1. Let $a<b<c<d<e$ be real numbers. Among the 10 sums of the pairs of these numbers, the least three are 32,36 and 37 while the largest two are 48 and 51 . Find all possible values of $e$.
1. We have 37 is either $a+d$ or $b+c$ and
$$
a+b=32, \quad a+c=36, \quad c+e=48, \quad d+e=51
$$

Thus $c-b=4, d-c=3$ and $d-b=7$. Therefore $(a+b)+(d-b)=a+d=39$. Hence $b+c=37$. We thus have $a=15.5, b=16.5, c=20.5, d=23.5$ and $e=27.5$.