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Let $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{121}=\frac{p}{q}$ where p,q are coprime integer couple. Prove $p\equiv 50 \pmod {121}.$

What im guessing about is that wolstenholme will do its job here but i don't exactly know how i apply it

Please give motivation of proof if possible if any...

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Note that the sum of the left hand side, when combined into one fraction, has a numerator that consists of products of denominators (all but one). That means that each of the terms of the sum in the numerator has a factor of 121, except for one: $120!$. So, All you need to show is that $120!\equiv50\pmod{121}$.

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  • $\begingroup$ I think that is not relevant idea here, lets see the case : $\frac{1}{2}+...+\frac{1}{4}=\frac{13}{12}$ and $13 \equiv 1 (mod4)$ but it is $3! \equiv 2 (mod4)$ $\endgroup$ – Solvable Potato Mar 12 at 22:00
  • $\begingroup$ I know what you are tryjng to say but this is different. You should aware of cases that numerator shares divisor with n!. $\endgroup$ – Solvable Potato Mar 12 at 22:07

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