Exercise 5. Let $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}<\mathrm{e}$ be real numbers. We calculate all possible sums of two distinct numbers among these five numbers. The three smallest sums are 32, 36, and 37, and the two largest sums are 48 and 51. Find all possible values of $e$.
Solution to exercise 5 The two smallest sums are $\mathrm{a}+\mathrm{b}<\mathrm{a}+\mathrm{c}$, so $\mathrm{a}+\mathrm{b}=32$ and $\mathrm{a}+\mathrm{c}=36$. Similarly, the two largest are $c+e<d+e$ so $c+e=48$ and $d+e=51$. The third smallest sum can be either $a+d$ or $b+c$. In the first case, we have $a+d=37$, so on the one hand $a+c+d+e=(a+c)+(d+e)=36+51=87$ and on the other hand $a+c+d+e=(a+d)+(c+e)=37+48=85$, which is absurd.

Therefore, we are in the second case, so $\mathrm{b}+\mathrm{c}=37$. We deduce that $2(\mathrm{a}+\mathrm{b}+\mathrm{c})=(\mathrm{a}+\mathrm{b})+(\mathrm{a}+\mathrm{c})+(\mathrm{b}+\mathrm{c})=105$ so $\mathrm{c}=\frac{105}{2}-32=\frac{41}{2}$, and then $\mathrm{e}=48-\frac{41}{2}=\frac{55}{2}$.