# Five integers are paired in all possible ways. The integers in each pair are added. The ten sums are $1,4,5,7,8,8,11,11,14,15$. What are the integers? [closed]

First, I know two things. There must be either a $$0$$ or a negative number. If there is a $$0$$, there must be a $$1$$ and no negative. If there is a negative, there must be only one negative and no $$0$$ or $$1$$.

• You also know that the five numbers sum to $21$. Commented May 17 at 19:47
• How can this question with only two lines (so without any context whatsoever ) receive $7$ upvotes and $2$ reopen-votes ? Commented May 19 at 12:37

The ten pairs sum to $$84$$. Each integer appears in exactly four of the pairs, so the five integers sum to $$21$$.

No two integers can be identical, or there would be at least three pairs of identical sums. Therefore, we're looking for five distinct integers, $$a \lt b \lt c \lt d \lt e$$.

We know that $$d+e=15$$ and $$a+b=1$$, so $$a+b+d+e=16$$ and $$c=5$$. Since $$14$$ is the largest remaining sum, it must be the case that $$c+e=14$$ so $$e=9, d=6$$. Since $$4$$ is the smallest remaining sum, it must be the case that $$a+c=4$$, so $$a=-1, b=2$$.

Summarizing, the integers are $$-1, 2, 5, 6, 9$$.

Note, by the way, that we never needed the fact that the five numbers are integers, so this analysis proves that we have found the only set of five real numbers with exactly these pairwise sums. Moreover, it's easy to see that if the ten pairwise sums of five complex numbers are all real, then the numbers themselves also must be real. Thus, we have found the only set of complex numbers that works.

Alternatively.

If the five numbers are $$a\le b\le c\le d\le e$$ we can order the sums.

We know $$a+b =1; a+c=4$$ so $$b=1-a$$ and $$c=4-a$$ and $$c=b+3$$

The next two numbers are $$a+d$$ (which is the least of any sum involving $$d$$ or $$e$$) and $$b+c$$ (which is the least of any sum not involving $$a$$) but we don't know which is which.

Either $$a+d = 5$$ and $$b+c=7$$ or $$a+d=7$$ and $$b+c=5$$

If $$b+c=7$$ then $$2b+3=7$$ and $$b=2;c=5$$ and $$a=1-b=-1$$ and $$d=5-a=6$$. so $$a,b,c,d= -1,2,5,6$$

Of if $$b+c = 5$$ then $$2b+3=5$$ so $$b=1;c=4$$ and $$a=1-b=0$$ and $$d=5-a=5$$ so $$a,b,c,d=0,1,4, 5$$.

We can do the same for the last two sums. We must have $$d+e = 15$$ and $$c+e=14$$ and so $$d=c+1$$ (which we already knew)

The next two lowest sums are $$c+d$$ (the highest that doesn't involve $$e$$) and $$b+e$$ (the highest that involve $$a$$ or $$b$$).

So we have $$c+d = b+e =11$$ and $$c+d=2c+1=11$$ so $$c=5; d=6$$. And $$e=15-d=14-c=9$$.

Which combined with result above means $$a,b,c,d,e = -1,2,5,6,9$$

The two sums we have not considered are $$a+e$$ and $$b+d$$ which must both be equal to $$8$$ which satisfies $$a+e = -1 + 9$$ and $$b + d = 2+6$$ (but does not satisfy $$a+e =0+9; b+d=1+5$$)