# Find the solutions of system of equivalences for modulo

$x \equiv 3 \pmod {13}$ and

$x \equiv 3 \pmod {17}$ and

$x \equiv 13 \pmod {23}$

Thank you!

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Hint: Use the Chinese Remainder Theorem. (CRT)

Note also that your moduli are all prime. That greatly simplifies matters.

The linked Wikipedia article explains in detail how to solve a system of linear congruences, and provides some illustrative examples, too, that help to demonstrate how to use the theorem.

See also other posts here at Math.SE: Search "Chinese Remainder Theorem" or "Solving a system of linear congruences":

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Hint: Using Chinese remainder theorem

Step-$1$: First note that $a_1=3,a_2=3,a_3=13$and $m_1=13,m_2=17,m_3=23$

Step-$2$: Find $M=13.17.23=5083$ and $M_1=17.23,M_2=13.23,M_3=13.17$

Step-$3$: Find the solution of equation $M_i x_i \equiv 1 \pmod {m_i}$where i=1,2,3

Step-$4$: After solving above three equation you will get $x_1,x_2,x_3$ put this value $x=M_1 x_1 a_1 +M_2 x_2 a_2 +M_3 x_3 a_3$

Step-$5$: Final answer is say $X \equiv x \pmod M$

$\rm 13,17\mid x\!-\!3 \iff 13(17)\mid x\!-\!\color{#C00}3,\:$ so $\rm\:mod\ 23\!:\ 13\equiv x\equiv \color{#C00}3+13(17)n\equiv 3+14n,\:$ so $\rm 14n\equiv 10,\:$ so $\rm\:n\equiv 10/14\equiv 5/7\equiv 28/7\equiv \color{#0A0}4.\,$ Thus $\rm\:x = 3\!+\!13(17)(\color{#0A0}4\!+\!23k)\equiv 887\!+\!13(17)23k$.