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Sudoku Mahabharat - Substitution and Neighbours (7^{th} - 12^{th} Feb) Score •Discuss

Puzzle Ramayan - Number and Object Placement(21^{st} - 26^{th} Feb) has started •Discuss

Puzzle Ramayan - Number and Object Placement(21

Sudoku HotPot - Incomplete Killer Solution Steps | |

LMI Essentials -> Solving Techniques | 9 posts • Page 1 of 1 • 1 |

purifire |
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Odd Even & Twisted Classics (SM 16/17) Author Posts: 458 Location: India | The single digit boxes can contain only 1 and 2. in Box 9 we have a cage (R7C9, R8C9) with sum 4 (1,3) hence in Box 6 we have 2 and in Box 4 we have 1 in the single cages. | ||

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Odd Even & Twisted Classics (SM 16/17) Author Posts: 458 Location: India | In Box 3, R1C9 and R2C9 has sum of 9. This can only be 4,5 since Column 9 already has used up 1,2,3. We can directly fill up single sums of 3 (R5C3) and 6 (r5C7). The middle row in Box 5 has to be 1,2,4 only for a sum of 7. R3C9 has to be 6 and for a sum of 9, R3C8 = 2 and R4C8 = 1. | ||

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Odd Even & Twisted Classics (SM 16/17) Author Posts: 458 Location: India | Now comes the tricky part... Box 5 already has a cage with sum 7. All the cages have to add up to 45, so the cage of (R4C4,R4C5,R4C6) (and R6C4,R6C5,R6C6) together add upto 38. Since the max of the missing sums is 20, one of this cage is 18 and the other cage is 20. The Row 4 cage cannot be 20 coz then R4C7 will become 2 which is already there is R4C9. Hence R4C7 is a 4 and R4C4,R4C5 and R4C6 has a sum of 18. This allows for R4C1,R4C2,R4C3 to be 5,6,9 for a sum of 20. R5C1,R5C2 has to be 5,8 and R5C8,R5C9 has to be 5 or 9. | ||

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Odd Even & Twisted Classics (SM 16/17) Author Posts: 458 Location: India | Since 5 is already there in either R1C9 or R2C9, we have R5C9 = 9 and R5C8 = 5. this allows us also to fill R9C8 = 4 and R9C9 = 8 and also R6C9 = 7. So till now we have used 1,2,14,15,18 and 20 of the missing sums. We are now left with 3,4,7,9,11,12,13,16,17,19. Edited by Rohan Rao 2012-05-27 5:21 PM | ||

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Odd Even & Twisted Classics (SM 16/17) Author Posts: 458 Location: India | in Box 9, we already have 2 cages with combined sum of 16. Hence the other 2 cages with missing sums add upto 29. From the remaining sums this 29 can either be 17,12 or 13,16. the pair of 12,17 is eliminated as the 2 cell cage cannot be 17 since 8 is already there in R9C9 and it cannot be 12 also since one of the numbers in 6. Hence the 2 cell cage has a missing sum of 13 (6,7) and the 3 cell cage has sum of 16 (2,5,9). so now we are left with missings sums of 3,4,7,9,11,12,17,19. I guess from here on the puzzle solves logically :) but if anyone wants me to continue I will be glad to do it. Rishi | ||

debmohanty |
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Location: India | thanks for the steps rishi. The puzzle is indeed very beautiful, both the concept and the particular puzzle. | ||

neerajmehrotra |
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Posts: 325 Location: India | this is grt!!!!!!!!! | ||

avermaful |
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Posts: 27 Location: india | rishi, hi.my name is dhaval.i m abig fan of yours.but this killer sudoku is easy as it contains one cell sum like 3 and 6 .can u explain a difficult one completely?please. i m a starter. also i find killer sudokus tough. so please can u explain a more difficult one and in easy langusge just like above one. it would be helpful to everyone.pls!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! | ||

avermaful |
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Posts: 27 Location: india | rishi. also please can u solve the sudoku completely which u left above and explain it.pls!!!!!!!!!!!!!!!!!! i know i m asking tooo much.ut plaese rishi sir!!!!!!!!!!!!!!!!!!!!!!!!! | ||

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