2
down vote
The first right thing to do is to simplify all coefficients and write them in the range of $0,1,2,3,4,5,6$, using Euclidian division.
$$151=21times7 +4 quad;quad 294 = 42times 7 quad;quad 44=7times 6 + 2$$
So the equation is the same as
$$4xequiv 2 pmod7$$
All we need to do is find the inverse of $4pmod7$. After a few tries, we find that it is $2$ since $2times 4 equiv 8 equiv 1 pmod7$. Hence, we get
$$x equiv 2times 2 equiv 4pmod7$$
as the answer of your equation.
answered Aug 26 at 19:33
Suzet
2,421527
add a comment |Â
up vote
2
down vote
The first right thing to do is to simplify all coefficients and write them in the range of $0,1,2,3,4,5,6$, using Euclidian division.
$$151=21times7 +4 quad;quad 294 = 42times 7 quad;quad 44=7times 6 + 2$$
So the equation is the same as
$$4xequiv 2 pmod7$$
All we need to do is find the inverse of $4pmod7$. After a few tries, we find that it is $2$ since $2times 4 equiv 8 equiv 1 pmod7$. Hence, we get
$$x equiv 2times 2 equiv 4pmod7$$
as the answer of your equation.
answered Aug 26 at 19:33
Suzet
2,421527
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The first right thing to do is to simplify all coefficients and write them in the range of $0,1,2,3,4,5,6$, using Euclidian division.
$$151=21times7 +4 quad;quad 294 = 42times 7 quad;quad 44=7times 6 + 2$$
So the equation is the same as
$$4xequiv 2 pmod7$$
All we need to do is find the inverse of $4pmod7$. After a few tries, we find that it is $2$ since $2times 4 equiv 8 equiv 1 pmod7$. Hence, we get
$$x equiv 2times 2 equiv 4pmod7$$
as the answer of your equation.
answered Aug 26 at 19:33
Suzet
2,421527
The first right thing to do is to simplify all coefficients and write them in the range of $0,1,2,3,4,5,6$, using Euclidian division.
$$151=21times7 +4 quad;quad 294 = 42times 7 quad;quad 44=7times 6 + 2$$
So the equation is the same as
$$4xequiv 2 pmod7$$
All we need to do is find the inverse of $4pmod7$. After a few tries, we find that it is $2$ since $2times 4 equiv 8 equiv 1 pmod7$. Hence, we get
$$x equiv 2times 2 equiv 4pmod7$$
as the answer of your equation.
answered Aug 26 at 19:33
Suzet
2,421527
answered Aug 26 at 19:33
Suzet
2,421527
answered Aug 26 at 19:33
Suzet
2,421527
answered Aug 26 at 19:33
Suzet
2,421527
2,421527
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2895415%2fsolving-151x-%25e2%2588%2592-294-equiv-44-pmod7%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time.
– Matt
Aug 26 at 19:33