If 𝐹(𝑛) denotes the set of factors of the natural
number, including 𝑛 but excluding 1, find a
number p such that 𝐹(12) ∩ 𝐹(18) = 𝐹(𝑝)
A. 18
B. 6
C. 2
D. 3
We begin by listing the factors of 12 and 18:
\[F(12)=\{2,3,4,6,12\}\]
\[F(18)=\{2,3,6,9,18\}\]
Their intersection is simply the common elements in the two sets, which are $\{2,3,6\}$. Therefore,
\[F(12)\cap F(18)=\{2,3,6\}=F(6)\]
So $\boxed{\textbf{(B)}\ 6}$ is our answer.
To find the number p such that 𝐹(12) ∩ 𝐹(18) = 𝐹(p), we need to find the common factors of 12 and 18.
First, let's find the factors of 12. The factors of 12 are 2, 3, 4, 6, and 12.
Next, let's find the factors of 18. The factors of 18 are 2, 3, 6, and 18.
The common factors of 12 and 18 are 2, 3, and 6.
Therefore, 𝐹(12) ∩ 𝐹(18) = {2, 3, 6}.
Now, let's find the number p such that 𝐹(p) = {2, 3, 6}.
To do this, we need to find the smallest common multiple of 2, 3, and 6.
The smallest common multiple of 2, 3, and 6 is 6.
Therefore, the number p = 6.
Therefore, the answer is B. 6.