Key Concepts and Formulas

• Complementary Counting: To find the number of elements satisfying "neither A nor B", we calculate the total number of elements and subtract those satisfying "A or B". $\text{Count}(\text{neither A nor B}) = \text{Total} - \text{Count}(\text{A or B})$
• Principle of Inclusion-Exclusion: For two sets A and B, the number of elements in their union is given by: $|A \cup B| = |A| + |B| - |A \cap B|$
• Counting Multiples in a Range: The number of multiples of an integer $d$ in a range $[a, b]$ can be found by counting the multiples of $d$ up to $b$ and subtracting the multiples of $d$ up to $a-1$. Alternatively, if the first multiple is $a_1$ and the last is $a_n$, the count is $\frac{a_n - a_1}{d} + 1$.

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Step-by-Step Solution

Step 1: Determine the Total Number of 4-Digit Integers

We need to find the total count of integers that have exactly four digits.

• Reasoning: This forms the universal set from which we will subtract numbers that meet the exclusion criteria.
• Calculation: The smallest 4-digit integer is $1000$ and the largest is $9999$. The total number of 4-digit integers is: $9999 - 1000 + 1 = 9000$ So, there are $9000$ four-digit numbers.

Step 2: Count 4-Digit Numbers Divisible by 3

Let $N_3$ be the number of 4-digit integers divisible by 3.

• Reasoning: We need to find the first and last 4-digit multiples of 3 and use the arithmetic progression formula.
• Calculation:
  • The smallest 4-digit number is $1000$. $1000 \div 3 = 333$ with a remainder of $1$. The first multiple of 3 is $1000 + (3-1) = 1002$.
  • The largest 4-digit number is $9999$. $9999 \div 3 = 3333$ with a remainder of $0$. The last multiple of 3 is $9999$.
  • Using the formula for the number of terms in an arithmetic progression ($n = \frac{a_n - a_1}{d} + 1$): $N_3 = \frac{9999 - 1002}{3} + 1 = \frac{8997}{3} + 1 = 2999 + 1 = 3000$ There are $3000$ four-digit numbers divisible by 3.

Step 3: Count 4-Digit Numbers Divisible by 7

Let $N_7$ be the number of 4-digit integers divisible by 7.

• Reasoning: Similar to Step 2, we find the first and last 4-digit multiples of 7.
• Calculation:
  • The smallest 4-digit number is $1000$. $1000 \div 7 = 142$ with a remainder of $6$. The first multiple of 7 is $1000 + (7-6) = 1001$.
  • The largest 4-digit number is $9999$. $9999 \div 7 = 1428$ with a remainder of $3$. The last multiple of 7 is $9999 - 3 = 9996$.
  • Using the formula for the number of terms in an arithmetic progression: $N_7 = \frac{9996 - 1001}{7} + 1 = \frac{8995}{7} + 1 = 1285 + 1 = 1286$ There are $1286$ four-digit numbers divisible by 7.

Step 4: Count 4-Digit Numbers Divisible by Both 3 and 7

Numbers divisible by both 3 and 7 are divisible by their Least Common Multiple (LCM). Since 3 and 7 are prime, their LCM is $3 \times 7 = 21$. Let $N_{21}$ be the number of 4-digit integers divisible by 21.

• Reasoning: We need to find the first and last 4-digit multiples of 21.
• Calculation:
  • The smallest 4-digit number is $1000$. $1000 \div 21 = 47$ with a remainder of $13$. The first multiple of 21 is $1000 + (21-13) = 1008$.
  • The largest 4-digit number is $9999$. $9999 \div 21 = 476$ with a remainder of $3$. The last multiple of 21 is $9999 - 3 = 9996$.
  • Using the formula for the number of terms in an arithmetic progression: $N_{21} = \frac{9996 - 1008}{21} + 1 = \frac{8988}{21} + 1 = 428 + 1 = 429$ There are $429$ four-digit numbers divisible by both 3 and 7.

Step 5: Count 4-Digit Numbers Divisible by 3 or 7

Using the Principle of Inclusion-Exclusion, the number of 4-digit integers divisible by 3 or 7 is $N_3 + N_7 - N_{21}$.

• Reasoning: This step calculates the count of numbers we want to exclude from our total set. We add the counts of multiples of 3 and 7, and subtract the count of multiples of 21 to correct for double-counting.
• Calculation: $N_{3 \text{ or } 7} = N_3 + N_7 - N_{21}$ $N_{3 \text{ or } 7} = 3000 + 1286 - 429$ $N_{3 \text{ or } 7} = 4286 - 429 = 3857$ There are $3857$ four-digit numbers divisible by 3 or 7.

Step 6: Calculate the Number of 4-Digit Integers Neither Divisible by 3 Nor by 7

We use complementary counting. The number of 4-digit numbers that are neither a multiple of 3 nor a multiple of 7 is the total number of 4-digit numbers minus those divisible by 3 or 7.

• Reasoning: This is the final step where we apply the complementary counting principle to answer the question.
• Calculation: $\text{Number (neither 3 nor 7)} = \text{Total numbers} - N_{3 \text{ or } 7}$ $\text{Number (neither 3 nor 7)} = 9000 - 3857 = 5143$

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Common Mistakes & Tips

• Off-by-One Errors: Be careful when calculating the first and last multiples of a divisor, and when calculating the total number of integers in a range. Always add 1 to the difference.
• Double Counting: Forgetting to subtract the intersection ($N_{21}$) in the Inclusion-Exclusion principle is a common error.
• LCM Calculation: Ensure you correctly identify the LCM for numbers divisible by two different integers. For coprime numbers, it's their product.

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Summary

To find the number of 4-digit integers that are neither multiples of 7 nor multiples of 3, we first determined the total count of 4-digit numbers. Then, we calculated the number of 4-digit integers divisible by 3, by 7, and by both 3 and 7 (i.e., by 21). Using the Principle of Inclusion-Exclusion, we found the total count of numbers divisible by 3 or 7. Finally, we subtracted this count from the total number of 4-digit integers to obtain the desired result.

The final answer is \boxed{5143}.