Since the numbers 3 5 7 are coprime, we use the Chinese remainder theorem
M0=m1m2m3=3∗5∗7=105
Mi=miM0
M1=m1M0=3105=35
M2=m2M0=5105=21
M3=m3M0=7105=15
Miyi=ai(mod mi)
35y1=2(mod 3)
33y1+2y1=2(mod 3)
2y1=2(mod 3)
y1=1
21y2=3(mod 5)
20y2+y2=3(mod 5)
y2=3(mod 5)
y2=8
15y3=2(mod 7)
14y3+y3=2(mod 7)
y3=2(mod 7)
y3=9
x=(M1y1+M2y2+M3y3)(mod M0)
x=(35∗1+21∗8+15∗9)(mod 105)
x=338(mod 105)
x=23
Answer: x=23
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