Let $\star$ be an operation defined in the set of nonnegative integers with the following properties: for any nonnegative integers $x$ and $y$, (i) $(x + 1)\star 0 = (0\star x) + 1$ (ii) $0\star (y + 1) = (y\star 0) + 1$ (iii) $(x + 1)\star (y + 1) = (x\star y) + 1$. If $123\star 456 = 789$, find $246\star 135$.
Problem
Source:
Tags: induction, algebra unsolved, algebra
chaotic_iak
13.10.2014 19:30
Let $0 \star 0 = c$. By induction, we can prove that $n \star 0 = 0 \star n = n+c$ for all nonnegative integer $n$. This is immediate: the base case $n = 0$ is trivial, and by using the induction hypothesis, if we have $k \star 0 = 0 \star k = k+c$ then $(k+1) \star 0 = (0 \star k) + 1 = k+c+1 = (k+1)+c$ and similarly for $0 \star (k+1)$. Now we will prove that $n \star (n+d) = (n+d) \star n = (0 \star d) + n = (n+d)+c$ for all nonnegative integers $n, d$, by induction on $n$. This is also immediate: the base case $n = 0$ is trivial, and by using the induction hypothesis, if we have $k \star (k+d) = (k+d) \star k = (k+d)+c$ then $(k+1) \star (k+d+1) = (k \star (k+d)) + 1 = (k+d)+c+1 = (k+d+1)+c$ and similarly for $(k+d+1) \star (k+1)$. Thus $x \star y = \max(x,y) + c$. Since $789 = 123 \star 456 = \max(123, 456) + c = 456 + c$, we have $c = 333$, and thus $246 \star 135 = \max(246, 135) + c = 246 + 333 = \boxed{579}$.
khan.academy
21.02.2019 07:02
Solution: We claim that the answer is $\boxed{579}$. We have that, $$ 123 \star 456 = (122+1) \star (455+1) \iff 789 = (122 \star 455) +1$$On further operation, the equation reduces to: $$ 789 = (0 \star 333)+123$$Now, using the second identity from the question, we get, $$ 789 = (0 \star 0) + 456 \iff (0 \star 0)= 333$$Now, $$ 246 \star 135 = (245+1) \star (134 +1) $$$$= (245 \star 134) +1$$On using the third identity and first identity, we get, $$246 \star 135 = (0 \star 0)+ 246 = 246 +333 = \boxed{579}$$
meet18
21.02.2019 08:37
chaotic_iak wrote: Let $0 \star 0 = c$. By induction, we can prove that $n \star 0 = 0 \star n = n+c$ for all nonnegative integer $n$. This is immediate: the base case $n = 0$ is trivial, and by using the induction hypothesis, if we have $k \star 0 = 0 \star k = k+c$ then $(k+1) \star 0 = (0 \star k) + 1 = k+c+1 = (k+1)+c$ and similarly for $0 \star (k+1)$. Now we will prove that $n \star (n+d) = (n+d) \star n = (0 \star d) + n = (n+d)+c$ for all nonnegative integers $n, d$, by induction on $n$. This is also immediate: the base case $n = 0$ is trivial, and by using the induction hypothesis, if we have $k \star (k+d) = (k+d) \star k = (k+d)+c$ then $(k+1) \star (k+d+1) = (k \star (k+d)) + 1 = (k+d)+c+1 = (k+d+1)+c$ and similarly for $(k+d+1) \star (k+1)$. Thus $x \star y = \max(x,y) + c$. Since $789 = 123 \star 456 = \max(123, 456) + c = 456 + c$, we have $c = 333$, and thus $246 \star 135 = \max(246, 135) + c = 246 + 333 = \boxed{579}$. Nice solution, Ivan koswara