**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.

WALKTHROUGH SECTION BELOW

**1/20) **A1 tells you that E1 and A4 are not numbers, meaning they are both X.

**3/20) **A1 tells you that E4 is a number. E1 tells you that for sure, D3, D4, E3, and E4 cannot be prime numbers. Since E4 is a number, it must follow the main rule, meaning it must be adjacent to the number that is the quantity of numbers to which it is adjacent. Since E4 is in a corner, it is only adjacent to three cells. So you know it must be adjacent to the 1, the 2, or the 3. But E1 tells you that nothing adjacent to E4 can be prime. So E4 cannot be adjacent to the 2 or 3. So it cannot be adjacent to two or three numbers, meaning it must be adjacent to just one number, meaning it must be adjacent to the 1. A4 tells you that Row 3 contains at least four numbers. D3 and E3 cannot both be a number, because then E4 would be adjacent to two numbers, but we just established that E4 must be adjacent to one number, and so it must be the 1. So since Row 3 has at least four numbers, you know the X is D3 or E3. This means the other one is the 1 and D4 must be X.

**4/20) **Since D3 or E3 is X, as established in step 3/20 above, you know that A3, B3, and C3 are all numbers, according to A4. D4 tells you that there is at least one X among A3, B3, and B4. Therefore you know that B4 is X.

**5/20) **You know that E4 is a number that is adjacent to the 1 and 2 X’s. You also know that E4 is not prime, according to E1. Since E4 is only adjacent to the 1, you also know that E4 is not even, according to B4. So if E4 is not prime or even, it can only be 1 or 9. But you know that it is adjacent to 1, which is D3 or E3. Therefore E4 is 9.

**6/20)** E2 can be adjacent to one or two or three numbers. (You know that D3 or E3 is X) If E2 is a number, it cannot be adjacent to two or three numbers because nothing in Columns D or E can be prime, according to E1 + E4 (explained in full in 6/20 below). So if E2 is a number, it must be adjacent to just one number, the 1 in D3 or E3. That would seem fine, but what numbers could E2 even be? Nothing prime. 9 is taken. 1 is in D3 or E3. That leaves only 4, 6, 8, and 10: all even numbers. But according to B4, all even numbers are adjacent to at least one prime number. E2 cannot be adjacent to a prime number, much less if the only number adjacent to it is the 1. Therefore there’s no way E2 can be a number; it is X.

**6/20) **E4 tells you which of the cases in E1 is true. You have the 1 narrowed down and the 9 is placed. This means all other odd numbers are prime. So either C3 or C4 is 3 or 5 or 7, according to E4. Either way, there’s a prime number in Row 3 or 4. So according to E1, all primes must be in Columns A-C. Now look at E3. If it is a number, you know it is the 1. Then D3 would be X. So then the 1 would have to be adjacent to two numbers or three numbers (can’t be just 1 because then it would have to be adjacent to the 1, but it is the 1), including the 2 or 3. But 2 and 3 are prime, and there’s nowhere E3 can be adjacent to a prime number, since E1 + E4 tell you that all primes are in Columns A-C. So E3 can’t be a number; it’s X.

**8/20) **The 9 must be adjacent to at least one number. With 2/3 of its adjacents being X, you know that D3 is 01.

**9/20) **All E3 tells you is that the 4 is adjacent to the 1. Now consider the scenario of D3. If A2 is a number, then B1, B2, and B3 would have to be numbers. You already know from A4 that A3 is a number. So if A2 is a number, then it would for sure be adjacent to four numbers. But it cannot be adjacent to the 4, since the 4 is adjacent to the 1. Therefore A2 is not a number, but X.

**10/20) **You may have realized already that A3 must be adjacent to the 2. You know from A4 that A3 and B3 are both numbers. But A3 must be adjacent to at least one more number, because the 1 is already at D3, not adjacent to A3. So you know that A3 is a number that is adjacent to the 2, the 2 being either B2 or B3. Now take a look at D1. It is adjacent to three cells that could be numbers: C1, C2, and D2. So if D1 is a number it must be adjacent to the 1 or 2 or 3. You just established that the 2 is B2 or B3. And the 1 is already at D3. So if D1 is a number, it must be adjacent to three numbers, one of them the 3. D2 cannot be the 3 because D2 cannot be prime according to E1 + E4. C1 and C2 also cannot be the 3, according to A2. This means D1 cannot be a number; it is X.

**11/20) **You have placed nine X’s. There’s only one left. A2 tells you that B1 is not adjacent to three numbers if it is a number. This means the final X is either B1 or something adjacent to it. Actually B2 cannot be X because A3 is definitely a number and cannot be adjacent to just one number (see step 9/20). So the final X is B1 or C1 or C2. If it’s B1, C1 and C4 would both have to be numbers that are adjacent to the 3. That cannot work. So B1 is not the X, meaning it is a number adjacent to two numbers: B2 and C1/C2. You also know that A3 is adjacent to two numbers, so the 2 must be B2.

**11/20) **Since the final X is C1 or C2, C4 is a number adjacent to the 3. If C1 is a number, C2 would be X and C1 would be adjacent to three numbers: B1, B2, and D2. C1 and C4 cannot both be adjacent to the 3, so C1 is not a number; it is X.

**13/20) **All X’s are placed. D2 and C4 are both adjacent to three numbers, so the 3 must be C3.

**14/20) **B2, B3, and D3 are all adjacent to five numbers. Therefore the 5 must be C2.

**15/20) **C3 tells you that B1 is not 7. You know that primes are not in Columns D or E, so D2 is not 7. The 3 is adjacent to six numbers, so the 6 must be D2, B3, or C4. The 5 is also adjacent to six numbers, meaning C4 cannot be the 6.

If the 6 goes in B3, the 7 would be A3 or C4, adjacent to the 6. C2 says that cannot be. Therefore D2 is 6.

**16/20) **E3 tells you that C4 must be 4.

**17/20) **C3 tells you that B1 must be 8.

**18/20) **Column B could add up to 20 if B3 is 10. Then A3 would be 7. So 2, 3, 5, and 7 would not all be adjacent to each of the others. So no column cad sum to 20, meaning B3 is not 10, but 7. And A3 is 10.

TRY HENRY820’S OTHER LOGIC PUZZLES!