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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?

RektTheNoob  Jan 12, 2018
 #1
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x = 506,629

506,629 mod 105 =2^2

Guest Jan 12, 2018
 #2
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(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

Guest Jan 12, 2018
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Thanks!

RektTheNoob  Jan 13, 2018

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