0
$\begingroup$

Since $53$ is prime, from Wilson's theorem, $52! \equiv -1\pmod{53}$,

i.e. $52 \times 51 \times 50 \times 49! \equiv -1\pmod {53}$

I don't understand how to take it from here.

The other form I have is $51! \equiv 1\pmod{53}$ but again I don't know how to use that.

$\endgroup$
2
$\begingroup$

Well, $52\cdot51\cdot 50\equiv (-1)\cdot(-2)\cdot(-3)\equiv -6$. Hence you know that $49!\cdot(-6)\equiv -1$ (or equivalently and from the second form $49!\cdot 6\equiv 1$. Know any $a$ with $6a\equiv 1\pmod{53}$?.

$\endgroup$
  • $\begingroup$ $a=9$ since $6\cdot9=54=1(mod53)$? $\endgroup$ – Diya Oct 10 '14 at 21:59
  • $\begingroup$ so finally $49! \equiv 9(mod43)$? $\endgroup$ – Diya Oct 10 '14 at 22:05
  • $\begingroup$ @Diya Yes ($\mod{53}$ of course) $\endgroup$ – Hagen von Eitzen Oct 10 '14 at 22:31
3
$\begingroup$

Hint: Find the multiplicative inverse, modulo $53$, of $51\cdot 50 \equiv (-2)(-3)=6$.

$\endgroup$
  • $\begingroup$ The multiplicative inverse is 9. So $52\times 9 \times 49= -9(mod 53)$. Then $(-1) \times 9 \times 49! = -9(mod53)$. So $9\times 49! = 9(mod 53)$. Then? $\endgroup$ – Diya Oct 10 '14 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.