**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?

**WALKTHROUGH SECTION BELOW**

1.) E2 tells you that A1 is X.

2.) A1 tells you that D1 is a number 3 or greater and that E1 is a number. You know that E1 is adjacent to either one or two numbers, depending on whether D2 is a number or an X. So E1 must then be adjacent to the 1 or 2. D1 can’t be either of those, according to A1, so that means that D2 must be the 1 or 2. But this means that E1 is adjacent to a second number; therefore D2 must be the 2.

3.) E4 is only adjacent to three cells, so according to D2, E3 and E4 must both be adjacent to one or two or three numbers. E3 is already adjacent to one number: the 2, and E2 tells you that E3 is a number, so E4 is also already adjacent to one number: whatever E3 ends up being. If they were both adjacent to just one number, it would have to be the 1.

If E3 and E4 are both adjacent to two numbers, they would both be adjacent to one of D3 or D4; the other would be X and E4 itself would be X.

If E3 and E4 are both adjacent to three numbers, D3 and D4 would have to be numbers, and E4 itself would have to be X. The bottom line is that any way you do it, E4 has to be X.

4.) E4 tells you that B1 is 3.

4.) At this point, only three cells can possibly be adjacent to eight numbers. All the rest are adjacent to fewer than eight total cells or otherwise are already adjacent to at least one X. So according to E4, the 4 must be in B3, C2, or C3. Now you know that E3 is a number from E2. You also know that the rest of the Row 1 is numbers. You still have four X’s to place that can only go in A2, A3, A4, B4, C4, and D4 If B3 or C3 were adjacent to eight numbers, that would force too many of the cells in the bottom row to be numbers; there wouldn’t be a place to put all of the remaining X’s. Therefore it is C2 that is adjacent to eight numbers, so C2 is the 4.

5.) Since the 4 is adjacent to 8 numbers, B2 is a number. A1 told you that C1 is a number. So B1 is adjacent to at least three numbers. It must be adjacent to the fourth, or it would need to be adjacent to the 3, but it is the 3. So A2 must also be a number. Since A2 is a number, it must also follow the main rule. It is not adjacent to the 4, so it cannot be adjacent to four numbers. You know it is adjacent to three numbers: B1, B2, and B3, since anything adjacent to the 4 is a number. So since A2 cannot be adjacent to a fourth number, A3 must be X.

6.) Row 1 has four numbers (A1). C2 is a number and is surrounded by all numbers (E4). A2 is a number (step 5 above). E3 must also be a number according to D2, because every cell adjacent to E4 is also adjacent to E3, except E3 itself. Since E3 is adjacent to a number that E4 is not adjacent to (D2), E4 must be adjacent to a number that E3 is not adjacent to in order for those two cells to be adjacent to the same number of numbers. So to recap: the following are all numbers or must be numbers: B1, C1, D1, E1, A2, B2, C2, D2, B3, C3, D3, and E3. That’s 12 of the 13 numbers. The only place for the final one is in the bottom row. D4 is adjacent to three numbers: C3, D3, and E3. So if the final number goes in C4, it would have to be the 4, but the 4 is already placed. So C4 is X.

7.) Now you know for sure that D4 is adjacent to exactly three numbers. The 3 is not adjacent to D4, so D4 cannot be a number; it is X.

7.) B4 is adjacent to two or three numbers: B3 and C3 (and possibly A4). Neither the 2 nor the 3 is adjacent to B4, so B4 cannot be a number; it is X.

8.) All the remaining cells are numbers, so the final one is A4. A4 is only adjacent to one cell, so that one cell must be the 1. Therefore B3 is the 1. Type “01.”

9.) C1 and C3 are both adjacent to five numbers, so they must both be adjacent to the 5. Therefore the 5 must go in B2.

10.) B2 tells you that the 6 and the 7 are both in Row 1 or both in Row 3. Therefore, according to B1, they cannot be in Row 3, for then they would both be adjacent to an X or 2 or 3. B2 is adjacent to six numbers, so one of them must be the 6. Therefore the 6 goes in C1.

11.) The remaining numbers are 7-13. C1 tells you that six of those are divided into equal pairs. Those pairs could equal 19: 7+12, 8+11, and 10+9 (and so 13 would go in D1). Those pairs could also equal 20: 7+13, 8+12, and 9+11 (and so 10 would go in D1). Or those pairs could equal 21: 8+13, 9+12, and 10+11 (and so 7 would go in D1).

If you use the first pair, no row could equal 33. It has to be Row 3 anyway, but if D3+E3=19, C3 would have to be 13, but in the first scenario, 13 is in D1.

If you use the second pair, 10 would go in D1, so 7 would go in E1, since it has to be in Row 1 according to B2. Then C3 would be 13 and D3+E3 would = 20. Row 33 would total 34, not 33.

Therefore you must use the last pair. So the pairs all add up to 21, and the unused 7 goes in D1.

12.) D3+E3=21, so in order for Row 3 to add up to 33, C3 must be 11.

13.) Since the pairs equal 21 and C3 is 11, E1 must be 10.

13.) The 4 is adjacent to eight numbers, so one of them must be the 8. Therefore D3 is 8.

14.) Since D3+E3=21, E3 is 13.

15.) Since 9 and 12 remain, Column A adds up to 21. If 12 goes in A2, then Row 2 adds up to 23 and would be third behind Row 3 (33) and Row 1 (26). Therefore, according to E3, the 9 goes in A2 so that Column A (21) exceeds Row 2 (20).

16.) So the 12 goes in A4.

TRY HENRY820’S OTHER LOGIC PUZZLES!