The base in which $1065 = 13 \cdot 54$ is true How can one find the base $r$ in which $1065 = 13 \cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.
$$r^{3}-5r^{2}-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.
Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
 A: The base is clearly less than $11$, since $13\cdot54\lt20\cdot54=T80\lt1000$ (where $T$ stands for Ten) in any base greater than $10$.  Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
A: Since $3\cdot4-5$ ends in a zero, your base divides $12-5=7$.  Therefore, your base is either $1$ or $7$.  Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1\cdot r+3$ and $54_r=5r+4$.  Therefore, by the distributive law, 
$$
13_r\cdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12\equiv12\pmod{r}.
$$
On the other hand, in base $10$,
$$
1065_r=1\cdot r^3+0\cdot r^2+6r+5\equiv 5\pmod{r}.
$$
Therefore, for equality, it must be that $5\equiv 12\pmod{r}.$. 
A: \begin{align}
   [1,3]_B \cdot [5,4]_B &= [1,0,6,5]_B \\
\end{align}
In base $B$:
\begin{array}{r}
      &          &  1 &  3 \\
      &  \times  &  5 &  4 \\
   -- &        --& -- & -- \\
      &          &  4 & 12 \\
      &        5 & 15 &  0 \\
   -- &        --& -- & -- \\
      &        5 & 19 & 12 \\
\end{array}
And, so $[5, 19,12]_B = [1,0,6,5]_B$
If we are to believe that $[1,0,6,5]_B$ is in normalized notation, then $B \ge 7$
It seems then that $12$ must be equal to $[1,5]_B$; which implies that $B = 7$. This is easily verified:
$[5,19,12]_7 = [5,20,5]_7 = [7,6,5]_7 = [1,0,6,5]_7$
A: Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5\Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5  =b^3+6b+5
$$
