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.
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
1 answer