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This is  a hard one see if you guys can solve it! Try to!

 

Find the least positive four-digit solution to the following system of congruences.

\(\begin{align*} 7x &\equiv 21 \pmod{14} \\ 2x+13 &\equiv 16 \pmod{9} \\ -2x+1 &\equiv x \pmod{25} \\ \end{align*}\)

 Dec 4, 2022
 #1
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The system reduces to

n = 7 mod 14

n = 6 mod 9

n = 2 mod 25

 

By the Chinese Remainder Theorem, the smallest four-digit number hat works is 2877.

 Dec 4, 2022
 #2
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I don't think 2877 is correct..

 Dec 7, 2022

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