1 to 9

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1 to 9
I wish I knew how to post diagrams in this place. This problem is somewhat difficult to describe without a diagram. Anyway, here goes...
Imagine a chessboard. Now imagine that you are singleing out a certain part of this board.
Start at the bottom left. Take the first three squares going up. Add the two first squares to the right of the top (third) square. Add the two first squares going up from the rightmost square. Add the two first squares to the right of the top square.
You should now have 9 squares in all. (Delete the background chessboard.)
Assume a coordinate system where the bottom left square is at postion (1,1).
The digits from 1 to 9 are now placed in the squares, with one digit in each square.
The sum of any three "connected" digits, vertically or horizontally, is 13.
Prove that the digit at position (3,3) is 4.
Imagine a chessboard. Now imagine that you are singleing out a certain part of this board.
Start at the bottom left. Take the first three squares going up. Add the two first squares to the right of the top (third) square. Add the two first squares going up from the rightmost square. Add the two first squares to the right of the top square.
You should now have 9 squares in all. (Delete the background chessboard.)
Assume a coordinate system where the bottom left square is at postion (1,1).
The digits from 1 to 9 are now placed in the squares, with one digit in each square.
The sum of any three "connected" digits, vertically or horizontally, is 13.
Prove that the digit at position (3,3) is 4.

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Thanks for the image, ceptimus. Could you clue me in on how you do this?ceptimus wrote:Here's an image.
http://www.mround.pwp.blueyonder.co.uk/ ... dd1to9.gif
I've proven it, but only by a bruteforce search. I'll look for something more elegant.
As far as your bruteforce method, well, I expect you know what I think of it. :)

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Well, now that the wonderful Girl 6 has added an attachment facility, it's easier than ever. The difficult part is in actually producing the image in the first place  you have to use a paint program, like photoshop, or PaintShop Pro, or the Gimp or (if you have nothing better) 'Paint' that comes as part of Windows. Then when you make a post, you will see the 'Add Attachment' thing next to the 'submit' button, and you just browse to the file you made, and attach it.DanishDynamite wrote:Thanks for the image, ceptimus. Could you clue me in on how you do this?ceptimus wrote:Here's an image.
http://www.mround.pwp.blueyonder.co.uk/ ... dd1to9.gif
I've proven it, but only by a bruteforce search. I'll look for something more elegant.
As far as your bruteforce method, well, I expect you know what I think of it. :)
.GIF images are one of the most compact ways of storing 'diagram' type images. The better paint programs allow you to make gifs with transparent backgrounds, so they look equally good (or bad) no matter what forum 'skin' a user has selected. Best to limit the gifs to about 300 or 400 pixels width when possible, so they can be displayed OK even by users with older, lowresolution displays.

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Thanks. This will come in useful when I provide my geometric proof of the "Pizza Slicing" case.ceptimus wrote:Well, now that the wonderful Girl 6 has added an attachment facility, it's easier than ever. The difficult part is in actually producing the image in the first place  you have to use a paint program, like photoshop, or PaintShop Pro, or the Gimp or (if you have nothing better) 'Paint' that comes as part of Windows. Then when you make a post, you will see the 'Add Attachment' thing next to the 'submit' button, and you just browse to the file you made, and attach it.DanishDynamite wrote:Thanks for the image, ceptimus. Could you clue me in on how you do this?ceptimus wrote:Here's an image.
http://www.mround.pwp.blueyonder.co.uk/ ... dd1to9.gif
I've proven it, but only by a bruteforce search. I'll look for something more elegant.
As far as your bruteforce method, well, I expect you know what I think of it. :)
.GIF images are one of the most compact ways of storing 'diagram' type images. The better paint programs allow you to make gifs with transparent backgrounds, so they look equally good (or bad) no matter what forum 'skin' a user has selected. Best to limit the gifs to about 300 or 400 pixels width when possible, so they can be displayed OK even by users with older, lowresolution displays.

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The easiest way, is to go and find another 'spoiler' answer (and it doesn't have to be in this thread) and hit the 'quote' button. Then you don't actually go through with the quote: select the text in the quote box, press CtrlC to copy to your clipboard, then use your browser's 'back' button to back out without posting, and come here. Start a reply and press CtrlV to paste the 'spoiler' code into your answer, then edit as appropriate before posting.Beleth wrote:How do you make the background white?
I have an answer but I don't want to spoil it.
============================================
Or here's how to do it by hand (you can cut and paste this, without quoting the message) if you prefer:
Put the 'Spoiler follows' text here (e.g. Spoiler, 'select' or CtrlA to view)
<table bgcolor="white"><tr><td>
text you want to hide goes here
(as many lines as you like)
</td></tr></table>

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I've got a simple answer.
<table bgcolor="white"><tr><td>
Label the squares a, b, c, d, e, f, g, h, i from the bottomleft up.
a+b+c+d+e+f+g+h+i = 45
a+b+2c+d+2e+f+2g+h+i = 52 (13*4)
c + e + g = 7
Since each number is only used once, the corner squares (c, e, g) must be 1, 2, and 4. If (3,3) were one or two, there would be a row with 1 at one end and 2 at the other, making it impossible for that row to sum to 13.
Hence (3,3) must be 4.</td></tr></table>
<table bgcolor="white"><tr><td>
Label the squares a, b, c, d, e, f, g, h, i from the bottomleft up.
a+b+c+d+e+f+g+h+i = 45
a+b+2c+d+2e+f+2g+h+i = 52 (13*4)
c + e + g = 7
Since each number is only used once, the corner squares (c, e, g) must be 1, 2, and 4. If (3,3) were one or two, there would be a row with 1 at one end and 2 at the other, making it impossible for that row to sum to 13.
Hence (3,3) must be 4.</td></tr></table>

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Thanks, ceptimus!
On rereading the problem, I see my solution was wrong. I proved that if you take any three adjacent numbers, that the number in the middle didn't have to be 4. But that's not what was asked.
I'll post another answer, or admit defeat, in a bit. I won't look at Cecil's answer until I do.
On rereading the problem, I see my solution was wrong. I proved that if you take any three adjacent numbers, that the number in the middle didn't have to be 4. But that's not what was asked.
I'll post another answer, or admit defeat, in a bit. I won't look at Cecil's answer until I do.

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I almost had to admit defeat but I thought of a new tack in a boring meeting yesterday.
<table bgcolor="white"><tr><td>Label the box at (1,1) "a", the box above it "b", etc. until you get to "i" at (5,5). So the problem becomes "Prove that e=4".
So, according to the problem,
a + b + c = 13
c + d + e = 13
e + f + g = 13
g + h + i = 13
Adding them all up:
a + b + 2c + d + 2e + f + 2g + h + i = 52
But since all the numbers from 1 to 9 are represented,
a + b + c + d + e + f + g + h + i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Subtracting the second equation from the first,
c + e + g = 7
The only three numbers in the set of integers we have to choose from that sum to 7 are (1,2,4). So c, e, and g are 1, 2, and 4, but not necessarily respectively.
c and e are in the same triplet, and so are e and g.
1 and 2 can NOT be in the same triplet (because, to sum to 13, the third number would have to be 10), so neither can be "e". So the only number that can possibly be "e" is 4.
Therefore e=4.
w^5.
</td></tr></table>
<table bgcolor="white"><tr><td>Label the box at (1,1) "a", the box above it "b", etc. until you get to "i" at (5,5). So the problem becomes "Prove that e=4".
So, according to the problem,
a + b + c = 13
c + d + e = 13
e + f + g = 13
g + h + i = 13
Adding them all up:
a + b + 2c + d + 2e + f + 2g + h + i = 52
But since all the numbers from 1 to 9 are represented,
a + b + c + d + e + f + g + h + i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Subtracting the second equation from the first,
c + e + g = 7
The only three numbers in the set of integers we have to choose from that sum to 7 are (1,2,4). So c, e, and g are 1, 2, and 4, but not necessarily respectively.
c and e are in the same triplet, and so are e and g.
1 and 2 can NOT be in the same triplet (because, to sum to 13, the third number would have to be 10), so neither can be "e". So the only number that can possibly be "e" is 4.
Therefore e=4.
w^5.
</td></tr></table>

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