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Find the integer between $0$ and $29\times 23$ $= 667$ that satisfies the two following congruences:

$x ≡ 15$ (mod $23$)
$x ≡ 1$ (mod $29$).

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  • $\begingroup$ Are you allowed to use guess and check? $\endgroup$
    – suomynonA
    Commented Oct 20, 2016 at 1:53

2 Answers 2

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You can just look for a pattern. Each time you add $29$ to a number, the remainder when divided by $23$ increases by $6$, so you can see that: $$1≡1\pmod {23}, \;30≡7\pmod {23}, \;59≡13\pmod {23},\;\;...\;\;291≡15\pmod {23}$$ and there's your answer.

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$29 = 6\pmod{23}\\ 6\cdot 4 = 1\pmod{23}\\ 1+2\cdot 6+2\cdot 1 = 15\\ 1+2\cdot29 + 8\cdot 29 \equiv 15\pmod{23}$

291

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