Question

Set A contains all 3 digit positive integers that are divisible by 5, set B contains all 3 digit positive integers that are divisible by 7 and set C contains all 3 digit positive integers divisible by 11. Then \(n(A\cup B\cup C)=\)

• A. \(320\)
• B. \(328\)
• C. \(331\)
• D. \(337\)

Hint:

For unions of three sets, always use Inclusion-Exclusion: \[ A+B+C-AB-BC-CA+ABC. \] This avoids double counting.

Answer 1

The Correct Option is C

Concept: Use the Principle of Inclusion and Exclusion: \[ n(A\cup B\cup C) = n(A)+n(B)+n(C) -n(A\cap B) -n(B\cap C) -n(C\cap A) +n(A\cap B\cap C) \]

Step 1: Count 3-digit multiples. Multiples of 5: \[ \left\lfloor\frac{999}{5}\right\rfloor - \left\lfloor\frac{99}{5}\right\rfloor = 199-19 = 180 \] Multiples of 7: \[ 142-14 = 128 \] Multiples of 11: \[ 90-9 = 81 \]

Step 2: Count pairwise intersections. Multiples of \(35\): \[ 28-2=26 \] Multiples of \(55\): \[ 18-1=17 \] Multiples of \(77\): \[ 12-1=11 \]

Step 3: Count triple intersection. LCM: \[ 5\times7\times11=385 \] Multiples of 385: \[ 2-0=2 \]

Step 4: Apply Inclusion-Exclusion. \[ 180+128+81 - 26 - 17 - 11 + 2 \] \[ 389-54+2 \] \[ 337 \] Therefore, \[ \boxed{337} \]