Suppose that $a,b,c,d,e$ are integers with the constraint that $a\leq b \leq c \leq d \leq e$.
Also, suppose that the sums of the $5 \choose 2$ = $10$ pairs (i.e. $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $b+e$, $c+d$, $c+e$, $d+e$) are in some order $25, 27, 28, 31, 32, 34, 35,36,38,42$ (please note that this order is not necessarily respective of the order of the paired sums, i.e. $a+b$ does not necessarily equal $25$, $a+c$ does not necessarily equal $27$, etc.)
Find all possibilities of $a,b,c,d,e$.
The only thing I knew to do was add all the values: $$(a+b) + (a+c) + \cdots + (d+e) = 4(a+b+c+d+e) = 25 + 27 + 28 + 31 + 32 + 34 + 35 + 36 + 38 + 42 = 328 \implies a+b+c+d+e = \frac{328}{4} = 82$$
After that, because of the given inequality constraint, I thought it'd make sense to assign the four smallest numbers (25, 27, 28, 31) to the first four pairs ($a+b, \ldots,b+c$). But, after solving for each variable, I ended up with $a$ being equal to $\frac{29}{3}$ which contradicts the condition that all 5 variables must be integers.
I'm getting overwhelmed by the number of possible combinations of ($a,b,c,d,e$) especially because of the inequality constraint, which only makes things more difficult. Can someone please show me a more systematic approach to solving this problem?
Appreciate any and all help.
Thanks
-A
Hint: As a start, note that $c=(a+b+c+d+e)-(a+b)-(d+e)$. You know how to find $a+b+c+d+e$. Also, $a+b=25$ and $d+e=42$.
Now everything collapses. We know $a+c$ and $c$, so we know $a$. And so on. Do check whether the solution you get meets all the conditions.