I've got my head around this now and can explain the 'method' as shown on the programme by Marcus de Sautoy.

Apologies if this was already obvious to you. (It wasn't for me!)

Since the grid is parallel to the axes it is possible to consider x distances and y distances separately.

I'll explain for x. y is done similarly.

Count how many students, N. If N is odd find the middle student. (In the 13 student case, look at student number 7.)

Choose this x coordinate as the x coordinate for the meeting place.

If you move this meeting place left by one square, then 6 (or less) students are now one square nearer, but 7 (or more) students are one square further away. So the distance total has gone up. Similarly, if you move right by one square.

The distance continues to increase if you move even further from the middle student position. So this position, which Marcus called the median, is the minimum position.

If N is even, do not do what is usually done for a median, namely take the value midway between. Rather choose both students as jointly occupying best positions. In the 14 student example that means considering the positions of the 7th and 8th students. In that case, we get 3 x coordinates that have equal minimum distances and similarly for y, giving the 9 equal solutions to the problem.

And that's it.

Thanks once again for helping me sort this out.

Bob

]]>I am trying to devise a method of my own.I am trying to see how the coordinates of the points we are looking for changes when we add a point or add 1 to an existing point.

]]>I need to examine his method more closely, now I have two examples. I expect you want to know what it is.

I'll do some work on it first and post later when I've got this clear in my head.

Bob

ps. I'm preparing a post on the Fundamental Theorem of Calculus for amberzak. So far I haven't got past my preamble so I think I may go quiet for a while.

]]>I got 51 at (5,5).I think the guys who said that might be correct.

]]>Bob

]]>I see it now there is only one solution now! Thanks for posting that.

]]>That's the corrected diagram.

Bob

]]>If you reduce the number of students at (3,7) to 4 you will get a unique answer.

]]>(i) I recorded the programme on my pvr so I don't need to wait.

(ii) There are 13 students as shown on the revised diagram below.

So sorry.

Bob

ps. Tt's a good programme if you can get it. There may be some U tube clips.

]]>Okay,tell us what happened.Maybe we can sue them.

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