Suppose the middle number is $n$. Then the other numbers are $n+12$ and $n-2$.

The three numbers add up to $100$, so $$\begin{align}n+(n+12)+(n-2)&=100\\

\Rightarrow n+n+12+n-2&=100\\

\Rightarrow 3n+10&=100\\

\Rightarrow 3n&=90\\

\Rightarrow n&=30\end{align}$$ So $n=30$, $n+12=42$ and $n-2=28$.

If the middle number was $50$, then the other two would be $50 + 12 = 62$ and $50 -2 = 48.$ $50 + 62 + 48 = 160.$ That is much too big!

If the middle number was $40$, then the other two would be $52$ and $38.$ $40 + 52 + 38 = 130.$ That is still too big!

If the middle number was $30$, then the other two would be $42$ and $28.$ $30 + 42 + 28 = 100.$

If we start by looking at numbers which are close together, $33 + 33 + 34 = 100.$

Then leaving the middle number as $33$, the smallest number would need to be $31$ and the largest should be $33 + 12 = 45$, which is the same as $34 + 11.$ So $31, 33, 45$ have the correct 'gaps'.

$33$ has gone down by $2$ to make $31$ and $34$ has gone up by $11$ to make $45$, so the sum has gone down by $2$ and up by $11$, which is the same as going up by $9$ (to $109$).

To reverse the increase of $9$, we can subtract $3$ from each of the numbers, leaving us with $28, 30, 42.$

You can find more short problems, arranged by curriculum topic, in our short problems collection.