An algebra problem by Alf-archie Sangkula
Let a < b < c < d < e be real numbers. Among the 10 sums of the pairs of these numbers, the least three are 32, 36 and 37 while the largest two are 48 and 51. Find all possible values of e
Hint: List out the pairwise sums in a systematic manner. It should be clear which sums are the smallest, 2nd smallest, largest, and 2nd largest.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
For a start, we could list out all 10 pairwise sums in a systematic way:
\begin{aligned} &a+b && && &&\ &a+c &&b+c && && \ &a+d &&b+d &&c+d && \ &a+e &&b+e &&c+e &&d+e \end{aligned}
Because we have listed these sums out systematically, we can make some comments about their relative sizes. In the table above, for any 2 sums in the same column, the one above is smaller than the one below. Also, for any 2 sums in the same row, the one that is further left is smaller than the one that is further right.
Hence, it is quite clear that a+b is the smallest sum, a+c is the 2nd smallest sum, d+e is the largest sum while c+e is the 2nd largest sum. (What is not obvious is what the 3rd smallest sum is: it could be either a+d or b+c.) We can then write the information given in the question in equations:
\begin{aligned} a + b &= 32 &&-(1) \ a + c &= 36 &&-(2) \ c+e &= 48 &&-(3) \ d+e &= 51 &&-(4) \end{aligned}
Some of the equations share a variable (e.g. equations 1 & 2), so by subtracting one from the other, we can tell how far apart some of the variables are (e.g. c must be 4 more than b). By looking through all possible pairs of the equations, we get the following relationships:
SMO 2013 JrRd2 1 01Letting the distance between a and b be denoted by x, and the distance between d and e be denoted by y, we can rewrite the 5 variables as
a, a+x, a+x+4, a+x+7, a+12.
(It turns out that the variable y wasn’t really needed at all!) Note that there are some restrictions on x because we must have a<b<c<d<e. In particular, 0 < x < 5. (Why?)
Let’s rewrite equations 1-4 with these variables to see if we can gain any further insight:
\begin{aligned} &(1): &&a + (a + x) = 32, 2a + x = 32 \ &(2): &&a + (a+x+4) = 36, 2a + x = 32 \ &(3): &&(a+x+4) + (a+12) = 48, 2a + x = 32 \ &(4): &&(a+x+7) + (a+12) = 51, 2a + x = 32 \end{aligned}
All 4 equations ended up being the same. Maybe at this point it’ll be good to check if we can determine which is the 3rd smallest pairwise sum:
\begin{aligned} a + d &= a + (a + x + 7) = 2a + x + 7, \ b + c &= (a+x) + (a+x+4) = 2a + 2x + 4. \end{aligned}
Which is bigger really depends on whether 7 or x+4 is bigger, i.e. whether 3 or x is bigger.
Case 1: \boldsymbol{x>3} In this case, the 3rd smallest sum is a + d. Hence,
\begin{cases} 2a + x + 7 &= 37 \ 2a + x &= 32 \end{cases}
A contradiction! Hence, x>3 is not valid.
Case 2: \boldsymbol{x \leq 3} In this case, the 3rd smallest sum is b + c. Hence,
\begin{cases} 2a + 2x + 4 &= 37 \ 2a + x &= 32 \end{cases}
This means that x = 1, and a = 15.5. From here, we can derive the 5 numbers
15.5, 16.5, 20.5, 23.5, 27.5
It is easy to check that these 5 numbers satisfy the given conditions in the question.
Hence, there is only 1 possible value for e, which is 27.5.