Let five numbers be $a<b<c<d<e$.

a) How many pairwise sums can be formed?

b) These ten sums are $21,26,35,40,49,51,54,60,65,79$. Find $a, b, c, d, e$.
a) We can form ten distinct sums: $a+b, a+c, a+d, a+e, b+c, b+d, b+e, c+d, c+e, d+e$.

b) The two smallest sums are necessarily: $a+b=21$ and $a+c=26$, and the two largest, necessarily $d+e=79$ and $c+e=65$. Furthermore, the sum of these ten sums is: $4(a+b+c+d+e)=480$, so $a+b+c+d+e=120$. Therefore, $c=(a+b+c+d+e)-(a+b)-(d+e)=20$, and $a=6, b=15, e=45$ and $d=34$. If we calculate the ten sums of the five numbers $6,15,20,34$ and 45, we indeed find the ten given numbers.