Grades 9 & 10 5-Star Challenge #2
Every possible pair within a set of five numbers is summed.
The sums are 0, 3, 11, 12, 20, 21, 23, 29, 32, 41. Find the five numbers.
In this problem, the answer may be obtained by trial and error, but there are some neat aspects to the problem if one works to solve it. Let us assume that the five numbers are distinct (as they would need to be if their ten pair-sums were distinct). Let us consider the sums as F < G < H < J < K < L < M < N < P < Q. Let the five numbers be a < b < c < d < e. The order of a, b, c, d, and e offers only an incomplete understanding of the order of the sums. We have
(a+b) < (a+c) < [(b+c),(a+d)] < [(a+e),(b+d)] < [(c+d),(b+e)] < (c+e) < (d+e).
We know that F=a+b and G=a+c, and that Q=e+d and P=c+e. We do not know if the third and fourth smallest sums represent a+d and b+c or b+c and a+d, respectively.
This problem can be approached by solving all of the little equations to establish which is greater of (b+c) and (a+d). Once b+c is known, the system of three equations a+b=F, a+c=G, b+c=(H or J) can be solved for the values of a, b, and c. The values of d and e can be found shortly thereafter by substitution.
There is a less tedious way to solve the problem. By identifying that each number is summed with each possible partner, we know that each number a < b < c < d < e appears four times in the sum F+G+H+J+K+L+M+N+P+Q, so the sum of the ten pair-sums must be equal to 4(a+b+c+d+e). By adding the first and last sum, we know that F+Q=a+b+d+e, and we can therefore find the value of c.
In our problem, the pair-sums are 0, 3, 11, 12, 20, 21, 23, 29, 32, 41, whose sum is 192, which is equal to 4(a+b+c+d+e). This means that 192/4=48=a+b+c+d+e.
We know that F+Q=0+41=a+b+d+e, so c=7.
If c=7, we know that G=3=(a+c)=(a+7), so a=-4. Then F=0=(a+b)=(-4+b) tells us that b=4. P=32=(c+e)=(7+e), so e=25. Since 48=a+b+c+d+e, d must be 16.