Let n is divisible by m, and we have to find $V=\frac{n}{m}(mod)p$...

In case if we know the value of $n(mod)p$,$m(mod)p$ not n,m how can we find $V$ ?????

I know we can find answer by evaluating this $(n(mod)p * m^{p-2}(mod)p)(mod)p$.but in this we should have value of m.

I also want to know how inverse modulo works.

for example in case $\frac{63}{9}(mod)5$

3=63(mod)5 , 4=9(mod)5

answer will be $(3(mod)5*4^{5-2}(mod)5)(mod)5$ = 12(mod) 5 = 2

  • $\begingroup$ Are you asking how to evaluate $\ m^{p-2}\pmod p,\ $ or are you asking for other ways to calculate modular inverses? $\endgroup$ – Math Gems Mar 28 '13 at 18:56
  • $\begingroup$ I want to know how $\frac{63}{9}(mod)5$ is equal to $(3(mod)5 * 4^{5-2} (mod) 5))(mod)5$ $\endgroup$ – user69910 Mar 29 '13 at 3:52

$$63=3\pmod5\;,\;\;9=4=-1\pmod 5\Longrightarrow \frac{63}{9}=$$

$$3\cdot 4^{-1}\pmod 5=3\cdot4=2\pmod 5$$

Or directly:

$$\frac{63}{9}=7\Longrightarrow \frac{63}{9}=7=2\pmod 5$$

Note, btw, that $\,4^{5-2}=4^3=(-1)^3=-1=4\pmod 4\,$ ...

  • $\begingroup$ Nicely Done, Don. $\endgroup$ – mrs Mar 30 '13 at 10:49

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