- #1

evinda

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I want to solve the following system of congruences:

$$x \equiv 13 \pmod{40} \\ x\equiv 5 \pmod{44} \\ x \equiv 38 \pmod{275}.$$I have thought the following:

$$x \equiv 13 \pmod{40} \Leftrightarrow x \equiv 13 \pmod{2^3 \cdot 5}$$

$$x \equiv 5 \pmod{44} \Leftrightarrow x \equiv 5 \pmod{2^2 \cdot 11}$$

$$x \equiv 38 \pmod{275} \Leftrightarrow x \equiv 38 \pmod{5^2 \cdot 11}$$$$x \equiv 13 \pmod{2^3 \cdot 5} \Leftrightarrow x \equiv 13 \pmod{2^3} \text{ and } x \equiv 13 \pmod{5} \ \ (1)$$

$$x \equiv 5 \pmod{2^2 \cdot 11} \Leftrightarrow x \equiv 5 \pmod{2^2} \text{ and } x \equiv 5 \pmod{11} \ \ (2)$$

$$x \equiv 38 \pmod{5^2 \cdot 11} \Leftrightarrow x \equiv 38 \mod{5^2} \text{ and } x \equiv 38 \pmod{11} \ \ (3)$$

$(1)$: $x \equiv 5 \pmod{2^3}$ and $x \equiv 3 \pmod{5}$

$(2)$: $x \equiv 1 \pmod{2^2}$ and $x \equiv 5 \pmod{11}$

$(3)$: $x \equiv 13 \pmod{5^2}$ and $x \equiv 5 \pmod{11}$Am I right so far?

How can we continue? Can we somehow apply the Chinese Remainder Theorem? (Thinking)