+22 Suppose that x is an integer that satisfies the following congruences:
3+x =2^2 (Mod 3^3)
5+x=3^2 (Mod 5^3)
7+x=5^2 (Mod 7^3)
What is the remainder when x is divided by 105?
+592 105=3x7x5
3+x=4(mod 27)
5+x=9(mod125)
7+x=25(mod343)
x=1(mod27)
x=4(mod125)
x=18(mod343)
27k+1=x
3(9k)+1=x
x=1(mod3)
for the same reasons:
x=4(mod5)
x=4(mod7)
x=4(mod 35)
possible sollutions lower than 105:
4, 39, 74.
the only one that has a remainder 1 when divided by 3 is $\boxed{4}$
