![]() ![]() In the same row (Y and Z) and two 1s in the same column (Y and EE), both of They must all be 1s, to make a total of 9 (6+1+1+1) but that would mean two 1s But if Z is 6 then what can the other squares in this cage be? We have shown in step 7 that each of Z and MM isġ, 3, 4 or 6. Possibilities even further? Yes we can! Look at the cage Y Z DD EE, which hasįour numbers adding up to 9. We can therefore be sure that Z and MM use But if Z+MM = 7 what are the possibilities?Ĭombination does not work for Z and MM. We are told that F+M+T = 9Īnd we know that FF is 5, so Z+MM must total 7 (the difference between 21 and 14 Information in the previous diagram? Do you remember how we used the total ofĢ1 for the bottom row to deduce (in step 2) that FF is 5? We can use the total Note that, because this cage must be 1, 2, 6 or 2, 3, 4, we Quick to prove that one of them does not work and therefore the other one must Of numbers are possible for a particular cage, because it can be relatively It is often very helpful to know that only two combinations Impossible here, because FF is 5, so only the two combinations 1, 2, 6 and 2,ģ, 4 are possible. Why is that? After all, we can make a total of 9 ![]() Should make a note of these two possibilities.Ĭage F M T, we can see that there are only two possible groups of three numbers KK, LL, MM is 3, 4, 6 and if GG HH JJ is 1, 3, 4, then KK LL MM is 2, 5, 6. Numbers that add to 8: 1, 2 and 5 or 1, 3 and 4. ![]() There are only two possible combinations of three different So both of these numbers are candidates in squares L and S.Īt this stage we can take stock of what we have learned soīased on the information in this diagram? There are a few things.Īll in the N U AA cage, so the remaining numbers in that column, those in A, GĪnd GG, are 1, 2 and 3.These three numbers are therefore candidates for allĭeeply into cage GG HH JJ. Numbers that are possible in this puzzle, that multiply together to give 20,Īre 4 and 5. Have only one possible combination of numbers: the L S cage. So AA can not be 5, it can only be 4 or 6. Therefore have the three “candidates” 4, 5 and 6. Of three different numbers from 1 to 6 will do. Use three different numbers in this puzzle to total 15 is 4+5+6. Now we look for any cages in which only one combination of Now we know that FF+KK+LL+MM =ġ8, and by subtraction we can see that FF must be 5 (18-13). Six numbers in the bottom row add up to 21 (1+2+3+4+5+6). We first fill in the given numbers: 2 in J and 3 in V.Īny squares for which only one number is possible. This tutorial, by David Levy, KENKEN’s technology expert, appears in KENKEN: Books 1 - 4: The New Brain-Training Puzzle Phenomenon, which HarperCollins in Britain published. Here, we present several of them as we walk through solving this 6 x 6 grid, step-by-step. So you’re breezing through the easier puzzles, huh? Ready to step up to the big leagues? Before you do, there are some more advanced solving techniques that will be very helpful to have in your arsenal. ![]()
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