A Congruence Problem for RMO basics

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Problem:

Find the solution of the congruence x^{24}+7x \equiv 2 \pmod {13}.

Solution:

It is clear that x \equiv 0 \pmod {13} is not a solution. So, let 1 \leq x \leq 12. Then, from Fermat’s Little Theorem, we have that x^{12} \equiv 1 \pmod{13} and this is why x^{24}= 1 \pmod{13}. The congruence to be solved can therefore be reduced to 7x \equiv 1 \pmod{13}which leads to solution x \equiv 2 \pmod{13}.

Nalin Pithwa

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