# Find all solutions, if any, to the system of congruences x ≡ 5(mod 6), x ≡ 3(mod 10), and x ≡ 8 (mod 15)

Using the Chinese Remainder Theorem:

$$m=6\times 10\times 15=900$$, $$M_1=\frac{900}{6} =150$$ , $$M_2=\frac{900}{10} =90$$ , $$M_3=\frac{900}{15} =60$$

i am trying to find inverse but $$150x_1 \equiv 1\pmod 6$$ and others have no inverse.

• The problem is that $6,10,15$ are not pairwise relative primes, which is required in order to use CRT.
– M.P
Commented Oct 2, 2019 at 17:57
• The Chinese remainder theorem applies when the moduli are relatively prime, and these are not, so you need to do a little more analysis. Commented Oct 2, 2019 at 18:00

$$x\equiv5\pmod 6$$ leads to $$x\equiv 1\pmod2$$ and $$x\equiv 2\pmod3$$.
$$x\equiv3\pmod {10}$$ leads to $$x\equiv 1\pmod2$$ and $$x\equiv 3\pmod5$$.
$$x\equiv8\pmod {15}$$ leads to $$x\equiv 2\pmod3$$ and $$x\equiv 3\pmod5$$.
Since these are all compatible, there is a solution, which is also the solution to $$x\equiv 1\pmod2$$ and $$x\equiv 2\pmod3$$ and $$x\equiv 3\pmod5$$. You should have better luck solving that system since the moduli are pairwise relatively prime.