Mixing
changing remainder $dfrac37$ to $4$ in mod $5$
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This may be a very easy question to answer but I can't seem to find a good source online explaining it. I want to know how they isolated for $j$. We can write $$7j+6equiv 4 text mod 5$$
$$7jequiv 3 text mod 5$$
and by the law of division for modular arithmetic we are allowed to divide both sides by $7$ since $gcd(7,5)=1$, which gives
$$jequiv dfrac37 text mod 5$$
So what did they do to change the remainder from $dfrac37$ to $4$?
Thanks
elementary-number-theory modular-arithmetic proof-explanation
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
asked Jul 15 at 6:59
john fowles
1,093817
add a comment |Â
up vote
0
down vote
favorite
Problem
This may be a very easy question to answer but I can't seem to find a good source online explaining it. I want to know how they isolated for $j$. We can write $$7j+6equiv 4 text mod 5$$
$$7jequiv 3 text mod 5$$
and by the law of division for modular arithmetic we are allowed to divide both sides by $7$ since $gcd(7,5)=1$, which gives
$$jequiv dfrac37 text mod 5$$
So what did they do to change the remainder from $dfrac37$ to $4$?
Thanks
elementary-number-theory modular-arithmetic proof-explanation
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
asked Jul 15 at 6:59
john fowles
1,093817
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem
This may be a very easy question to answer but I can't seem to find a good source online explaining it. I want to know how they isolated for $j$. We can write $$7j+6equiv 4 text mod 5$$
$$7jequiv 3 text mod 5$$
and by the law of division for modular arithmetic we are allowed to divide both sides by $7$ since $gcd(7,5)=1$, which gives
$$jequiv dfrac37 text mod 5$$
So what did they do to change the remainder from $dfrac37$ to $4$?
Thanks
elementary-number-theory modular-arithmetic proof-explanation
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
asked Jul 15 at 6:59
john fowles
1,093817
Problem
This may be a very easy question to answer but I can't seem to find a good source online explaining it. I want to know how they isolated for $j$. We can write $$7j+6equiv 4 text mod 5$$
$$7jequiv 3 text mod 5$$
and by the law of division for modular arithmetic we are allowed to divide both sides by $7$ since $gcd(7,5)=1$, which gives
$$jequiv dfrac37 text mod 5$$
So what did they do to change the remainder from $dfrac37$ to $4$?
Thanks
elementary-number-theory modular-arithmetic proof-explanation
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
asked Jul 15 at 6:59
john fowles
1,093817
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
edited Jul 15 at 7:05
Siong Thye Goh
77.8k134796
77.8k134796
asked Jul 15 at 6:59
john fowles
1,093817
asked Jul 15 at 6:59
john fowles
1,093817
asked Jul 15 at 6:59
john fowles
1,093817
1,093817
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Note that $7j equiv 2j bmod 5$ so you want $2jequiv 3equiv 8$ (it is convenient that $5$ is odd so you know that adding once gives an even number). Then you can simply divide by $2$ since $5$ and $2$ have no common factor.
In general terms if you want to divide by $a$ modulo $p$ with $(a,p)=1$ you can find $b,c$ with $ab+pc=1$ (Euclid's algorithm will do this for you). This is the same as $abequiv 1 bmod p$
So to solve $$aqequiv r bmod p$$ multiply by $b$ to get $$aqb=(ab)qequiv qequiv rb bmod p$$
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
add a comment |Â
up vote
2
down vote
We have to figure out what is the inverse of $7$.
$$7cdot 3 equiv 21 equiv 1 pmod 5.$$
Hence $$7^-1 equiv 3 pmod 5$$
$$7^-1 cdot 3 equiv 3 cdot 3 equiv 9 equiv 5+4 equiv 4 pmod5.$$
In general, to find the inverse for modulo arithmetic, we use Euclidean algorithm.
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
add a comment |Â
up vote
0
down vote
$$3≡3+5(5)≡28bmod5$$
Now we can divide both sides by 7 and get $j≡4bmod5$.
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $7j equiv 2j bmod 5$ so you want $2jequiv 3equiv 8$ (it is convenient that $5$ is odd so you know that adding once gives an even number). Then you can simply divide by $2$ since $5$ and $2$ have no common factor.
In general terms if you want to divide by $a$ modulo $p$ with $(a,p)=1$ you can find $b,c$ with $ab+pc=1$ (Euclid's algorithm will do this for you). This is the same as $abequiv 1 bmod p$
So to solve $$aqequiv r bmod p$$ multiply by $b$ to get $$aqb=(ab)qequiv qequiv rb bmod p$$
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
add a comment |Â
up vote
1
down vote
accepted
Note that $7j equiv 2j bmod 5$ so you want $2jequiv 3equiv 8$ (it is convenient that $5$ is odd so you know that adding once gives an even number). Then you can simply divide by $2$ since $5$ and $2$ have no common factor.
In general terms if you want to divide by $a$ modulo $p$ with $(a,p)=1$ you can find $b,c$ with $ab+pc=1$ (Euclid's algorithm will do this for you). This is the same as $abequiv 1 bmod p$
So to solve $$aqequiv r bmod p$$ multiply by $b$ to get $$aqb=(ab)qequiv qequiv rb bmod p$$
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $7j equiv 2j bmod 5$ so you want $2jequiv 3equiv 8$ (it is convenient that $5$ is odd so you know that adding once gives an even number). Then you can simply divide by $2$ since $5$ and $2$ have no common factor.
In general terms if you want to divide by $a$ modulo $p$ with $(a,p)=1$ you can find $b,c$ with $ab+pc=1$ (Euclid's algorithm will do this for you). This is the same as $abequiv 1 bmod p$
So to solve $$aqequiv r bmod p$$ multiply by $b$ to get $$aqb=(ab)qequiv qequiv rb bmod p$$
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
Note that $7j equiv 2j bmod 5$ so you want $2jequiv 3equiv 8$ (it is convenient that $5$ is odd so you know that adding once gives an even number). Then you can simply divide by $2$ since $5$ and $2$ have no common factor.
In general terms if you want to divide by $a$ modulo $p$ with $(a,p)=1$ you can find $b,c$ with $ab+pc=1$ (Euclid's algorithm will do this for you). This is the same as $abequiv 1 bmod p$
So to solve $$aqequiv r bmod p$$ multiply by $b$ to get $$aqb=(ab)qequiv qequiv rb bmod p$$
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
answered Jul 15 at 7:36
Mark Bennet
76.7k773171
76.7k773171
add a comment |Â
add a comment |Â
up vote
2
down vote
We have to figure out what is the inverse of $7$.
$$7cdot 3 equiv 21 equiv 1 pmod 5.$$
Hence $$7^-1 equiv 3 pmod 5$$
$$7^-1 cdot 3 equiv 3 cdot 3 equiv 9 equiv 5+4 equiv 4 pmod5.$$
In general, to find the inverse for modulo arithmetic, we use Euclidean algorithm.
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
add a comment |Â
up vote
2
down vote
We have to figure out what is the inverse of $7$.
$$7cdot 3 equiv 21 equiv 1 pmod 5.$$
Hence $$7^-1 equiv 3 pmod 5$$
$$7^-1 cdot 3 equiv 3 cdot 3 equiv 9 equiv 5+4 equiv 4 pmod5.$$
In general, to find the inverse for modulo arithmetic, we use Euclidean algorithm.
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have to figure out what is the inverse of $7$.
$$7cdot 3 equiv 21 equiv 1 pmod 5.$$
Hence $$7^-1 equiv 3 pmod 5$$
$$7^-1 cdot 3 equiv 3 cdot 3 equiv 9 equiv 5+4 equiv 4 pmod5.$$
In general, to find the inverse for modulo arithmetic, we use Euclidean algorithm.
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
We have to figure out what is the inverse of $7$.
$$7cdot 3 equiv 21 equiv 1 pmod 5.$$
Hence $$7^-1 equiv 3 pmod 5$$
$$7^-1 cdot 3 equiv 3 cdot 3 equiv 9 equiv 5+4 equiv 4 pmod5.$$
In general, to find the inverse for modulo arithmetic, we use Euclidean algorithm.
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
answered Jul 15 at 7:03
Siong Thye Goh
77.8k134796
77.8k134796
add a comment |Â
add a comment |Â
up vote
0
down vote
$$3≡3+5(5)≡28bmod5$$
Now we can divide both sides by 7 and get $j≡4bmod5$.
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
add a comment |Â
up vote
0
down vote
$$3≡3+5(5)≡28bmod5$$
Now we can divide both sides by 7 and get $j≡4bmod5$.
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$3≡3+5(5)≡28bmod5$$
Now we can divide both sides by 7 and get $j≡4bmod5$.
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
$$3≡3+5(5)≡28bmod5$$
Now we can divide both sides by 7 and get $j≡4bmod5$.
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
edited Jul 17 at 2:42
Parcly Taxel
33.6k136588
33.6k136588
answered Jul 17 at 2:04
judith Khan
22117
answered Jul 17 at 2:04
judith Khan
22117
answered Jul 17 at 2:04
judith Khan
22117
22117
add a comment |Â
add a comment |Â
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