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$ x ≡ 5 (mod 11) $

$x ≡ 6 (mod 9) $

Which integers $x$ is between $0-300$ that fulfills this two congruences?

I think I got this right but I am not quit sure how to find all x's.

This is my solution:

$x ≡ a(mod m1): m1=11$

$x ≡ b(mod m2): m2=9$

$11u+9v=1$

$gcd(11,9)$

        | u  | v  |    |    | u  | v  |
        | 1  | 0  | 11 |  9 | 0  | 1  |
        | 1  | -1 | 2  |  9 | 0  | 1  |
        | 1  | -1 | 2  |  7 | -1 | 2  |
        | 1  | -1 | 2  |  5 | -2 | 3  |
        | 1  | -1 | 2  |  3 | -3 | 4  | 
        | 1  | -1 | 2  |  1 | -4 | 5  |

$u=-4, v=5$

$x=a*m2*v+b*m1*u$

$x=5*9*5+6*11*(-4) =-39$

And here is where I'm not certain if I am doing it right.

$m1*m2 = 11*9 = 99$

And to find more x's:

$k=$constant just number of times I add $m1*m2$

$-39+k*(99)$

So the x's are:

$-39+1*99=60$

$-39+2*99=159$

$-39+3*99=258$

Answer: The integers x, between $0-300$ are $60, 159, 258$.

Is this a valid answer? Is this a anwer that should work universally?

Anton Svarén
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1 Answers1

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That's correct. By CRT the solution is $\,x \equiv -39\pmod{99}\,$ so your method is valid. If you don't know CRT, then note $\,9,11\mid x+39\iff 9\cdot 11\mid x+39\,$ since $\,{\rm lcm}(9,11) = 9\cdot 11.\,$ Alternatively we can use the mod Distributive Law $\, ab\bmod ac = a(b\bmod c)\,$ as below

$\,x\!-\!5\bmod{99} = 11\underbrace{\left[\dfrac{x\!-\!5}{11}\bmod 9\right]}_{\large x\ \equiv\ 5\pmod{\!11}}\! = 11\underbrace{\left[\dfrac{\color{#c00}6\!-\!5}{2}\bmod 9\right]}_{\large x\ \equiv\ \color{#c00}6\pmod{\!9}}\!= 11\left[\dfrac{10}{2}\bmod 9\right] = 11\left[5\right]$ $\,\Rightarrow\, x=60\!+\!99k$

Bill Dubuque
  • 272,048