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I just need to know is my calculation here correct. I have

$$ 4m-2 \equiv 1,2,5,6 \pmod{10} $$

and I want to find the value of m. Is it correct if I divide through by 2, I will have

$$ 2m-1 \equiv \dfrac{1}{2},1,\dfrac{5}{2},3 \pmod{5} $$

Then,

$$2m \equiv \dfrac{1}{2}+1,(1+1),\dfrac{5}{2}+1, (3+1) \pmod{5}$$

And finally,

$$m \equiv 2,2,3,4 \pmod{5}$$

Is this correct?

shahrina ismail
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  • for m = 0:10 mod(4*m-2,10) end ans = 8 ans = 2 ans = 6 ans = 0 ans = 4 ans = 8 ans = 2 ans = 6 ans = 0 ans = 4 ans = 8 –  Nov 01 '16 at 00:36
  • Hi, shahrina. I notice that you've asked several questions on this site, and the community welcomes your participation. At some point, however, please read the tutorial on MathJax formatting and start using it in your questions. It is the standard way of formatting mathematical notation on this website, and using it will make your questions easier to read and more likely to get quick answers. –  Nov 01 '16 at 00:38
  • Thank you for pointing that out @MikeHaskel. I have edited the question. – shahrina ismail Nov 01 '16 at 00:49
  • @user236182 Thank you. that was a typo. I have fixed it. – shahrina ismail Nov 01 '16 at 00:53
  • How do you figure 1/2 +1 =1 or 5/2 +1 =3. Note 4x3-2=0 not 5 and 4x2-2=6 not 1 ot 2 so clearly you are doing something wrong. – fleablood Nov 01 '16 at 01:20
  • @fleablood. Thank you for pointing that out. – shahrina ismail Nov 01 '16 at 01:22

1 Answers1

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Is $m$ supposed to be an integer? If it is, note that $$4m-2\ \equiv1,2,5,6\ \pmod{10}$$ means the last digit of $4m-2$ is 1, 2, 5, or 6.* But $4m-2$ is an even number and so its last digit cannot be 1 or 5. Therefore you can eliminate these and just consider $$4m-2\ \equiv\ 2,6\pmod{10}$$

*If $4m-2$ is negative, its last digit is 9, 8,5, or 4; e.g. $-14\equiv6\pmod{10}$.

George Law
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  • yes $m$ is an integer – shahrina ismail Nov 01 '16 at 00:52
  • In that case $4m-2\equiv1,5\pmod{10}$ is impossible, so you should just be considering $4m-2\equiv2,6\pmod{10}$. – George Law Nov 01 '16 at 01:01
  • So if $4m-2 \equiv 2,6 \pmod{10}$, my concern is how am I to solve for $m$. Is it I can divide through by 2, and obtain $2m-1 \equiv 1,3 \pmod{5}$? And finally get answer as $m \equiv 1,2 \pmod{5}$? – shahrina ismail Nov 01 '16 at 01:04
  • Sorry, but you can't divide by 2 here. Note that 22=4 and also 72=4, so 4/2 could be either 2 or 7. Also $2x=3$ had no solutions mod 10. This is because 2 is a 0-divisor. 2*5=0 (and no, you can't divide by 5 either!). Also, when you divide (or multiply) in this sort of equation it doesn't change the modulus. You're doing your math mod 10, and it stays mod 10. – user383778 Nov 01 '16 at 04:30
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    @ScottBurns You can divide the whole equation by 2. $2x\equiv4\pmod{10}$ $\iff$ $2x=4+10k\ (k\in\mathbb Z)$ $\iff$ $x=2+5k$ $\iff$ $x\equiv2\pmod5$. There is nothing in the question that says you have to work only in mod 10. – George Law Nov 01 '16 at 08:46
  • @shahrinaismail $m\equiv1,2\pmod5$ is correct. – George Law Nov 01 '16 at 08:48
  • Thank you for the feedbacks. It has been really helpful. – shahrina ismail Nov 01 '16 at 09:03