### Always Two

Find all the triples of numbers a, b, c such that each one of them plus the product of the other two is always 2.

### Not Continued Fractions

Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?

### Surds

Find the exact values of x, y and a satisfying the following system of equations: 1/(a+1) = a - 1 x + y = 2a x = ay

# Alien Currency

##### Age 14 to 16 Short Challenge Level:

Answer: $19$

Elimination multiplying equations by $3$ and $8$
\begin{align}3g+8b=46\quad\Rightarrow\quad&24g+64b=368\\ 8g+3b=31\quad\Rightarrow\quad&\ \ \underline{24g+9b=\ \ 93}\\ \text{subtracting gives:}\quad&\ \ \ \qquad 55b=275\\ &\ \ \ \Rightarrow\ \ 11b=55\\ & \ \ \ \Rightarrow\quad\ \ b=5\end{align}
And so $3g+40=46\Rightarrow g=2$
$\therefore 2g+3b= 4+15=19$

Elimination by combining equations more than once
\begin{align} 3g+8b=46&\\ \underline{+\qquad 8g+3b=31}&\\ 11g+11b=77&\\ \Rightarrow \quad g+b=\ \ 7&\\ \Rightarrow 3g+3b=21\end{align}
\begin{align} 3g+8b=46&\\ \underline{-\qquad 3g+3b=21}&\\ 5b=25&\\ \Rightarrow \quad b=\ 5\ &\end{align}              \begin{align} 8g+83=31&\\ \underline{-\qquad 3g+3b=21}&\\ 5g=10&\\ \Rightarrow \quad g=\ 2\ &\end{align}

Therefore $2g + 3b = 19$.

Getting $2g+3b$ without finding $g$ and $b$
\begin{align} 3g+8b=46&\\ \underline{+\qquad 8g+3b=31}&\\ 11g+11b=77&\\ \Rightarrow \quad g+b=\ \ 7& \end{align}

$3g+8b$
$+ \quad (\ g\ +\ b\ )\times$ some    to get blues and greens in ratio $2:3$
difference will be $5$ since $8$ and $3$ have difference of $5$
$10:15 = 3+7:8+7$

\begin{align} 3g+8b=46&\\ \underline{+\qquad 7g+7b=49}&\\ 10g+15b=95&\\ \Rightarrow \ \ 2g+3b=19& \end{align}

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.