Solving Systems of Congruences

Lesson · Intermediate

Number Theory

ar(modm)ar(modn)    ar(modlcm(m,n)) \begin{aligned} a &\equiv r \pmod m\\ a &\equiv r \pmod n \end{aligned} \iff a\equiv r\pmod{\operatorname{lcm}(m,n)}

Approach 1 — List Values for Both Moduli

a2(mod5)a3(mod11)a?(mod55) \begin{aligned} a&\equiv2\pmod5\\ a&\equiv3\pmod{11} \end{aligned} \Longrightarrow a\equiv?\pmod{55}

List the congruent values until a repetition occurs.

a=2,7,12,17,22,27,32,37,42,47(mod5)a=2,7,12,17,22,27,32,37,42,47\pmod5
a=3,14,25,36,47,58(mod11)a=3,14,25,36,47,58\pmod{11}
a47(mod55)\Longrightarrow a\equiv47\pmod{55}

Approach 2 — List Values for the Larger Modulus

a3(mod5)a2(mod41)a?(mod205) \begin{aligned} a&\equiv3\pmod5\\ a&\equiv2\pmod{41} \end{aligned} \Longrightarrow a\equiv?\pmod{205}

Select the larger mod (here 41) and list the congruent values until it satisfies the other mod.

a3(mod5)a\equiv3\pmod5
a=2,43(mod41)a=2,43\pmod{41}
a43(mod205)a\equiv43\pmod{205}

Approach 3 — Substitution

a5(mod19)a11(mod47)a?(mod893) \begin{aligned} a&\equiv5\pmod{19}\\ a&\equiv11\pmod{47} \end{aligned} \Longrightarrow a\equiv?\pmod{893}

Select the larger mod.

a11(mod47)a=47k+11(1)a\equiv11\pmod{47}\Longrightarrow a=47k+11\qquad\qquad\text{(1)}

Substitute in the other mod.

a5(mod19)47k+115(mod19)a\equiv5\pmod{19}\Longrightarrow47k+11\equiv5\pmod{19}
47k6(mod19)9k6(mod19)\Longrightarrow47k\equiv-6\pmod{19}\Longrightarrow9k\equiv-6\pmod{19}
3k2(mod19)3k21(mod19)\Longrightarrow3k\equiv-2\pmod{19}\Longrightarrow3k\equiv-21\pmod{19}
k712(mod19)k=19q+12(2)\Longrightarrow k\equiv-7\equiv12\pmod{19}\Longrightarrow k=19q+12\qquad\qquad\text{(2)}
(1), (2)a=47(19q+12)+11=893q+575\text{(1), (2)}\Longrightarrow a=47(19q+12)+11=893q+575
a575(mod893)\Longrightarrow a\equiv575\pmod{893}