7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 99 = sum of all 3 rows

$\therefore$ sum of each row is 33

$\begin{align}26-2n\ \ +\ \ 19-n\ \ +\ \ 12\ \ &=33\\

26+19 + 12-33&=3n\\

24&=3n\\

8&=n\end{align}$

7 + 14 = 21, 12 is missing to make 33

Say $n$ + 12 + $x$ + 12 = 33,

then $n$ + 14 + ($x-$2) = 33

Same idea gives $x-$4 at the bottom

$\therefore$ 3$x$-6=33, so $x$ = 13

And $n$ + $x$ + 12 = 33 $\Rightarrow n$ + 13 + 12 = 33 $\Rightarrow n$ = 8

7 + 14 = 21, 12 is missing to make 33

33 $-$ 14 = 19

19 = 7 + 12

19 = 8 + 11

19 = 9 + 10

no more options so these 4 numbers are 8, 9, 10, 11

$n$ + middle + 14 = bottom + middle + 12

difference between $n$ and bottom = difference between 12 and 14

This gives 2 options:

This problem is taken from the UKMT Mathematical Challenges.

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