Question

A list of six positive integers, p, q, r, s, t, u satisfies p < q < r < s < t < u. There are exactly 15 pairs of numbers that can be formed by choosing two different numbers from this list. The sums of these 15 pairs of numbers are: 25, 30, 38, 41, 49, 52, 54, 63, 68, 76, 79, 90, 95, 103, 117. Which sum equals r + s?

Answer 1

The Big Idea: Because p, q, r, s, t, and u, are given to us in order
of size, we should actually be able to deduce whivh sum corresponds with which pair in at least a few ases; if we're lucky, we'll be able match enough sums to pairs to determine which sum equals r + s. (Of ourse, since r and s are the middle" numbers, then we'd expect their sum to be pretty much in the middle, so 63 is probably a good guess; sin e this is a multiple choice problem, 63 is almost certain to be the wrong answer.)
Let's get started. Can we tell whi h of the pairs should give the smallest
sum? Yes! Sin e p and q are the smallest two numbers in the list, their sum must be the smallest of all of the sums. In other words, p + q = 25.
Can we tell whi h of the pairs should give the largest sum? Yes! Sin e
t and u are the largest two numbers in the list, then their sum should be the largest of all of the sums. In other words, t + u = 117.
Next, we'll use the fa ts that q = 25 − p and t = 117 − u to do a little
housekeeping and redu e the number of unknowns in our list to four. We thus rewrite our list as p, 25 − p, r, s, 117 − u, u.
Are there other pairs that we an tra k down easily? We had some
su ess starting at the ends of the list of pairs, so let's keep trying to work inwards from the ends of the list of sums. Can we tell which of the pairs should give the se ond smallest sum? It's not p + q, since we've already used this one. We might guess q + r or p + r. In fa t, p + r is smaller than

q + r, which we can ac ctually deduce by using a chart:

p + q
p + r q + r
p + s q + s r + s
p + t q + t r + t s + t
p + u q + u r + u s + u t + u
In the chart, when we move down a fixed column, the sums increase since the first summand stays the same and the second increases; when we move to the right along a xed row, the sums increase because the first summand increases and the second stays the same. Therefore, p + r is smaller than q + r (since it's immediately to the left) and is in fact smaller than any other sum except p+q. Therefore, p+r is the second smallest sum, so p+r = 30.
Similarly, the second largest sum must be s + u, so s + u = 103.
We can now do some more housekeeping, writing r = 30 − p and
s = 103 − u to make our list p, 25 − p, 30 − p, 103 − u, 117 − u, u.
Now we come to a fork in the road. The third smallest total, acording to our snazzy chart, is either p + s or q + r. (Can you see why?) So either

p+s = p+103−u = 38 (that is, p = u−65) or q+r = 25−p+30−p = 38.
In the second case, 55 − 2p = 38 or 2p = 17. But p is an integer, so this is not the case. Therefore, p = u − 65, and our list bec omes
p = u − 65 , q = 90 − u , r = 95 − u , s = 103 − u , t = 117 − u , u .
We have written each unknown in terms of one variable, which is undoubtedly a good thing.

p + q = (u − 65) + (90 − u) = 25 ; q + u = (90 − u) + u = 90 ;
p + r = (u − 65) + (95 − u) = 30 ; r + u = (95 − u) + u = 95 ;
p + s = (u − 65) + (103 − u) = 38 ; s + u = (103 − u) + u = 103 ;
p + t = (u − 65) + (117 − u) = 52 ; t + u = (117 − u) + u = 117 .
Let's redraw our chart, but this time expressing the remaining pairs in
terms of u only:
25
30 185 − 2u
38 193 − 2u 198 − 2u
52 207 − 2u 212 − 2u 220 − 2u
2u − 65 90 95 103 117
Phew! The remaining numerical sums that are not yet visible in the
chart are 41, 49, 54, 63, 68, 76, and 79.