Hi All am a bit stuck on some revision that I am trying to do.
Firstly (part a) I must calculate the inverse of 11 modulo 41, which I have done and believe it to be 15.
The next part is to: Now use your result from part (a) to find x where x^11 mod 42 = 10
I have been trying all sorts but don't seem to be getting anywhere.
Thanks
On
You are trying to compute $10^\frac{1}{11}$:
$10^\frac{1}{11}=10^{15}=(10^3)^5$.
But $10^3=1000\equiv 16\ (mod\ 41)$
So $(10^3)^5=16^5=16\cdot16^2\cdot16^2=16\cdot 100=16\cdot 18=288\equiv 7\ (mod\ 41)$ since $16^2\equiv10\ (mod\ 41)$, $100\equiv 18\ (mod\ 41)$, and $288\equiv7\ (mod\ 41)$.
So the solution to $x^{11}\equiv 10\ (mod\ 41)$ is $x=7$.
$4\cdot 11\equiv 3\pmod{41}$, hence $$8\cdot 11\equiv 6\pmod{41}$$ and by multiplying both sides by seven we get that $8\cdot 7\equiv 56\equiv 15\equiv 11^{-1}.$
By the Chinese theorem, $x^{11}\equiv 10\pmod{42}$ is equivalent to: $$\left\{\begin{array}{rcl} x^{11} &\equiv& 0\pmod{2},\\ x^{11} &\equiv& 1\pmod{3},\\x^{11}&\equiv& 3\pmod{7},\end{array}\right.$$ or to: $$\left\{\begin{array}{rcl} x &\equiv& 0\pmod{2},\\ x &\equiv& 1\pmod{3},\\x^{-1}&\equiv& 3\pmod{7},\end{array}\right.$$ and since the inverse of $3\pmod{7}$ is $5$, from: $$\left\{\begin{array}{rcl} x &\equiv& 0\pmod{2},\\ x &\equiv& 1\pmod{3},\\x&\equiv& 5\pmod{7},\end{array}\right.$$ we get the solution $x\equiv 40\equiv -2\pmod{42}$.
I cannot see how the inverse of $11\pmod{41}$ is involved here: $42$ is not a prime!