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Sunday Puzzle-36

User Rating: / 4
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Written by Administrator   
Sunday, 28 April 2013
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Puzzle courtesy of Barry R. Clarke, columnist for The Daily Telegraph and international puzzle expert

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At the livestock market, the sheep were secured in 36 pens arranged in a 6x6 grid, the number in each pen shown left. "I want to make a tidy sum out of it," said the farmer. So, in line with the farmer's wishes, the buyer bought all of the sheep in 12 of the pens. Delete 12 of the 36 numbers to leave four numbers in each row and column, so that each of the six horizontal lines and each of the six vertical lines totals 20.

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Comments (3)
1. Written by Takanome on 08:50 28 2013 .
 
 
i think the key to solve this is double numbers in each row and column..we have to take in account the possible combinations they offer in accordance with each row and column containing the same number..and that we must eliminate one of the double numbers to obtain a sum of 20 by respecting the rules. 
there's also row 3 which has 2 double numbers ,making it's only possible combo : 4+5+8+3= 20 and removing the other 4 and 5.(thus we should know which 4 and 4 to remove given the other columns ) 
however, there's one row and one column wich they don't contain a double number, these are row : 4 and column:1, so i think the start must be from there,they exclude the possibility of having at least two combinations of sums, since they have many possible answers giving this, we must AVOID counting them and keeping them last to solve since other lines and columns will be already solved, giving us the remaining numbers to remove via elimination process starting from row 3 
so i guess removing one of the (4,5 ) pair is the key since the 3rd column itself has a double number to use (unlike column 1 ). 
i may be wrong with my methodology , i think i may have overlook something, but regardless, this is the effort i could pull off..
 
2. Written by Takanome on 09:25 28 2013 .
 
 
there's a little mistake , not all of them must have a double number eliminated, since column 4 has BOTH doubles removed,and row 6 has BOTH left in the count  
it seems only row 3 is the key to start this.. testing either of the 4 and 5 pair to remove
 
3. Written by This e-mail address is being protected from spam bots, you need JavaScript enabled to view it on 00:02 29 2013 .
 
 
Simply a delightful puzzle, Barry! 
705062 
810650 
084305 
468020 
100478 
053705 
 
I went at it this way (turns out easy): 
row 1 = 28, so 2 numbers adding to 8 must be removed; 
so number 7 must stay, since it can't be used to make 8. 
Similarly in row 2: 8,6,5 must stay (big bonus!) 
And so on...
 

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