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.

• Hint: think of the numbers as being expressed in base $b$ instead of base $10$. You are being told that $5(b+2) = b^2+4$. So $b$ is .... Commented Jan 21, 2014 at 0:21
• @CameronWilliams He needs to find the right base Commented Jan 21, 2014 at 0:21
• Titles are for titles; question bodies are for question bodies. When you read an article in the paper, do the first sentence say "The news is in the title", or is the body of the article self-contained? Commented Jan 21, 2014 at 0:23

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!

Provided the base is 2 or larger, you have $10\times 11 = 110$.

• It also works if the base is equal to 2.
– Dan
Commented Jan 21, 2014 at 0:38
• Yerright, Dan. What's interesting is that the hypothesis is entirely superfluous. Commented Jan 21, 2014 at 0:40
• Well, it's really just $b(b+1)=b^2+b$... Commented Jan 21, 2014 at 0:53
• Multiply by 2 in binary (10) is just left shift or appending zero at the end, in the same way here also it is 10 x 11 = 110, we don't need to calculate the base in the given problem Commented Jan 21, 2014 at 12:26

$5\times 12\ =\, 104\$ in base $\,b\,$ means
$5(b\!+\!2) = b^2\!+4\, \Rightarrow\, \color{#0a0}{b^2 = 5b+6}$
$10(11) = \color{#c00}b(b\!+\!1)\, =\, \color{#0a0}{b^2}\!+b = \color{#c00}6(b\!+\!1)\,\Rightarrow \color{#c00}{b = 6}$