Key Concepts and Formulas

• Princ of Inclusion-Exclusion: For two sets $P_1$ and $P_2$, the number of elements in their union is $|P_1 \cup P_2| = |P_1| + |P_2| - |P_1 \cap P_2|$.
• Counting Integers in a Range: The number of integers in an inclusive range $[a, b]$ is $b - a + 1$.
• Counting Multiples: The number of multiples of $k$ in the range $[a, b]$ can be found by determining the first multiple of $k$ greater than or equal to $a$ and the last multiple of $k$ less than or equal to $b$, and then using the arithmetic progression formula: Number of terms $= \frac{\text{Last term} - \text{First term}}{k} + 1$. Alternatively, it can be calculated as $\lfloor \frac{b}{k} \rfloor - \lfloor \frac{a-1}{k} \rfloor$.
• LCM: If a number is a multiple of both $x$ and $y$, it must be a multiple of their Least Common Multiple (LCM).

Step-by-Step Solution

The problem asks us to find the number of integers in the range $[100, 700]$ that are neither a multiple of 3 nor a multiple of 4. We will use the Principle of Inclusion-Exclusion.

Let $U$ be the set of all integers in the range $[100, 700]$. Let $M_3$ be the set of integers in $U$ that are multiples of 3. Let $M_4$ be the set of integers in $U$ that are multiples of 4.

We want to find the number of elements in $U$ that are not in $M_3$ and not in $M_4$. This is given by $|U| - |M_3 \cup M_4|$. Using the Principle of Inclusion-Exclusion, $|M_3 \cup M_4| = |M_3| + |M_4| - |M_3 \cap M_4|$. Therefore, the number of elements in $A$ is $|U| - (|M_3| + |M_4| - |M_3 \cap M_4|)$.

Step 1: Calculate the Total Number of Integers in the Range ($|U|$) The range is $[100, 700]$, inclusive. The total number of integers is $700 - 100 + 1$. $|U| = 700 - 100 + 1 = 601$

Step 2: Calculate the Number of Multiples of 3 in the Range ($|M_3|$) We need to find the number of multiples of 3 between 100 and 700, inclusive. The first multiple of 3 $\ge 100$: $100 \div 3 = 33$ remainder 1. So, $3 \times 34 = 102$. The last multiple of 3 $\le 700$: $700 \div 3 = 233$ remainder 1. So, $3 \times 233 = 699$. The multiples of 3 form an arithmetic progression: $102, 105, \ldots, 699$. The number of terms is $\frac{699 - 102}{3} + 1 = \frac{597}{3} + 1 = 199 + 1 = 200$. Alternatively, using the floor function: $|M_3| = \lfloor \frac{700}{3} \rfloor - \lfloor \frac{100-1}{3} \rfloor = \lfloor \frac{700}{3} \rfloor - \lfloor \frac{99}{3} \rfloor = 233 - 33 = 200$.

Step 3: Calculate the Number of Multiples of 4 in the Range ($|M_4|$) We need to find the number of multiples of 4 between 100 and 700, inclusive. The first multiple of 4 $\ge 100$: $100 \div 4 = 25$. So, 100 is the first multiple. The last multiple of 4 $\le 700$: $700 \div 4 = 175$. So, 700 is the last multiple. The multiples of 4 form an arithmetic progression: $100, 104, \ldots, 700$. The number of terms is $\frac{700 - 100}{4} + 1 = \frac{600}{4} + 1 = 150 + 1 = 151$. Alternatively, using the floor function: $|M_4| = \lfloor \frac{700}{4} \rfloor - \lfloor \frac{100-1}{4} \rfloor = \lfloor \frac{700}{4} \rfloor - \lfloor \frac{99}{4} \rfloor = 175 - 24 = 151$.

Step 4: Calculate the Number of Multiples of Both 3 and 4 ($|M_3 \cap M_4|$) An integer is a multiple of both 3 and 4 if and only if it is a multiple of their LCM. LCM(3, 4) = 12. We need to find the number of multiples of 12 between 100 and 700, inclusive. The first multiple of 12 $\ge 100$: $100 \div 12 = 8$ remainder 4. So, $12 \times 9 = 108$. The last multiple of 12 $\le 700$: $700 \div 12 = 58$ remainder 4. So, $12 \times 58 = 696$. The multiples of 12 form an arithmetic progression: $108, 120, \ldots, 696$. The number of terms is $\frac{696 - 108}{12} + 1 = \frac{588}{12} + 1 = 49 + 1 = 50$. Alternatively, using the floor function: $|M_3 \cap M_4| = \lfloor \frac{700}{12} \rfloor - \lfloor \frac{100-1}{12} \rfloor = \lfloor \frac{700}{12} \rfloor - \lfloor \frac{99}{12} \rfloor = 58 - 8 = 50$.

Step 5: Apply the Principle of Inclusion-Exclusion The number of elements in $A$ is $|U| - (|M_3| + |M_4| - |M_3 \cap M_4|)$. $|A| = 601 - (200 + 151 - 50)$ $|A| = 601 - (351 - 50)$ $|A| = 601 - 301$ $|A| = 300$

Common Mistakes & Tips

• Off-by-One Errors in Range: Always remember to add 1 when calculating the total number of integers in an inclusive range $[a, b]$, as the formula is $b - a + 1$.
• Incorrectly Finding First/Last Multiples: When finding the first or last multiple of a number $k$ within a range $[a, b]$, ensure you are using the correct ceiling or floor operations, or the arithmetic progression method. For example, the first multiple of $k \ge a$ is $k \times \lceil a/k \rceil$.
• Confusing "OR" and "AND": For "multiples of 3 AND multiples of 4", you need the LCM. For "multiples of 3 OR multiples of 4", you use the union formula.

Summary We are looking for integers in the range $[100, 700]$ that are neither multiples of 3 nor multiples of 4. We first determined the total number of integers in the range. Then, we calculated the number of multiples of 3, the number of multiples of 4, and the number of multiples of both 3 and 4 (which are multiples of 12) within the given range. Finally, we applied the Principle of Inclusion-Exclusion to subtract the numbers that are multiples of 3 or 4 from the total number of integers, yielding the count of integers that satisfy neither condition.

The final answer is $\boxed{300}$ which corresponds to option (A).