If $5 \times 12 = 104$, how much is $10 \times 11$? Question is in the title, this is for my analysis course. I don't know where to begin.
 A: Provided the base is 2 or larger, you have $10\times 11 = 110$.
A: $ 5\times 12\  =\, 104\ $ in base $\,b\,$ means
$5(b\!+\!2) = b^2\!+4\, \Rightarrow\, \color{#0a0}{b^2 = 5b+6}$
$ 10(11) = \color{#c00}b(b\!+\!1)\, =\, \color{#0a0}{b^2}\!+b = \color{#c00}6(b\!+\!1)\,\Rightarrow \color{#c00}{b = 6} $
A: Since 5 and 2 appear in the 'ones place' of the first equation, we know that the base is higher than both these numbers. Then:
$$10\cdot11=110$$
...just like in our usual number system.
Let $b$ denote the number base. Now in base b, the number $10$ represents $b$, the number $100$ represents $b^2$, and so on.
Now in base 10, we can re-write our equation as: $$5\times 12 = 104 \Rightarrow 5\cdot(b+2)=b^2+4$$
$$5b+10=b^2+4$$
$$b^2-5b-6=0$$
This polynomial's only positive root is at $b=6$, so that must be the base we are working in. 
Now we can rewrite $10\times11$ in base-10:
$$10\times 11 =b\times(b+1)\Rightarrow6\cdot(6+1)=6\cdot7=42$$

Now to make sure, we can can convert that back to base 6 and make sure it looks like $110$:
$$42=36+6+0=b^2+b\Rightarrow110$$
Great, it works out!
