1. Let $a<b<c<d<e$ be real numbers. All possible pairwise sums of these 5 numbers are calculated. Of these 10 sums, the three smallest are 32, 36, 37, while the two largest are 48 and 51. Determine all possible values that $e$ can take.
SOLUTION: It is first evident that the two smallest sums, 32 and 36, are respectively $a+b$ and $a+c$. Similarly, the two largest sums, 48 and 51, are respectively $c+e$ and $d+e$. Thus, $a+b=32$, $a+c=36$; subtracting the first equation from the second yields

$$
c-b=4 .
$$

Similarly, from $d+e=51, c+e=48$ we get

$$
d-c=3
$$

and finally, subtracting (1) from (2), we get

$$
d-b=7 \text {. }
$$

It remains to determine which sum corresponds to the value 37. A priori, there are two possibilities: $37=a+d$ or $37=b+c$. If $37=a+d$, then by (3) we would have $37=a+d=(a+b)+(d-b)=$ $32+7=39$, which is absurd. Therefore, the only possibility left is $37=b+c$. At this point, it is easy to calculate the values of $a, b$, and $c$. Indeed, $2 a=(a+b)+(a+c)-(b+c)=32+36-37=31$, from which $a=15.5, b=32-a=16.5$ and $c=37-b=20.5$. Knowing the value of $c$, we can also calculate $e=48-c=27.5$, from which $d=23.5$. With these values of $a, b, c, d$, and $e$, it is indeed the case that the three smallest sums are 32, 36, and 37, and the two largest sums are 48 and 51, so the only possible value for $e$ is 27.5.