Is the function $f(x) = 7x + 11 \pmod{10}$ invertible for all positive integers?
I don't know how to go about it, mainly because I have never dealt with functions such as this one.
Is the function $f(x) = 7x + 11 \pmod{10}$ invertible for all positive integers?
I don't know how to go about it, mainly because I have never dealt with functions such as this one.
Working modulo $\,10\,$ all the time:
$y=7x+11\Longrightarrow 7x=y-11=y+9\Longrightarrow x=\frac{y+9}{7}=(y+9)\cdot 3=3y+7\Longrightarrow$
the inverse function is $\,g(x)=3x+7\,$
Notes:
$(1)\;\;\;\;\;\;\;\;-11=9\pmod{10}\Longleftrightarrow -11-9=-20=0\pmod{10}$
$(2)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{7}=3\pmod{10}\Longleftrightarrow 3\cdot 7=21=1\pmod{10}$
This function cannot be invertible because it is not injective. Note that f(0) = f(10), for example.
Hope this helps!