Let A = {n in [100, 700] cap mathbb{N} : n te | vedclass

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(A) The total number of integers in the set $[100, 700]$ is $700 - 100 + 1 = 601$.
Let $S_3$ be the set of multiples of $3$ in $[100, 700]$. The mult

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$300$

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(A) The total number of integers in the set $[100, 700]$ is $700 - 100 + 1 = 601$.
Let $S_3$ be the set of multiples of $3$ in $[100, 700]$. The multiples are $102, 105, \dots, 699$. Using $T_n = a + (n-1)d$,we get $699 = 102 + (n-1)3$,which gives $n = 200$.
Let $S_4$ be the set of multiples of $4$ in $[100, 700]$. The multiples are $100, 104, \dots, 700$. We get $700 = 100 + (n-1)4$,which gives $n = 151$.
Let $S_{12}$ be the set of multiples of both $3$ and $4$ (i.e.,multiples of $12$) in $[100, 700]$. The multiples are $108, 120, \dots, 696$. We get $696 = 108 + (n-1)12$,which gives $n = 50$.
By the Principle of Inclusion-Exclusion,the number of elements that are multiples of $3$ or $4$ is $n(S_3 \cup S_4) = n(S_3) + n(S_4) - n(S_{12}) = 200 + 151 - 50 = 301$.
The number of elements in $A$ is the total number of elements minus those that are multiples of $3$ or $4$: $601 - 301 = 300$.

Let $A = \{n \in [100, 700] \cap \mathbb{N} : n \text{ is neither a multiple of } 3 \text{ nor a multiple of } 4\}$. Then the number of elements in $A$ is

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