Pairwise Puzzler - Harvey's solution

Thank you to Harvey for sending us his thoughts on Pairwise Puzzler


Pairwise puzzler- strategy for original four numbers

I did this puzzle by using some of the strategy for the original five numbers puzzle.

The five numbers can be calculated simply by using simultaneous equations.

Let's call the lowest number a, the second lowest b etc... and we'll place the sums from smallest to largest in series s.
We know straight away that a + b = s1, as the smallest sum must be made up of the smallest pair.
Also the largest sum is clear: d + e = s10
we can also work out that a + c = s2, as this is the second smallest sum; a and c is the smallest two numbers which haven't yet been paired.
Similarly, c + e = s9 (the second largest).

For the four original pairwise puzzlers I was given six possible numbers 5, 9,10,12,13 and 17 and I need to find out what they are. The sums I have to find are:







The question marks are supposed to be (not in correct order): a+b, a+c ,a+d, b+c ,b+d and c+d. 

So as in the original s1 will be the two lowest added together a+b=5 then s6 will be the two biggest c+d=17.

After that you can find the second smallest a+c will=9 then the second largest b+d=13.

After this I used trial and error by seeing if a=2 because if a=2 then b would=3 and it did work out to be 2.

When trying this I made c=7 so when I did a+c it would=9 to find d I did d=17-c so it was 10 and by doing the other equations it worked out that: a=2 b=3 c=7 d=10

By doing that I worked out the other sums so it was: 







Also while I was doing my trial and error I worked out another way of finding the answer with the two we didn’t know being the other way round

I worked it out as the same A being the smallest and was 1 then b=4 c=8 and d=9 I did the maths and it all worked out but a+d being 10 not like the last one which was 12 and b+c=12 not 10.

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