I need help fast
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 $10$.
@oobleck, there is a larger value to the equeation
No, in inputted it into my thing, and it said wrong
was something wrong with the solution I proposed before? Here it is again.
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.
To solve this problem, we need to convert the numbers from base $7$ to base $10$ and from base $11$ to base $10$.
Let's start by converting $\overline{ABC}_7$ into base $10$. In base $7$, each digit represents a power of $7$. The rightmost digit, $C$, represents $C \times 7^0 = C \times 1$. The middle digit, $B$, represents $B \times 7^1$, and the leftmost digit, $A$, represents $A \times 7^2$.
So, in base $10$, the value of $\overline{ABC}_7$ is $C + 7B + 49A$.
Similarly, let's convert $\overline{CBA}_{11}$ into base $10$. In base $11$, each digit represents a power of $11$. The rightmost digit, $A$, represents $A \times 11^0 = A \times 1$. The middle digit, $B$, represents $B \times 11^1$, and the leftmost digit, $C$, represents $C \times 11^2$.
So, in base $10$, the value of $\overline{CBA}_{11}$ is $A + 11B + 121C$.
Now, we need to find a positive integer $n$ that satisfies both equations. In other words, we need to find values of $A$, $B$, and $C$ such that:
$C + 7B + 49A = A + 11B + 121C$
Simplifying the equation, we get:
$48A - 4B - 120C = 0$
Since we want to find the largest possible value for $n$, let's maximize the values of $A$, $B$, and $C$.
To maximize $A$, we set $A = 6$. Then the equation becomes:
$288 - 4B - 120C = 0$
To maximize $B$, we set $B = 11$. Then the equation becomes:
$288 - 44 - 120C = 0$
Simplifying further, we get:
$244 - 120C = 0$
Solving for $C$, we find $C = \frac{244}{120} = \frac{61}{30}$. But since $C$ must be a positive integer, we round up to the nearest integer, which is $C = 2$.
Finally, substituting the values of $A$, $B$, and $C$ into the equation, we get:
$6 + 11 \times 2 + 121 \times 2 = 6 + 22 + 242 = 270$
Therefore, the largest possible value of $n$ in base $10$ is $\boxed{270}$.