Find $5$ numbers where the sum of the pairs gives the results $110, 112, 113, 114, 115, 116, 117, 118, 120, 121$
I have been trying on this question for some time and it seems easy at first glance, but later on, I was unable to start on the question. I tried turning these into equations with $a_1, a_2, a_3, a_4, \text{ and }a_5$ but couldn't as it is not determined which $2$ numbers give which sum. I was, however, able to achieve that $a_1+a_2+a_3+a_4+a_5=289$ (which is obvious). Now is there anyway I could continue without guessing. Thank you.
Best Answer
As there are no duplicates among the pair sums, the five numbers are distinct. So wlog. $a_1<a_2<a_3<a_4<a_5$. Then $110=a_1+a_2$, $112=a_1+a_3$, which makes $a_3=a_2+2$. Similarly, we find $a_4=a_3+1$ from the last two entries $120,121$. So far we have found that the numbers are $$a_1, a_2, a_2+2, a_2+3, a_5.$$ This also tells us that $a_1+a_4=113$ and that $a_2+a_5=118$. Now that the three pair sums $a_2+a_3=2a_2+2$, $a_2+a_4=2a_2+3$, $a_3+a_4=2a_2+5$ must occur among $114,115,116,117$. The only possible match is that $a_2+a_3=114$, i.e., $a_2=56$. The rest follows easily from here.