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Using the multiplicative inverse of 5 (which I used $8$), and multiplying it to $5x \equiv 4$ mod $13$, I get $x \equiv 6$ mod $13$. From that I get $x=13k+6$. Plugging that into $4x \equiv 5$ mod $17$ and simplifying, I got $k\equiv 15$ mod $17$. And plugging that back into $x=13k+6$, I get $x=13\cdot 15 + 6 = 201$ as the smallest natural number.

Is $201$ the smallest natural number that is possible? How do I know I got the smallest natural number? I'm pretty sure what I did was correct but $201$ seems like a really high number.

Bill Dubuque
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cypherfry
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    Any two solutions would need to be congruent $\pmod {13\times 17}$ and $13\times 17=221>201$ so you got the least positive solution. – lulu Aug 31 '23 at 21:13
  • Deleted my answer as this question is a dupe – Mike Aug 31 '23 at 21:27
  • Scaling each by inverses of the coef of $x$ yields an equivalent system in standard CRT form, and by Easy CRT in the dupe we know that solutions are unique mod $13\cdot 17 = 221,,$ so your solution $201$ is the unique nonnegative solution $< 221\ \ $ – Bill Dubuque Aug 31 '23 at 21:29
  • To elaborate: if there were also a smaller solution $,x_1,,$ i.e. $,0\le x_1< 201,,$ then this yields two incongruent solutions $!\bmod 221,,$ contra uniqueness. Here "solutions are unique $!\bmod n$" means that if $,x_1,$ is a solution then every other solution $,x_2,$ is congruent to it $,x_2\equiv x_1\pmod{!n}.\ \ $ – Bill Dubuque Aug 31 '23 at 21:44
  • Easier: $\ 5^2\equiv -1\pmod{!13},\ 4^2\equiv -1\pmod{!17},$ so scaling the 1st by $,-5,$ and the 2nd by $,-4,$ yields $,x \equiv -20\pmod{ ! 13\ &\ 17}$ $\iff x \equiv -20\equiv 201\pmod{!221=13(17)},$ by CCRT $\ \ $ – Bill Dubuque Aug 31 '23 at 22:03
  • @BillDubuque if this question is a dupe then why answer again in the comments. Doesn't that just encourage more dupes? – Mike Aug 31 '23 at 22:13
  • @Mike For abstract dupes it is usually helpful to say a bit about how the general result specializes (here CRT uniqueness). Note that what you wrote in your answer is not enough. We have to know that the transformation to CRT form yields an equivalent system, which is true here because scaling a congruence by an invertible yields an equivalent congruence – Bill Dubuque Aug 31 '23 at 22:31
  • We can solve it directly your way (w/o CRT) if we keep careful track of equivalences between the transformed systems. Briefly: $,x,$ satisfies 1st congruence $\iff x=6+13j,\ j\in \Bbb Z.,$ which satisfies the 2nd congreunce $\iff \color{#c00}{j = -2! +! 17k},,$ so both initial congruences hold true $\iff x = 6!+!13:!\color{#c00}j = 6!+!13(\color{#c00}{-2!+!17k})=-20!+!221k,\ k\in\Bbb Z$ $\iff x\equiv -20\equiv 201\pmod{!221}$ – Bill Dubuque Aug 31 '23 at 23:06

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