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Show that: $$[6]_{21}X=[15]_{21}$$

I'm stuck on this problem and I have no clue how to solve it at all.

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What is making you stuck? Also, are you asked to find an $X$ that solves it? – Tobias Kildetoft Mar 18 '13 at 14:56
solving it is not the same as showing it. if you where asked to show it it would mean it is true for all x or a given set of numbers. – Carry on Smiling Mar 18 '13 at 14:57

Well, we know that $\gcd\ (6,21)=3$ which divides $15$. So there will be solutions: $$6x \equiv 15 \ (\mod 21)$$ $$2x \equiv 5 \ \ (\mod\ \ 7),$$

because that $2\times 4\equiv 1 \ \ (\mod \ \ 7)$, thus: $$x \equiv 4\times 5 \ \ (\mod\ \ 7)$$ $$ \equiv 6 \ \ (\mod\ \ 7)$$

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I downvoted because I disagree with giving a full solution to a homework question where the OP has not clarified where he is stuck. – Tobias Kildetoft Mar 18 '13 at 15:07
I too downvoted for the same reason as Tobias. – Jayesh Badwaik Mar 18 '13 at 15:08
Well, I'm upvoting to partially balance those two downvotes. Certainly a full solution for a homework question is not advisable, yet the solver is a brand new participant and I feel like we veterans should be more lenient and not to rush to downvote. A simple, well intentioned comment to the poster can equally do the job . – DonAntonio Mar 18 '13 at 19:17
@DonAntonio I see what you mean. However, as there is no consensus to this (as can be seen from the number of upvotes), I would not feel right in writing a comment along the lines of "please don't do this". Herpderp: Please don't take this as discouraging you to answer questions in general, just homework questions where the OP has not had time to clarify the precise problem. – Tobias Kildetoft Mar 19 '13 at 0:48
Just trying to be helpful, but ya sure, got it! – user67258 Mar 19 '13 at 2:13

Hint $\rm\,\ 21\mid 6x\!-\!15 = 6x\!+\!6\!-\!21 \iff 21\mid 6(x\!+\!1)\iff 7\mid 2(x\!+\!1)\iff 7\mid x\!+\!1$

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One way is just to try all the choices for $X$, the integers from $0$ to $20$. That isn't very many.

Another is to look for representatives in the class of $15$, which are numbers of the form $15+21k$, for one that is a multiple of $6$.

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You write "look for one...", but there is more than one solution mod $21$. – Math Gems Mar 18 '13 at 15:11

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