# 1-12 Numbers Logic Puzzle Walkthrough

EXPLANATION OF RULE IN HOW TO PLAY

Each cell that contains a number is adjacent to the number that corresponds to how many numbers are around it.

In the crude diagram below, you see that the 4 is adjacent to 5 cells. Two of them are X’s. If you find out that the other three have to be numbers, then one of them must be the 3.

|X|  |  |

|4|  |  |

|X|  |  |

In the next diagram, suppose you know that all the X’s are already on the grid and that the top row is all prime numbers. What would the next move be?

|X|  |2|  |  |  …

|4|  |7|  |  |  …

|X|  |3|  |5|  …

It would be to put the 6 to the right of the 7 because the 7 is adjacent to six numbers and 6 is not prime so cannot go in the top row. Got it? Remember that this rule is not in play for X’s.

(Back to puzzle)

WALKTHROUGH SECTION BELOW

(1/20)
There are 20 cells and 12 numbers. So there are 8 X’s. D2 says that B3 is adjacent to half X’s. If it’s talking about half the total X’s, that would be 4. If it’s talking about half the cells adjacent to B3, that would also be 4. So B3 is adjacent to 4 numbers, and therefore also 4 X’s. Since all numbers are adjacent to the number that is the quantity of numbers adjacent to it, B3 would need to be adjacent to the 4. But it’s not. This means that B3 cannot be a number, so it is X.

(2/20)
D2 tells you that A1, B2, C3, and D4 are all numbers, not X’s. B3 tells you that A3 is a prime number. You know from D2 that B3 is adjacent to four numbers. Now you know that three of them are at A3, B2, and C3.
Take a look at A4. If A4 were to be a number, it would be the final of the four numbers adjacent to B3. Those four would be A3, A4, B2, and C3. This would make B4 an X. So A4 would be a number that is adjacent to only one number, meaning it would have to be adjacent to the 1, according to the main rule. So A3 would have to be the 1, but A3 is prime, so it cannot be 1. So A4 cannot be a number; it is X.

(3/20)
If B4 were to be a number, the four numbers adjacent to B3 would be A3, B2, B4, and C3. A2, C2, and C4 would be X’s. This would mean that both A3 and B4 would be adjacent to two numbers, so they would both have to be adjacent to the 2. A3 and B4 only share two adjacents: A4 and B3; both are X’s. So A3 and B4 cannot both be adjacent to the 2, meaning B4 cannot be a number; it is X.

(4/20)
If B1 is a number and A2 is a number, A2 would have to be adjacent to the 4, since you know that A1, B2, and A3 are all numbers. The 4 is too far away, so if B1 is a number, A2 is not. Then A3 would be adjacent to one number, so B2 would be 1. A1 would be adjacent to two numbers, so B1 would be 2. Then B1 would have to be adjacent to a third number, but not a fourth because the 4 is too far away. So C1 and C2 would be 3 and X or vice versa. So with the 1, 2, 3, and 4 not adjacent to D4, D4 would never be able to be adjacent to the number that is the quantity of numbers adjacent to it. Five would not work either because that would place the 3 at D3, since C4 and E4 would both be adjacent to three numbers, but 3 is used elsewhere in this scenario. So B1 is definitely not a number, but X.

(5/20)
Applying similar logic, if A2 is a number, it would have to be adjacent to the 3. A1 and A3 would both be adjacent to two numbers, so B2 would be 2. Again, D4 is a number and the 2, 3, and 4 would be too far away. The only hope would be for C3 to be the 1 and D3, E3, C4 and E4 all be X’s, but there are not that many X’s, because if A2 is a number, then C2 would be X, according to D2. So A2 cannot be a number either; it is X.

(6/20)
You know from B3 that A3 is a number. It is clearly only adjacent to one number, meaning it has to be the 1. B2 is 01.

(7/20)
B2 gives you E3 as X.

(8/20)
You learn from B2 that E4 is a number and gather that it must be adjacent to two numbers: D3 and D4. You have known from D2 that C3 and D4 are numbers. So D4 is adjacent to at least three numbers. D4 is not adjacent to the 4, so it cannot be adjacent to a fourth number, meaning C4 is X.

(9/20)
Now you know that the four numbers adjacent to B3 (according to D2) are A3, B2, C2, and C3. At this point you also have only one more X to find. The only cells that you do not know for sure to be numbers are C1, D1, E1, and E2. You can test them to see that most of them would work as a number. But if E1 were a number, it would need to be adjacent to the 2 or 3. You know that the 2 must be adjacent to E4, too far from E1. You also know that D4 must be adjacent to the 3, also too far from E1. So E1 cannot be a number; it is the final X.

(10/20)
The remaining cells are all numbers. D4 and E2 are both adjacent to three numbers, so the 3 must go in D3.

C2, D2, and D3 are all adjacent to six numbers. The only cell adjacent to all three of them is C3, so C3 is 6.

(12/20)
E4 is adjacent to two numbers. D3 is not 2, so D4 must be 2.

(13/20)
C3 is adjacent to five numbers, so the 5 must go in C2.

(14/20)
D4 tells you that the 7 is in Column E. You’ve now placed all prime numbers but 11. So according to B3, 11 must go in A3.

(15/20)
Right now, the sum of the numbers in Column C is 11 and the sum of the numbers in Column D is 9. This means, according to C2, that D1 is two greater than C1. The remaining numbers are 7, 8, 9, 10, 12. This means your options for C1 and D1 are 8 and 10 or 10 and 12. So 10 is in C1 or D1 for sure.
A3 tells you that E2 + E4 = an odd number. You know from D4 that one of those numbers is 7, so the other number must be even: 8 or 12. So C1 and D1 are some combination of 10 and 8/12. E2 and E4 are some combination of 7 and 8/12. This means the only remaining cell that can be 9 is A1.

(16/20)
As just established, E2 is 7 or 8 or 12. According to A1, if E2 is 8 or 7, then E4 would have to be less than that. 7 is the lowest number remaining, so if A1 is greater than E2, E2 is 8 and E4 is 7. What if A1 is not greater than E2? 10 has to be C1 or D1, so if A1 is not greater than E2, E2 would be 12. That would make C1 and D1 8 and 10 and so E4 would be 7. Either way, E4 is 7.

(17/20)
E4 tells you that the 12 is in Row 1. So it must be D1, with C1 being 10, according to C2, and E2 being the remaining 8.

(Back to puzzle)

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