Solve $3x \equiv 17 \pmod{2014}$

Question

Solve

$$3x \equiv 17 \pmod{2014}$$

So first I suppose $3^{-1} \pmod{2014}$

$2014 = 671(3) + 1 \implies 1 = 2014 - 671(3)$

But this gives $3^{-1} = 1 \pmod{2014}$

which is incorrect?

take a look at extended euclidian algorithm to compute multiplicative inverse — shcolf, Jan 25 '15 at 11:42
You have the multiplicative inverse the wrong way around... you need to look for $3x\equiv 1 \bmod 2014$ not $3x\equiv -1 \bmod 2014$ — Joffan, Jan 25 '15 at 11:43

score 2 · Accepted Answer · answered Jan 25 '15 at 11:52

2

$$3x=17\implies x=3^{-1}\cdot17 $$

Now, you can see at once that $\;2014=1\pmod 3\;$ , so

$$2\cdot2014+1=4029\implies 4029\div3=1343\implies 3^{-1}=1343\pmod{2014}$$

and thus

$$x=1343\cdot17\pmod{2014}=677\pmod{2014}$$

answered Jan 25 '15 at 11:52

Timbuc

34,191

(+1) Sir! But why are you doing $2\cdot 2014 + 1$ ?? – Amad27 Jan 25 '15 at 11:54
No but I got, $3^{-1} = 1 \pmod{2014}$? ??? – Amad27 Jan 25 '15 at 11:55
Because since $;2014=1\pmod 3;$ , then $;2\cdot2014=2\pmod 3;$ , and then $;2\cdot2014+1=0\pmod 3;$ – Timbuc Jan 25 '15 at 11:59
@Amad27 That is obviously wrong, right?! $;x^{-1}=1\pmod m\iff x=1\pmod m;$ , for any integer $;m;$ and coprime $;x;$ ! – Timbuc Jan 25 '15 at 12:00
Wait, so let me ask. What is $3^{-1} \pmod{2014}$ ? – Amad27 Jan 25 '15 at 12:00
@Amad27 $;3^{-1}=t\pmod{2014}\iff 3t=1\pmod{2014};$ . As shown above, we get $;3^{-1}=1343\pmod{2104};$ . – Timbuc Jan 25 '15 at 12:02
Here it is:
From Euclids, algorithm I get.

$$2014 = 3(671) + 1$$

Then $\implies 1 = 2014 - 3(671)$

So $3^{-1} \pmod{2014} = -671$

Then

$$3x \equiv 17 \pmod{2014}$$

$$x \equiv (17)(-671) \pmod{2014} \equiv -11407 \pmod{2014}$$

$$-11407 = 5(2014) - 1337 \equiv 677$$
– Amad27 Jan 25 '15 at 12:05
@Amad27 Yes, that is correct . Observe that $;1343=-671\pmod{2104};$ and thus we get the same ! – Timbuc Jan 25 '15 at 12:07
Except the sign though. There must be something wrong there. – Amad27 Jan 25 '15 at 12:49
@Amad27 You've already asked quite a few question about modular arithmetic. I think you're lacking some basic understanding of this subject, and I think it'd be good for you to go the basics again before you attempt to go on anymore. In this case, we have that $;-671\pmod{2014};$ is exactly the same as $;1343\pmod{2014};$ , so I don't understand what you say about "except the sign"! The sign doesn't really matter here. We also have that $;3=-3=-18=12\pmod 3;$ , or $;4=-3=11=-31\pmod 7;$ , and etc. – Timbuc Jan 25 '15 at 13:18
I meant the answer is suppose to be $x \equiv 677$ not $-677$ – Amad27 Jan 25 '15 at 15:12

score 1 · Answer 2 · answered Jan 25 '15 at 11:42

1

$3^{-1}=-671\bmod 2014$, so the solution is $x=-671\times 17\bmod 2014$. You should get $677$ at the end.

answered Jan 25 '15 at 11:42

Bernard

175,478

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

0

Hint $\,\ {\rm mod}\,\ 3n\!+\!1\!:\,\ 3(-n)\equiv 1\,\Rightarrow\, 3^{-1}\equiv -n$

Alternatively $ $ note $\,\ \dfrac{1}{3} \,\equiv\, \dfrac{n}{3n}\,\equiv\, \dfrac{n}{-1}\equiv -n\ $ by Gauss's algorithm.

edited Apr 13 '17 at 12:21

Community

1

answered Jan 25 '15 at 18:39

Bill Dubuque

272,048

score 0 · Answer 4 · answered Jan 25 '15 at 20:30

0

$3x=2014~k+17=3~\underbrace{(671~k+6)}_x+\underbrace{(k-1)}_0\quad=>\quad k=1~$ and $~x=677$.

answered Jan 25 '15 at 20:30

Lucian

48,334
2
83
154

Solve $3x \equiv 17 \pmod{2014}$

4 Answers4