B5. The diagram shows three triangles, $A B C, P Q R$ and $X Y Z$, each of which is divided up into four smaller triangles. The diagram is to be completed so that the positive integers from 1 to 10 inclusive are placed, one per small triangle, in the ten small triangles. The totals of the numbers in the three triangles $A B C, P Q R$ and $X Y Z$ are the same.
Numbers 1, 2, 4 and 10 have already been placed.

In how many different ways can the diagram be completed?
Solution
Label the empty small triangles in the second row as $a, b, c$ and $d$ from left to right and the empty small triangles in the third row as $e$ and $f$ from left to right.
These triangles may only contain the integers $3,5,6,7,8$ or 9 .
Since the totals in the triangles $A B C, P Q R$ and $X Y Z$ are the same and the sum of the integers 1 to 10 is 55 , we can deduce that
$$
55+2 b \text { is a multiple of } 3 \text {. }
$$

Therefore, $b=7$, as it is the only remaining integer that satisfies (1). Consequently, the total in each triangle $A B C, P Q R$ and $X Y Z$ is 23 .
Considering the totals in the triangles $A B C, P Q R$ and $X Y Z$ gives
$$
\begin{aligned}
a+b+11 & =23 \\
b+c+d+2 & =23 \\
\text { and } b+e+f+4 & =23
\end{aligned}
$$
respectively. Equation (2) gives $a=5$. The only remaining integers now are 3, 6, 8 and 9. From equations (3) and (4), $c+d=14$ and $e+f=12$. To satisfy these equations $c$ and $d$ must be 8 and 6 in any order and $e$ and $f$ must be 3 and 9 in any order. Therefore, there are only 4 different ways in which the diagram can be completed and these are shown below: