17 of 100: Sorting Numbers

• A = {9,15,21,27,...} (Note: Set A cannot be a prime or an even number)
• B = {3}
• C = {} (Note: Set C is empty because there is no such positive integer which can be prime, divisible by 3 and even all at the same time)
• D = {6,12,18,24,...}
• E = {5,7,11,13,...} (Note: Set E cannot be divisible by 3 or an even number)
• F = {2}
• G = {4,8,10,14,...} (Note: Set G cannot be divisible by 3 or a prime number)
• H = {1,25,35,49,...}

As we can clearly see that only set C is empty, the answer is 1.

To show that a region represented by a particular letter is not empty, all we have to do is list a single element of that set.

$9 \in A$ (divisible by 3, odd, not prime)

$3 \in B$ (divisible by 3, odd, prime)

$6 \in D$ (divisible by 3, even, not prime)

$5 \in E$ (not divisible by 3, odd, prime)

$2 \in F$ (not divisible by 3, even, prime)

$4 \in G$ (not divisible by 3, even, not prime)

$25 \in H$ (not divisible by 3, odd, not prime).

Region $C$ is the only one left. For it not to be empty, there would have to be a prime number divisible by both $2$ and $3$ . But a prime number, by definition, may only be divisible by $1$ and by itself, so it may not be divisible by two numbers different from $1$ and from each other. The set $C$ is therefore empty.

• A ={ 3,6,9,12 ...} (Multiple of 3)
• B ={ 3 } (Prime and multiple of 3)
• C ={}. [ the set C is the only null set as there is no positive integer which is prime and also divisible by {(3×2)=6} ]
• D ={ 6,12,18 ...} (Multiple of 6)
• E ={ 2,3,5,7,11,13,17 ...} (Prime)
• F ={ 2 }. (Prime and even)
• G ={ 2,4,6,8,10 ...}. (Even)
• H ={ 1,25,35,49,55,65 ,...}. (Except prime and multiple of 3 and 2) .
So, the answer is 1 .

BC = Prime number which is DIvisible by 3. Only 3 is a prime number and divisible by 3. Other multiples of 3 like 6, 9, 12.... are not prime-number. So, 3 is the only member in BC region.

CF = Prime number which is an even number. Only 2 is the single EVEN prime number. Other even numbers like 4, 6, 8, 10, 12.... are not prime-number. So, 2 is the unique member in CF region.

CD = Even number which is Divisible by 3. So, the numbers will be divisible by 2 ( for the even number) and 3 = 2X3 = 6. So, the members will be 6, 12, 18, 24.... etc

C = This Region has to satisfy all conditions from BC CF CD regions. There is no common element among BC=3, CF=2, CD= 6, 12, 18.... So, C region will be the empty one.

The number of Empty Region is 1.

Only set C is empty as it's not possible for being prime and having 2 and 3 as it's factors,as it itself contradicts the definition of a prime.

The definition of prime is that it is only divisible by 1 and itself. It can be divisible by 3 or by 2 but not both.

Just start looking at the positive integers, and before you arrive at ten you'll have all categories filled-- except for one, $\boxed{1}$ .

$\begin{array}{r|ll} \hline 1 & H & \\ 2 & F & \text{ This is the only even prime number. } \\ 3 & B & \text{ This is the only prime number divisible by two.} \\ 4 & G & \\ 5 & E & \text { All prime numbers from here on go here. } \\ 6 & D & \text{ All multiples of 6 go here. } \\ 7 & E & \\ 8 & G & \\ 9 & A & \text { Together with all odd multiple of 3 from here on.}\\ \vdots & \vdots & \end{array}$

We only have nothing yet for category $C$ . To prove it is actually empty, consider that if a number is even and divisible by 3, it has at least the prime factors $2$ and $3$ . It is therefore not a prime number.

There are plenty of prime numbers, even numbers, and numbers divisible by three. There is one even prime number, which is two. A prime number divisible by three is three. An even number divisible by three is six. Even Prime numbers divisible by three would have to be empty because it could only be multiples of six, which are not prime. The number one is not prime, even, or divisible by three

I suppose that the answers all have to be integers correct? Because I would argue that 2 would fill both C and F. It didn't specify that the results had to be integers just that values you choose have to be integers.