nafistiham wrote:
$9.$ Consider a family of functions $f_{m}:N\cup \left \{ 0 \right \}\rightarrow N\cup \left \{ 0 \right \}$ that follows the relations:
$f_{m}(a+b)=f_{m}((f_{m}(a)+f_{m}(b)))$
$f_{m}(km)=0$
Here, $m$ is any positive integer apart from $1$. Find the number of elements in the range of the function $f_{2012}$ .
Well, the given function looks a lot like $"mod"$ function.
Suppose, $g_m(x)=g_m(c)$, where, $c\equiv x(\text{mod}m)$
So, if $a\equiv c_1(\text{mod} m)$, and, $b\equiv c_2(\text{mod}m)$, then, $a+b\equiv c_1+c_2(\text{mod}m)
\Rightarrow a+b\equiv g_m(a)+g_m(b)(\text{mod}m)$.
$\Rightarrow g_m(a+b)=g_m(g_m(a)+g_m(b))$ (This matches with the 1st property.)
And, $km\equiv 0(\text{mod}m)$. So, $g_m(km)=g_m(0)$.(This matches with the second property.)
So, $g_m $ and $f_m$ treat $x$ similarly. This means that they should have the same no. elements in their range.
Now, no. of elements in the range of $g_m$ is equal to the no. of elements in the complete residue system of $m$. Therefore, no. of elements in the range of $f_{2012}$ is $\boxed{2012}$.
I am sorry that my reasoning might be a little hazy....
![Confused :?](./images/smilies/icon_e_confused.gif)
, but, you get the idea.
![Rolling Eyes :roll:](./images/smilies/icon_rolleyes.gif)