**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.) D1 tells you that E2 is an X and that rows 1 and 2 have no X’s other than the two already known.

2.) You know from D1 that E1 is a number and not an X. Clearly the only number E1 is adjacent to is D2. Therefore E1 is adjacent to one number, so it must be adjacent to the number 1, which must go in D2. Remember to type it as 01.

3.) You know that any number in a corner must be adjacent to the 1, the 2, or the 3. The 1 is already used, and you know from D1 that A2, B1, and B2 are numbers. So A1 must be adjacent to the 3. This means that if the other two corners are numbers, they must be adjacent to the 2, but they cannot be, according to D2. Therefore A4 and E4 cannot be numbers, but must be X’s.

4.) A4 and D1 tell you that A3, B3, B4, C3, and C4 contain two X’s. All the rest must be in D3, E3, and E4. Since there must be 9 X’s, you know that *all* of D3, E3, and E4 must be X’s.

5.) E2 tells you that C3 is not an X, so now you know that both C1 and D2 are adjacent to four numbers. So they must both be adjacent to the 4. Therefore the 4 must be in C2.

6.) C2 tells you that the last two X’s are adjacent to A4. If B3 and B4 are both X’s, then C4 would have to be a number and it would be adjacent to one number. So C3 would have to be the 1, but the one is already in D2. Therefore, if B3 and B4 cannot both be X’s but A4 is adjacent to two X’s, A3 must be an X.

7.) A2 is currently adjacent to four unknown cells. It cannot be adjacent to four numbers because it is clearly not adjacent to the 4, which is in C2. Therefore the final X must go in B3, since it must also be adjacent to A4.

8.) Since all the X’s are now known, you know that every other cell is a number. You also know how many numbers they are adjacent to. B4 and C4 are both adjacent to two numbers, and so must be adjacent to the 2. The only place for it to go is C3.

8.) C3 is clearly adjacent to five numbers. B1 is also adjacent to five numbers. Therefore the 5 must be in B2.

9.) A1 and A2 are both adjacent to three numbers. Therefore the 3 must go in B1.

10.) The ways to make 15 with the given numbers are 4+11, 5+10, 6+9, 7+8, and all their reverses. At this point you know that 6 is in A1 or A2 (B2 is adjacent to six numbers) and that C1 is 7 or 11 (E2 says it is prime). If C1 is 7, there’s no way to form 7+8 or 11+4. 9+6 and 10+5 would both still work, but only one of those combinations, either 9+6 in A1 and A2 or 10+5 in A1/A2 and B2. So if 7 in C1 would prevent B1’s clue from being true, C1 must be 11.

11.) C1 tells you that 2 is adjacent to 4 (it already is), 3 is adjacent to 9 (A1 or A2, the other being 6) and 2 is adjacent to 8. This means that E1 must be 7 or 10, but it cannot be 7 because D4 says it is greater than 7. Therefore E1 must be 10.

12.) You know that A1 and A2 are 9 and 6 and that B4 and C4 are 7 and 8. So at this point two columns add up to a multiple of 5: A (15) and E (10). Column D adds up to 1, and E1 says that at least three columns add up to a multiple of 5. This means that B or C or B and C must add up to a multiple of 5. Since B4 and C4 can only be 7 or 8, either both columns add up to a multiple of 5 (15 and 25), or neither do (16 and 24). Therefore both must, in order for “most” columns to add up to a multiple of 5. B4 is 7 and C4 is 8.

13.) B4’s “Most X’s” means at least 5. At this point the following X’s meet the requirement: D1, D3, E3, and D4. So one more X must be adjacent to the number that is the quantity of numbers to which it is adjacent. A3 is adjacent to three numbers, but the 3 is already in B1. B3 is adjacent to six numbers and the 6 has not yet been placed. B3 must be adjacent to the 6; therefore A2 is 6.

14.) The only number remaining is 9; it goes in A1.

TRY HENRY820’S OTHER LOGIC PUZZLES!