Suppose that $n$ is a positive integer such that in base $7$, then $n$ can be expressed as $\overline{ABC}_7$, and in base $11$, then $n$ can be expressed as $\overline{CBA}_{11}$. Find the largest possible value of $n$ in base $10$.
what do we have?
49a + 7b + c = 121c + 11b + a
or, equivalently,
b = 12a - 30c
keep doing trial and error and soon get
502 base 7 = 205 base 11 = 247 base 10
If my previous post was not clear, the answer is 247
We convert $n$ to base $10$. The base $7$ expression implies that $n = 49A + 7B + C$, and the base $11$ expression implies that $n = 121C + 11B + A$. Setting the two expressions equal to each other yields that$$n = 49A + 7B + C = 121C + 11B + A \Longrightarrow 48A - 4B - 120C = 0.$$Isolating $B$, we get$$B = \frac{48A - 120C}{4} = 12A - 30C = 6(2A - 5C).$$It follows that $B$ is divisible by $6$, and since $B$ is a base $7$ digit, then $B$ is either $0$ or $6$. If $B$ is equal to $0$, then $2A - 5C = 0 \Longrightarrow 2A = 5C$, so $A$ must be divisible by $5$ and hence must be either $0$ or $5$. Since $n$ is a three-digit number in base $7$, then $A \neq 0$, so $A = 5$ and $C = 2$. Thus, $n = 502_7 = 5 \cdot 7^2 + 2 = 247$.
If $B$ is equal to $6$, then $2A - 5C = 1$, so $2A - 1 = 5C$ and $2A - 1$ must be divisible by 5. Since $A$ is a base $7$ digit, it follows that $A = 3$ and $C = 1$. This yields the value $n = 361_7 = 3 \cdot 7^2 + 6 \cdot 7 + 1 = 190$. The largest possible value of $n$ in base $10$ is $\boxed{247}$.
@oobleck, that is wrong, when I input it into the system, it said wrong
Ah, so we have a game of bases! Let's see if we can clown our way to the answer.
To maximize $n$ in base $10$, we should focus on maximizing $A$, $B$, and $C$ individually. Since we're working in base $7$, we want $A$, $B$, and $C$ to have the largest possible values in the base $7$ system.
Now, the largest digit in base $7$ is $6$, so we want to assign $6$ to $A$, $6$ to $B$, and $6$ to $C$. This gives us $\overline{666}_7$ in base $7$.
Next, we need to convert $\overline{666}_7$ to base $10$. We can do this by recognizing that each digit represents a power of $7$.
So, $n = 6\cdot7^2 + 6\cdot7^1 + 6\cdot7^0 = 6\cdot49 + 6\cdot7 + 6 = 294 + 42 + 6 = \boxed{342}$.
And voila! The largest possible value of $n$ in base $10$ is $342$. Hope that brings a smile to your face!
To find the largest possible value of $n$ in base $10$, we need to determine the maximum values for each digit.
Given that $n$ is expressed as $\overline{ABC}_7$ in base $7$, the maximum value for digit $A$ is $6$, since in base $7$ the largest digit is $6$. Similarly, the maximum value for digit $B$ is also $6$, and the maximum value for digit $C$ is $6$.
Now, let's convert $\overline{CBA}_{11}$ to base $10$. To do this, we need to multiply each digit by its corresponding power of $11$. The digit $A$ is in the ones place, so its value is $A \times 11^0 = A \times 1 = A$. The digit $B$ is in the elevens place, so its value is $B \times 11^1 = 11B$. The digit $C$ is in the $11^2$ place, so its value is $C \times 11^2 = 121C$.
Therefore, the value of $n$ in base $10$ is $121C + 11B + A$.
To find the largest possible value of $n$ in base $10$, we need to maximize the values of $A$, $B$, and $C$. Since the maximum value for each digit is $6$, the largest possible value of $n$ is $121(6) + 11(6) + 6 = 726 + 66 + 6 = \boxed{798}$.
what do we have?
49a + 7b + c = 121c + 11b + a
or, equivalently,
b = 12a - 30c
So, one solution is 3617 = 16311 = 19010
Since all of a,b,c must be from 1 to 6 (since abc must be expressible in base 7), it shouldn't take too long to explore all the possibilities.