Find a solution $X$ of the system of four congruences list below

Question

Find a solution $x$ of the system of four congruences

$X\equiv1\pmod{5}$

$X\equiv3\pmod{6}$

$ X\equiv5\pmod{11}$

$ X\equiv10\pmod{13}$

score 1 · Accepted Answer · answered Oct 14 '15 at 06:07

1

Let $$x = 5a +1$$ Now $$5a+1 = 3 \bmod 6 \implies 5a = 2 \bmod 6 \implies a = 4 \bmod 6 \implies x = 5(6b+4)+1 = 30b+21$$

Again $$30b+21 = 5 \bmod 11\implies 4b = 3 \bmod 11 \implies b = 9 \bmod 11 \implies x = 30(11c +9)+21 = 330c +291$$

Finally $$330c +291 = 10 \bmod 13 \implies c = 1 \bmod 13 \implies x = 330(13d+1)+291 = 4290d + 621 = 621 \bmod 4290$$

answered Oct 14 '15 at 06:07

PTDS

May I ask how you get a=4mod6 from 5a=2mod6 ? Thx – Leon K Oct 14 '15 at 16:48
Do you know how to calculate 5^{-1} mod 6? – PTDS Oct 14 '15 at 18:34
Em, I don't quite get it... – Leon K Oct 15 '15 at 03:48
In this case, you can calculate it mentally... Can you find a number from the set {0, 1, 2, 3, 4, 5} which satisfies 5x = 1 mod 6? – PTDS Oct 15 '15 at 04:06
x=5 works, I don't understand how it become x=4 mod 6 – Leon K Oct 15 '15 at 19:31
Okay, you agree that 5^{-1} mod 6 = 5, right? Now consider the equation 5a=2 mod 6. Multiply both the sides by (5^{-1} mod 6). So you have 5^{-1} 5a = a = (5^{-1}) (2) mod 6 = 10 mod6 = 4 mod 6. Understood? – PTDS Oct 15 '15 at 22:21

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