Pair sums

Many people correctly noticed that it won't help to take two of the same number because then there won't be enough different totals. Tyler, Harry and others started with ways of making 0. Having decided that taking 0 + 0 = 0 won't work, they realised that a negative number and its positive value are needed. Hannah started with -1 and 1, then found what else is needed to make the rest of the totals. Using trial and improvement gives something like this:

Can't take 2 because -1 + 2 = 1 is not allowed.

Must take 3 because -1 + 3 = 2 and 1 + 3 = 4 are needed.

Can't take 4 because -1 + 4 = 3 is not allowed

Must take 5 because -1 + 5 = 4, 1 + 5 = 6 and 3 + 5 = 8 are needed.

At this point you need 10 to make 15, and need to check that all the possible totals are reached, and nothing extra.

Well done to everyone who used a method like this to get the right answer! It's lucky that -1 and 1 were in the solution, or else you'd need to try -2 and 2, -3 and 3, ...

Woo from Hampton School gave an algebraic solution; since it's often useful to use letters to represent unknown numbers this sounds like a good idea!

The five numbers can be thought of as a, b, c, d and e with

a < b < c < d < e

Then we can order the totals too. There is a grid of inequalities

\begin{array}{ccccccc}

a+b \\

{}\wedge{}\\

a+c &<& b+c\\

{}\wedge{} && {}\wedge{}\\

a+d &<& b+d &<& c+d\\

{}\wedge{} && {}\wedge{} && {}\wedge{}\\

a+e &<& b+e &<& c+e &<& d+e\\

\end{array}

Therefore, smallest combinations are as follows in order:

a+b < a+c < b+c or a+d < b+d...

And the totals are 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 so

0 = a+b

2 = a+c

both b+c and a+d can be the third or fourth total, but both are 4, hence

4 = b+c

4 = a+d

And treating the first three equations as a system of simultaneous equations, they can be solved to find a. This gives -2 = 2a. So a = -1, b = 1, c = 3 and d = 5. There is one last number to find; e. We know the largest total is 15 = d+e; so must have e = 10.

Well done Woo, this is very sophisticated. Sometimes it's hard to know when trial and improvement won't work so it's useful to have an algebraic method that doesn't rely on trials.