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Question
I'm thinking of five distinct integers: a<b<c<d<e. When I sum all possible pairs of these numbers, I get the values 2,3,6,7,9,10,11,13,14,17. What are the original numbers?

Answer
The pairs are (a+b), (a+c), (a+d), (a+e), (b+c), (b+d), (b+e), (c+d), (c+e), and (d+e).
Summing them:
(a+b) + (a+c) + (a+d) + (a+e) + (b+c) + (b+d) + (b+e) + (c+d) + (c+e) + (d+e) = 2+3+6+7+9+10+11+13+14+17
4a + 4b + 4c + 4d = 92
a+b+c+d+e = 23

Since a<b<c<d<e,
a+b = 2
a+c = 3
b = c-1

Similarly,
d+e = 17
c+e = 14
d = c+3

Substitution for b and d:
2+c+17 = 23
c = 4

b = c-1 = 3
d = c+3 = 7
a = 2-b = -1
e = 17-d = 10

The numbers are -1, 3, 4, 7, and 10.

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