Intermediate Maths Challenge

The numbers 1 to 25 are to be placed on a black and white 5x5 grid so that each number, except 1, is next to (horizontally or vertically), the number one less than itself. eg.

http://i11.tinypic.com/6badc07.jpg

a) Explain why number 1 must be placed on a black square
b) Explain why the numbers 1, 3, 5, 7, 9 can not all be placed on one long diagonal
c) Show that in a completed numbering the sum of the numbers on a long diagonal must be at least 33 and that 33 can occur.

Any help is appreciated!
Thanks!

Have you even attempted any of these?

a) How many odd numbers are there between 1 and 25? How many even numbers? Now look at how many black and white squares there are.

b) Given that you have to place consecutive numbers next to each other, what is the only arrangement in which 1, 3, 5, 7, 9 could be on the diagonal? What does that mean for half of the remaining squares?

c) Following on from the previous question, once the diagonal has been completed all of the squares in one half must be filled in. Including the diagonal, how many black squares is this? Since they are all odd (see part a), what is the lowest sum you can get? You know that 4 of these aren't on the diagonal, by looking at the picture. And you know that 17 must be the largest, because that's the point at which you have to start filling in the other half of the board. So what numbers do you need to make the lowest score?
Show that it is possible by an example. Since you know exactly which numbers you need, and you can see that 4 of them are consecutive odd numbers, you will have to start in one corner with 1, work down the diagonal, fill in the rest of that half of the board and end up in the opposite corner with 17. It won't take you long to work it out.



zaza

ok this is what i got, but its really random and may not really answer your question. Im also doing these problems..its hard

i sort of got a) but its really bad
a)
The number one is a odd number and must be placed on the black squares because all the odd numbers are placed on this. As well it is a 5x5 grid and this will have less even numbers making the even numbers going on a white square as there is 12 white squares and 13 black squares. To sum this up numbers have to be vertical or horizontally next to each other not diagonally as it shows a black square surrounded by white square/s or vice versa.

b)if you try this, it will show that you have filled in exactly one half of the 5x5 square but then you are unable to get to the other side. Even if you jumble the numbers around on the long diagonal it still does not work as you have to have the numbers next to each other eg 1 and 2, 2 and 3 and so on.
if you dont understand nor do i :sick:

c)the previous person stated to put the number 1 and 17 in the the long diagonal corners , i tried this but it didn't work , so im still trying to work this out.

a) More or less.
b) Your first sentence is basically the point.
c) Start with 1 in the top left corner. You now need to get 3 in the next black diagonal square - you have two ways of doing it. Then 5 in the next - another two ways. Then 7 in the next - two more ways. Then you need to go back and fill in the rest of that half before winding up with 17 in the bottom right, and you'll find that if you've chosen badly on the previous decisions you've made it impossible to do this.

Quote:

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Originally Posted by zaza

c) Start with 1 in the top left corner.

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That will never work to complete the grid with a 1-3-5-7-17 diagonal. Once you have 3 on the second square of the diagonal, you will only be able to complete the grid if you put 4 in the other white square adjacent to 1 in the corner. Thus, you will not be able to place 5 on the third square of the diagonal.

The solution to a 33-sum diagonal is 17-1-3-5-7, as in the attached image.

Quote:

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Originally Posted by Logophobic

That will never work to complete the grid with a 1-3-5-7-17 diagonal. Once you have 3 on the second square of the diagonal, you will only be able to complete the grid if you put 4 in the other white square adjacent to 1 in the corner. Thus, you will not be able to place 5 on the third square of the diagonal.

The solution to a 33-sum diagonal is 17-1-3-5-7, as in the attached image.

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Whoops, I stand corrected.

Wow thanks so much =]
and yes I did attempt these for many long hours with no results that made sense...i'm not the best at maths =[

Hey thx this also hlpd me out alot.:thumb: :eek2: