Suppose that x is an integer that satisfies the following congruences:
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
What is the remainder when x is divided by 105?
x = 506,629
506,629 mod 105 =2^2
(3 + x) mod 3^3 = 2^2
(5 + x) mod 5^3 = 3^2
(7 + x) mod 7^3 = 5^2
By simple iteration:
A*27 + 4 - 3=B*125 + 9 - 5 =C*343 + 25 - 7
A=18,764, B =4,053, C=1,477, Therefore:
x=18,764 *27 + 1 =506,629
506,629 mod 105 =4 =2^2
Thanks!
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