# 1-10 Numbers Logic Puzzle (#2) 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)
E4 tells you that A4 is X.

(2/20)
According to E4, C1 cannot be adjacent to more than four numbers. A4 tells you that C1 is not adjacent to the 1, the 2, the 3, or the 4. This means that C1 cannot be adjacent to the number that is the quantity of numbers adjacent to it. So C1 is X.

(3/20)
E2 x D3 = C4. The only numbers that could work are
2 x 3 = 6
3 x 2 = 6
2 x 4 = 8
4 x 2 = 8
2 x 5 = 10
5 x 2 = 10
The last two actually don’t work because they require both multiples of 5, at least one of which is adjacent to C1, according to A4. But the bottom line is 2 is going to be used in either E2 or D3. Now take a look at A1. According to E4 it can be adjacent to up to two numbers, meaning, if it is a number, it will need to be adjacent to the 1 or 2. You just established that the 2 is nowhere near. A4 says that 1 cannot be in Column A. It also says that neither B1 nor B2 is any number under 5. So A1 cannot be adjacent to the 1 or the 2, meaning it cannot be a number; it is X.

(4/20)
Apply similar logic as the previous step to B1. If it is a number, it must be adjacent to the 1, the 2, or the 3. No number adjacent to C1 is less than 5 and the A column has no odd number. And since the 2 is either E2 or D3, B1 cannot be adjacent to the number that is the quantity of numbers adjacent to it. So B1 is X.

(5/20)
B1 tells you that Row 3 is all numbers. E4 tells you that D4 is a number. C1 tells you that C4 is a number. Therefore, according to E4, B4 must be X or else C4 is not adjacent to any X.

(6/20)
You know from B1 that A3 and B3 are numbers. So A3 must be adjacent to the 1, the 2, or the 3.
If A3 is only adjacent to one number, then it is adjacent to the 1 and it would have to be B3, since B3 is definitely a number.
A3 cannot be adjacent to two numbers because then it would be adjacent to the 2, but the 2 is E2 or D3, according to C1.
A3 cannot be adjacent to three numbers because then it would be adjacent to the 3, but A2 cannot be odd (A4) and B2 and B3 cannot be prime (B4).
So A3 is adjacent to just one number, and you know B3 is a number so it must be the 1. A2 and B2, of course, are both X.

(9/20)
Consider B3. You know that A3, C3, and C4 are numbers. C2 could be, so B3 is either adjacent to three numbers or four numbers, meaning it must be adjacent to the 3 or 4. C4 is either 6 or 8, according to C1. C2 is either X or 5 or greater, according to A4. So the number that is the quantity of numbers adjacent to B3 is definitely in C3. So if C3 is 3, C2 would be X. E3, which you know is a number would not be adjacent to the 3, so would have to be adjacent to four numbers and to the 4. So D2 would be a number and the final X would be D1 or E1. But then both C3 and D2 would be numbers that are adjacent to five numbers, so they would both have to be adjacent to the 5. C3 and D2 are both adjacent to C2 and D3. In this scenario C2 is X. D3 has been established as 2 or 3 or 4. So the scenario does not work. C3 cannot be 3, so B3 must be adjacent to four numbers, meaning C3 is 4.

(10/20)
C1 tells you that E2 x D3 = C4. Since 4 is out, it  must be 2 x 3 = 6 or 3 x 2 = 6. So C4 is 6.

E3 is a number (E4) that is adjacent to four unknowns that could be numbers. But E3 is not adjacent to the 4, so it must be adjacent to at least one unknown X. E4 tells you that  D3 and D4 are definitely numbers and C1 tells you that E2 is a number. So the only box adjacent to E3 that could be X is D2.

(12/20)
You have placed nine X’s. You know B3 is adjacent to four numbers, so C2 is a number. C1 and E4 tell you that E2, D3, E3, and D4, are all numbers. So the last remaining X is D1 or E1. If D1 is the final X, then E1 would be adjacent to only one number, but E1 clearly is not adjacent to the 1. So D1 is not the final X; E1 is.

(13/20)
Now every unknown box must be a number since all X’s are placed.

D1 is adjacent to two numbers, so it must be adjacent to the 2. C2 cannot be 2, according to A4, so E2 is 2.

Since D1 and C2 are both numbers, one of them must be 5, according to A4. There’s no way 10 can be the smallest number adjacent to C1 unless it is the only number adjacent to C1. C3 is adjacent to five numbers, so C2 must be 5.

(15/20)
E2 being 2 means, according to C1, that D3 is 3.

(16/20)
A3 is not an odd number, according to A4. 2, 4, and 6 are already placed. D3 tells you that the 10 is either in D1 or E3. So A3 must be 8.

D3 says the 10  is either in D1 or E3. C2 says the 10 is not in D1, so it is in E3.

(18/20)
The 8 is already known, so E3 must be referring to 3+7. So D4 is 7.

(19/20)
The only remaining number is 9. It goes in D1.

(Back to puzzle)

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