Ten Brilliant Mathematical Problems

[Andrew Camp]

http://www.dailymail.co.uk/home/moslive ... zzles.html

I love number 3.

[Alexander Hardwick]

Bearing in mind number 2, should this really have been posted in the "Not Chess!" section anyway?

[Alex Holowczak]

5) is part of the D1 syllabus for A Level Maths. It has by far the most simple solution of the problems listed there!

[Richard Thursby]

5) is part of the D1 syllabus for A Level Maths. It has by far the most simple solution of the problems listed there!

I would have thought that would be number 4, especially since they have basically written out the solution.

Bearing in mind number 2, should this really have been posted in the "Not Chess!" section anyway?

One might consider renaming this board "Not (Necessarily) Chess!"

[Alex Holowczak]

5) is part of the D1 syllabus for A Level Maths. It has by far the most simple solution of the problems listed there!

I would have thought that would be number 4, especially since they have basically written out the solution.

Yes, good point. I solved #4 when asked it a few months ago (by a chessplayer, no less) in a few seconds... That's probably GCSE level.

[Keith Arkell]

Another chess connection: Sam Loyd's 1914 version of number 7(the fifteen puzzle) is referred to. This is presumably the chess composing,chess playing Sam loyd!

[Matthew Turner]

Keith,
Yes it is. A lot of people have heard about his chess puzzles and a lot of people have heard about his maths puzzles, but surprisingly few have heard about both.

[Nick Thomas]

5) is part of the D1 syllabus for A Level Maths. It has by far the most simple solution of the problems listed there!

I would have thought that would be number 4, especially since they have basically written out the solution.

Bearing in mind number 2, should this really have been posted in the "Not Chess!" section anyway?

One might consider renaming this board "Not (Necessarily) Chess!"

Quite right Richard. Every schoolboy/girl with a reasonable grasp of GCSE Maths should be capable of solving this. Alex. Was it me who tested you on this?

[Alex Holowczak]

Alex. Was it me who tested you on this?

It was indeed. I came up with the formula 2*pi*r = 2*pi*(r+1), and you told me the answer as I was cancelling it down in my head to get to 2pi...

[Michael Jones]

5) is part of the D1 syllabus for A Level Maths. It has by far the most simple solution of the problems listed there!

I would have thought that would be number 4, especially since they have basically written out the solution.

4 is certainly the easiest solution, and 2 and 3 probably aren't any harder than 5 either - summing a series and solving a recurrence. I once worked out a proof of 7 for another forum, which it's my pleasure to reproduce here:

The disorder parameter, call it D, is defined as the number of pairs of numbers which are in the wrong order. Clearly, for the correct arrangement D = 0 and for that with only one pair swapped, D = 1.

In any given position, there are at most four possible moves: move a block up, down, left or right into the empty space (obviously if the empty space is at an edge or corner then there will be fewer than four). Consider the effect each of these moves has on the value of D. A left or right move doesn't change the order of the blocks, and thus doesn't change D; so we only need to consider up and down moves. A down move takes one block and moves it to a position after three other blocks which it had previously been before, whilst leaving the orders of all other blocks unchanged. Consider the four possibilities:

i) All 3 blocks should have been after the block which was moved; in this case 3 new "wrong pairs" have been created and none removed, so D increases by 3.

ii) 2 blocks should have been after the block which was moved and 1 should have been before; in this case 2 new "wrong pairs" have been created and 1 has been removed, so D increases by 1.

iii) 1 block should have been after the block which was moved and 2 should have been before; in this case 1 new "wrong pair" has been created and 2 have been removed, so D decreases by 1.

iv) All 3 blocks should have been before the block which was moved; in this case 3 "wrong pairs" have been removed and none created, so D decreases by 3.

The main point to note is that in each case D changes by an odd number. Since an up move is just the reverse of a down move, the above argument applies in reverse, so for an up move D also changes by an odd number. Clearly if the empty space ends up in the same row in which it started, there must have been the same number of up moves as down moves. Since the net result of one up move and one down move is an even change in the value of D, the same is true of any number of up moves and the same number of down moves.

Conclusion: any arrangement which is derived from the correct one and which has the empty space in the bottom row must have an even value of D. The arrangement with only 14 and 15 swapped has the empty square in the bottom row and an odd value of D, therefore it cannot be derived from the correct one. QED.

The proof can be generalised: if you have a m x n rectangle with mn-1 numbered blocks and one empty square, such that all the blocks are in numerical order from left to right and top to bottom except for mn-2 and mn-1 being reversed, then no sequence of up, down, left and right moves can transform that arrangement to the one with all the blocks in the correct order and the empty square in the same place.