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How many ordered triples of integers $(a,b,c)$ which are between 0 and 10 inclusive do we have if:
$a * (b+c) = a * b +c$

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  • $\begingroup$ Does $'*'$ imply normal multiplication ? $\endgroup$ Nov 28, 2013 at 6:37

3 Answers 3

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$a*b+a*c=a*(b+c)=a*b+c$ 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\}$.

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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
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  • $\begingroup$ Great catch, and an egregious error on my part. I have incorporated this into my post. This would have been good as a comment, but I know you are unable to comment due to rep. Thanks for the look! $\endgroup$ Nov 28, 2013 at 8:14
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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.

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