How many ordered triples of integers which are between 0 and 10 inclusive do we actually have if $a * (b+c) = a * b +c$ How many ordered triples of integers $(a,b,c)$ which are between 0 and 10 inclusive do we have if:
 $a * (b+c) = a * b +c$
 A: $a*b+a*c=a*(b+c)=a*b+c$ if and only if $a*c=c$ by cancellation. We see thus that if $c=0$, it works for all $a$ and $b$. We also see that if $c\neq 0$, it works only for $a=1$ (since $a*c=c$ implies $a=1$.
Thus it works for $\{(a, b, c)|c=0$ or $ a=1\}$.
A: We have that
$$a(b+c)=ab+c \Rightarrow ab+ac=ab+c.$$
If $a=1$, then we have $b+c=b+c$, and if $c=0$ then $ab=ab$.
When $a=1$ there are $11 \cdot 11=121$ ways to write two of the eleven numbers from $0$ to $10$ inclusive for $b,c$ with repeat, and when $c=0$, there are again $11 \cdot 11=121$ ways to write two of the eleven numbers from $0$ to $10$ inclusive for $a,b$ with repeat. 
These two sets have eleven duplicates when $a=1$ and $c=0$, since $b$ is one of the same eleven values in both cases.
Hence there are $2 ( 11 \cdot 11)-11=231$ ordered triples $(a,b,c)$ that satisfy your equation.
A: J W Perry is almost right, but he's double counting the cases where a = 1 and c = 0. There are 11 of those, so the right answer is 242 - 11 = 231.
To check:
*Main> length [ (a,b,c) | a <- [0..10], b <- [0..10], 
                          c <- [0..10], a * (b + c) == a * b + c ]
231

