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Recreational Maths
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=17&t=21
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Author:  Para [ Sun May 25, 2008 10:15 am ]
Post subject:  Re: Recreational Maths

Yeah, yeah. I was just looking for the answer. Not really trying to prove why that had to be the answer. My maths education has stopped since high school. Just found it an interesting problem to tackle :).

Author:  udosuk [ Mon May 26, 2008 7:26 am ]
Post subject:  Re: Recreational Maths

Okay, having received a little bit more of Maths education :study:, here is a complete walkthrough to "logically, manually and arithmetically" solve this problem (while proving uniqueness of the answer):
Hidden Text:
First notice that we need to get 10 numbers {10..19} from the sums & products of the 3 rows, 3 columns & 2 diagonals. That means we need at least 2 products.
Any product without 1 is at least 2*3*4=24.
Any product without {23} is at least 1*4*5=20.
=> We can only use products with 1 and at least one of {23}.
Also note that the corresponding sum of a product must be from {10..19}.
The only possible triple with 1 & 3 is {136}, giving 1+3+6=10 & 1*3*6=18.
And then we need another triple, which must be with 1 & 2, to generate a sum and a product from {11..17,19}
The only possible triple is {128}, giving 1+2+8=11 & 1*2*8=16.
And with these no more products are available, the remaining 6 numbers must all be given by the sums.

Now we have the 8 sums as 10,11,12,13,14,15,17,19 (16 & 18 have been taken care by the products).
45-rule: all 3 rows sum to 45, so as all 3 columns.
So the total of the sums of the 3 rows & 3 columns = 45+45 = 90.
From the above list of 8 numbers they must be {12,13,14,15,17,19}, in other words 10 & 11 must be presented as sums of diagonals.
Therefore {128} & {136} are the diagonals, forcing 1 to be the centre cell.

Without loss of generality, the square must be of the following pattern (all other cases are reflection/rotation of it):
Code:
2     4579  3
4579  1     4579
6     4579  8

To give the sum of 19, r3c123 must be [658].
To give the sum of 17, r123c1 must be [296].
To give the sum of 15, r123c3 must be [348].
Thus r1c123=[273] to give the sum of 12.
r2c123=[914] to give the sum of 14.
r123c2=[715] to give the sum of 13.

Hence the square must be:
Code:
2 7 3
9 1 4
6 5 8

Pretty easy compared to a typical Killer Sudoku, don't you think? ;)

Author:  Glyn [ Mon May 26, 2008 6:56 pm ]
Post subject:  Re: Recreational Maths

Of course at least 4 of the values must be sums:-

Hidden Text:
11,13,17 and 19 are all prime

Author:  udosuk [ Wed May 28, 2008 5:39 am ]
Post subject:  Re: Recreational Maths

Glyn wrote:
Of course at least 4 of the values must be sums:-

Hidden Text:
11,13,17 and 19 are all prime

Following that line here is a different approach:
Hidden Text:
First establish that at most 2 products can be used (the 1st from 1,2,x and the 2nd from 1,3,y).

So all 8 sums must be used (along with the 2 products) to complete the set {10,19}.

Obviously 11,13,17,19 can't be from products, they must be from sums.

Thus 10,12,14,15,16,18 come from 4 sums, 2 products.

10 = 1*2*5
12 = 1*2*6 or 1*3*4
14 = 1*2*7
15 = 1*3*5
16 = 1*2*8
18 = 1*2*9 or 1*3*6

Sums must be at least 10
=> 1+2+5, 1+2+6, 1+3+4, 1+3+5 can't be the sums
=> 10,12,15 can't be from products, must be from sums

Hence 14,16,18 come from 1 sum, 2 products.

14 = 1*2*7
16 = 1*2*8
18 = 1*2*9 or 1*3*6

1*3*6 must be used to give one of the 2 products.
=> 10=1+3+6 and not 1+2+7
=> 14 can't be from a product, must be from a sum
=> 16 must be from a product 1*2*8

The rest is the same as my earlier walkthrough.


I posted some other similar challenges :arrow: here. But most of them requires a brute force approach using computers (I guess).
:ugeek:

Author:  udosuk [ Mon Jun 16, 2008 2:29 am ]
Post subject:  Re: Recreational Maths

rcbroughton wrote:
One of my very favourite Mathematical puzzles is one that my father showed to me about 20 years ago - and it was old, even then.

If you haven't come across it before, it's a real gem. It just doens't look like you have enough information to make any headway but very rewarding when you crack it.

Image

Code:
Across
1   Area in square yards of the Paddock
5   Age of Martha, Squire Root’s aunt
6   Difference in yards between the length and the breadth of the Paddock
7   Number of Roods in the Paddock times 8 down
8   The year the Root family acquired the Paddock
10   Squire Root’s age
11   Year of Mary’s birth
14   Perimeter in yards of the Paddock
15   Cube of Squire Root’s walking speed
16   15 across minus 9 down


Down
1   Value in shillings per rood of the Paddock
2   Square of the age of  Squire Root’s mother-in-law
3   Age of Mary, Squire Root’s daughter
4   Value in pounds of the Paddock
6   Age of Ted, Squire Root’s son, who will be twice the age of his sister, Mary in 1945
7   Square of the breadth of the Paddock
8   Time in minutes it takes Squire Root to walk 1+1/3 times around the Paddock
9   10 down divided by 10 across
10   10 across multiplied by 9 down
12   Sum of the digits of 10 down plus 1
13   Number of years the paddock has been in the Root family


Useful Information
The Paddock is a rectangular plot of land
The year is 1939
1 acre = 4840 square yards = 4 roods
1 mile = 1760 yards
1 pound = 20 shillings

Finally worked out the solution. :ugeek:
The solution:
Code:
38720.5
4.91.44
0.2.384
.1110..
72.1918
9...792
27.16.9

I believe the little typo in the "8 Down" clue (1+1/3=4/3 and 11/3=3+2/3 are very different :rambo:) made it impossible for everyone else to find the solution. :evil:

Otherwise, quite enjoyable once you got everything worked out. ;)

Author:  Bigtone53 [ Mon Oct 06, 2008 10:58 pm ]
Post subject:  Re: Recreational Maths

udosuk,

I think that I may have mentioned elsewhere my enthusiasm for Gardner, ever since I started seeing his Scientific American articles over 40 years ago.

A real polymath (still apparently with us at 94). His interest in impromptu magic and the deep research into annotating things like Alice in Wonderland also touched nerves. A good man :salute:

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