Posts: 148
 
Location: France | It is really easy, actually.
1) We know that the values of these two cages end with the same digit; hence they are both even or both odd.
2) The sum of two even numbers is even. The sum of two odd numbers is also even. So, whatever may be the value of X, the sum of the values of these two cages is even.
3) The sum of the digits from 1 to 9 is equal to 45; so, the value of R2C5 is equal to 45 minus an even number; hence, an odd number; hence, 9.
I hope it is clear enough.
Bastien |
Posts: 25
Location: United States | sena - 2014-01-19 9:40 AM
In No X is alive, author has given the clue that the sums of both cages in block 2 must be even. why? How to manipulate that?
Could you please explain? When the remaining cell can be 2,4,6,8 or 9 how do you choose 9?
First, the idea that the sum of the two cages in block 2 (top, center 3x3 region) must be even:
Note that it is not that each individual cage total must be even, just that their sum is even. They both
end with the same digit ("X"), so their sum is even.
Those two cages are everything in block 2 except the center square, and the sum of all digits in a block is 45. Thus
the remaining digit must be odd (45 minus some even number).
The remaining digit must be 9 since we know it is odd, and 1,3,5 and 7 are all eliminated by the placement of those
digits in the same row/column of the one we're looking for.
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