How to calculate this congruency?
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Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.
modular-arithmetic congruence-relations
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Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.
modular-arithmetic congruence-relations
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.
modular-arithmetic congruence-relations
Let's say I have this linear congruency: $2x + 1234 = 7 mod 17$.
Without "$+1234$" I would've used the following formulas: $x = x_0 + k(dfracmgcd(a, m))$, whereas $ax_0 + my_0 = b$. But I don't know what to do with $1234$? Thank you in advance.
modular-arithmetic congruence-relations
edited Aug 17 at 7:53
Jack M
17.2k33473
17.2k33473
asked Aug 17 at 7:49
ponikoli
31
31
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2 Answers
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I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
$$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
$$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$
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No worries, just solve it like you would solve a normal equation.
$$2x+1234=7$$
$$2x=7-1234$$
and then apply your usual method.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
$$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
$$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$
add a comment |Â
up vote
0
down vote
accepted
I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
$$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
$$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
$$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
$$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$
I would calculate everything mod. $17$: $;1234equiv 10mod 17$, so the equation becomes
$$2x+10equiv 7mod17iff 2xequiv-3equiv14mod 17. $$
Now as $2$ is a unit mod. $17$, we may apply the cancellation law:
$$2xequiv 14=2cdot 7mod17iff xequiv 7mod 17.$$
answered Aug 17 at 8:07
Bernard
111k635103
111k635103
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add a comment |Â
up vote
1
down vote
No worries, just solve it like you would solve a normal equation.
$$2x+1234=7$$
$$2x=7-1234$$
and then apply your usual method.
add a comment |Â
up vote
1
down vote
No worries, just solve it like you would solve a normal equation.
$$2x+1234=7$$
$$2x=7-1234$$
and then apply your usual method.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No worries, just solve it like you would solve a normal equation.
$$2x+1234=7$$
$$2x=7-1234$$
and then apply your usual method.
No worries, just solve it like you would solve a normal equation.
$$2x+1234=7$$
$$2x=7-1234$$
and then apply your usual method.
answered Aug 17 at 7:52
Jack M
17.2k33473
17.2k33473
add a comment |Â
add a comment |Â
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