Question

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 $10$.

Answer 1

what do we have?
49a + 7b + c = 121c + 11b + a
or, equivalently,
b = 12a - 30c
So, one solution is 361_₇ = 163_₁₁ = 190_₁₀

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.

Answer 2

@oobleck, that is wrong, when I input it into the system, it said wrong

Answer 3

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

Answer 4

If my previous post was not clear, the answer is 247

Answer 5

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 $\boxed{247}$.