Monday, 9 February 2009

A Blog of Two Problems (with apoligies to C. Dickens)

One of my students sent me a nice mathematical puzzle from The Number Warrior by Jason Dyer from the US Puzzle Championship in 2001. The origninal problem was stated this way, "
You have a set of five numbers. When adding together each of the ten possible pairs, you get the following sums:

0, 6, 11, 12, 17, 20, 23, 26, 32, 37

What are the five original numbers?"

I am going to give the way I solved this below so don't read on too much more if you want to work this out by yourself. After the solution I have a couple of variations that occurred to me that I don't have worked out yet, but first I wanted to post a problem Al Harmon sent me from Mr. Milanovich, a teacher at Edgren HS where I used to teach in Misawa, Japan


A right triangle has a perimeter of 60 and the altitude to the right angle is 12. Find the area. ...

I actually was stumped by this problem and had to buy a clue, so if you want the clue, I will put it in the comments, and I will give a specific and general answer to this in my next blog.

Now for the five number problem above.. It is obvious from the fortuitous zero as one of the sums that one of the numbers is negative, and that it is the opposite of the smallest of the four positive values that remain, otherwise there would be one of the sums that was negative.

The next big jump was a statistics idea. I keep trying to remind my students that the mean of the sampling distribution is the same as the mean of the population. A sampling distribution is the set of all the possible measures you could get by taking every possible set of n values from the population. The numbers above represent the sampling distribution of sums from samples of size two. If we add up all ten numbers we get a total of 184, so the average of these "samples of size two" is 18.4; from which we can conclude that the average of the five original numbers is 9.2, and so they must total to 46. But the smallest two add up to zero. From this we can extract that the three largest (call them a, b, and c) add up to 46 also. We know that the two largest add up to 37, so 46-37=9 must be the middle number of the five. At this point I assumed that the 32 represented a+c, the sum of the first and third largest numbers, so if I subtracted 9, I would get the largest number, 23. If that were correct, I would be left with 23+b=37, so the second largest number must be 14. Now looking at the second smallest total, 6, I figured that it had to be the sum of the middle value, 9, and the smallest number. Subtracting 9 from 6 gave -3 for the smallest number, and 3 for its additive opposite.


I then had {-3, 3, 9, 14, 23} for the solution set, and needed only to check that they produced ten distinct sums as given above.

Variations and extenstions:

It came to me as I worked with this problem that it was not necessary for a collection of five distinct numbers to produce ten distinct sums when paired. for example if the set included 1, 3, 4, 6 we would have two sums of 7. It occurred to me to wonder first, what is the probablilty that a set of five distinct integers had ten distinct sums when paired in every possible way. To be answerable, the problem must have some domain for the original values, so I would assume 1-30 would be good for a start. It would seem as if the probability grew as the domain interval grew wider, but I have not actually calculated that out yet. I have not worked an answer for this out yet, and in truth, do not yet have a good idea for an attack other than writing a program to count them.. Anyone who has suggestions (or an answer)... please respond.

I also wondered aobut the probability that any increasing sequence of ten numbers being the sums of five distinct numbers. I am also sort of lost and where to attack this problem, but think it must have something to do with differences in the sums. Again, if you have an idea about a way to attack this, please advise.

Ignore this, it is just a test to see if the equation editor on google will copy to format
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