Question

The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is _________.

Hint:

To find the number of multiples of \( k \) between \( L \) and \( R \), use \( \lfloor \frac{R}{k} \rfloor - \lfloor \frac{L-1}{k} \rfloor \). It avoids common manual counting errors.

Answer 1

Correct Answer: 5143

Step 1: Understanding the Concept:
We use the Principle of Inclusion-Exclusion. To find the count of numbers that are neither multiples of 7 nor 3, we subtract the count of multiples of 7, multiples of 3, and add back the multiples of their LCM (21) from the total count of 4-digit numbers.
Step 2: Key Formula or Approach:
Total 4-digit numbers \( N = 9999 - 1000 + 1 = 9000 \).
Let \( A \) be the set of multiples of 3, and \( B \) be the set of multiples of 7.
Required count \( = N - |A \cup B| = N - (|A| + |B| - |A \cap B|) \).
Step 3: Detailed Explanation:
1. Multiples of 3 (\( |A| \)): Numbers from 1002, 1005, ..., 9999.
\( n_A = \frac{9999 - 1002}{3} + 1 = 3000 \).
2. Multiples of 7 (\( |B| \)): Numbers from 1001, 1008, ..., 9996.
\( 1001 = 7 \times 143 \); \( 9996 = 7 \times 1428 \).
\( n_B = 1428 - 143 + 1 = 1286 \).
3. Multiples of 21 (\( |A \cap B| \)): Numbers from 1008, 1029, ..., 9996.
\( 1008 = 21 \times 48 \); \( 9996 = 21 \times 476 \).
\( n_{A \cap B} = 476 - 48 + 1 = 429 \).
Now, calculate \( |A \cup B| \):
\( |A \cup B| = 3000 + 1286 - 429 = 3857 \).
The required number of 4-digit numbers:
\( 9000 - 3857 = 5143 \).
Step 4: Final Answer:
The number of such 4-digit numbers is 5143.