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for this question I used the Euclidean Algorithm in order to get an equation with the format of: $6v+27w=1$. I got there eventually and my answer was $x\equiv 7\pmod{27}$.

I was told that this was wrong and the correct answer was supposed to be: 6, 15 and 24 modulo 27.

I do not understand how to get to this result and any help would be appreciated.

Bill Dubuque
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walterman
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    "...in order to get an equation with the format of $6v+27w=1$. I got there eventually" You can't get to that equation since it is false for all integer values of $v,w$. The left hand side is a multiple of $3$. The right hand side is not a multiple of $3$. Contradiction. – JMoravitz Oct 19 '21 at 16:03

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If $6x\equiv9\pmod{27}$, then $6x=27n+9$.
Divide both sides by three. And you get $2x=9n+3$.
$2x\equiv3\pmod{9}$, then multiple both side by 5. $x\equiv10x\equiv15\equiv6\pmod{9}$.
The answer is $x\equiv6\pmod{9}$.
6, 15, 24 are equivalent modulo 9.

Arturo Magidin
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MH.Lee
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  • Hi, thank you so much for your response. Can you explain how I could find out that from x≡6 mod 9, that 15 and 24 are equivalent and a part of the answer as well? Sorry if that's a dumb question. – walterman Oct 19 '21 at 16:02
  • @walterman It is obvious that 15, 24 are equivalent to 6 modulo 9 because 15-6=9, 24-6=18... – MH.Lee Oct 19 '21 at 16:03
  • Thank you so much – walterman Oct 19 '21 at 16:14
  • If you are going to edit the original, please use proper MathJax code. \mod gives the wrong spacing; it's not meant to be either the operator or the relation symbol. For the binary operator, use \bmod; for the parenthetical congruence version, use \pmod. Compare $a\equiv b\mod c$ (a\equiv b\mod c), $a\equiv b \bmod c$ (a\equiv b\bmod c), and $a\equiv b\pmod{c}$ (\equiv b \pmod{c}). – Arturo Magidin Oct 19 '21 at 16:33
  • @ArturoMagidin Oooooo Thx A lot... I'm not used to MathJax... – MH.Lee Oct 19 '21 at 16:36