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Source : Burton, Elementary number theory (7th edition)

Show that any integer of the form $6k + 5$ is also of the form $3 j + 2$, but not conversely.

As to the $(\implies)$ part :

$n= 6k+5$

$\implies n = 3(2k')+5 \space $ for some $ k' \in \mathbb Z$

$\implies n = 3(2k')+3+2 $

$\implies n = 3(2k'+1)+2 $

$ \implies n = 3j +2 \space $ with $ j = 2 k'+1 $

As to the " not conversely part".

It seems to suffice to find a number of the form $3j+2$ that does not leave a remainder of $5$ when divided by $6$.

Consider number $20 = 3(j)+2 $ where $j=6$.

We have also $ 20 = 6(3) +2$ .

My question is : is the counterexample method adequate and, if so, did I apply it correctly? Is stating that $20$ can be written with a remainder of $2$ when divided by $6$ sufficient to show that it cannot be written with a remainder of $5$.

J. W. Tanner
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Vince Vickler
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    Yes, because $20=6m+5$ and $20=6m'+2$ would imply $6$ divides $5-2$, which is a contradiction – J. W. Tanner Feb 28 '24 at 02:15
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    A counter example is perfectly acceptable. It just takes one example to prove a statement so... throw one if you want. However if you want to really "grok" why the converse isn't true not that $6k + 5 = 3(2k +1) + 2=3j+2$ so the converse is only true for odd values $j$. Conversely $3j + 2 = 3(j-1) + 5 = 6(\frac {j-1}2) + 5$ only holds true if $\frac {j-1}2$ is an integer (i.e. if $j$ is odd). – fleablood Feb 28 '24 at 02:20
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    By the dupe: $,x\equiv 2\pmod{!3}\iff x\equiv 2,$ or $,5\pmod{!6},,$ which yields a more precise claim than you seek, and clarifies the relationship between the two congruences. – Bill Dubuque Feb 28 '24 at 02:21
  • I think OP is asking if it's possible that $6\mid x-2$ and simultaneously $6\mid x-5$; OP doesn't use congruences – J. W. Tanner Feb 28 '24 at 02:23
  • It's best to delete this question once all is clear, since we already have too many examples of this sort (which complicates searching for answers). – Bill Dubuque Feb 28 '24 at 02:23
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    Actually, on rereading you example. I don't think you adequately showedd that $20$ can't be $6j + 5$. You should $20 = 3\dot 6 + 2$ so it's of the form but to show $20\ne 6j + 6$ we must show .... If $20 = 6j + 5$ then $15 = 6j$ and $j = \frac {15}2$ which is not an integer.... But once you do that, you have one counter example and that absolutely shows beyond all doubt that the converse is not true and you don't need to do anything more. – fleablood Feb 28 '24 at 02:26
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    @J.W.T That too is a dupe, viz. $,2\equiv x\equiv 5\pmod{!6},$ contradicts the uniquness of remainders. $\ \ $ – Bill Dubuque Feb 28 '24 at 02:27
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    argj!!! I typed "It just takes one example to prove a statement " .... that was a typo. I meant "It just takes one example to prove a statement wrong". I'd like to say it was "obviously" a typo but.... – fleablood Feb 28 '24 at 02:30

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