. 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) The number of sums that can be formed is equal to the number of ways to choose 2 elements out of 5, which is $\frac{5 \times 4}{2}=10$.

b) We necessarily have $a+b=21, a+c=26, c+e=65$ and $d+e=79$. Furthermore, the sum of all the numbers $21+26+35+40+49+51+54+60+65+79=480$ is equal to $4(a+b+c+d+e)$, hence $a+b+c+d+e=120$. We have $c=120-(a+b)-(d+e)=120-21-79=20$. From this, we deduce that $a=6, b=15, e=45$ and $d=34$.