Solve $x \equiv 2$ mod 6 and $x \equiv 8$ mod 9

Question

We have the following system of congruences: $$x \equiv 2 \, \text{mod 6}\\ x \equiv 8 \, \text{mod 9}$$ How do we solve this when $6$ and $9$ are not coprime? This excludes the use of CRT, unless we can rewrite the congruences in some clever fashion?

I hope you can help.

Possible duplicate of How to solve for multiple congruence's what aren't relatively prime. You can see the general strategy here, at least. — Ian Coley, Sep 27 '13 at 06:50
That is an old Brahmagupta problem which I've solved, simply throwing away the congruences that are equivalent (keeping only one). But here I cannot apply CRT without reducing. Does $x \equiv 2$ mod 6 imply $x \equiv 0$ mod 2, and does $x \equiv 8$ mod 9 imply $x \equiv 2$ mod 3?
I think it does. This way $(2,3)=1$ and we can use CRT?

Thanks for your reply by the way! — Numbersandsoon, Sep 27 '13 at 07:01
That sounds right to me, but I couldn't quote you a theorem. — Ian Coley, Sep 27 '13 at 07:06
By CCRT $,x\equiv 8\pmod{!6\ &\ 9}\iff x\equiv 8\pmod{!18},\ $ by $18 ={\rm lcm}(6,9)$. Or apply general Easy CRT formula. — Bill Dubuque, Sep 02 '23 at 07:54

drhab · Accepted Answer · 2013-09-27T12:41:28.033

5

You have integers $n,k$ such that $x=6n+2=9k+8$. Then $2(n-1)=3k$ resulting in $n\equiv1\text{ mod }3$. Setting $n=3m+1$ gives you $x=18m+8$, i.e. $x\equiv8\text{ mod }18$.

edited Sep 27 '13 at 12:41

answered Sep 27 '13 at 08:07

drhab

151,093

@DonAntonio Thanks. A typo. Restored now. – drhab Sep 27 '13 at 12:42
Now it's fine. +1 – DonAntonio Sep 27 '13 at 12:43
Makes perfect sense. Thanks! – Numbersandsoon Sep 27 '13 at 15:19
@drhab Credit where credit is due. (+1) – Git Gud Dec 15 '13 at 16:56

Solve $x \equiv 2$ mod 6 and $x \equiv 8$ mod 9

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