Question 1
How many numbers from 1 to 11 are divisible by 3 or 5?
Step-by-Step Solution:
Concept: Inclusion-Exclusion Principle - when counting elements in the union of sets, add individual counts and subtract overcounted intersections.
Formula: |A ∪ B| = |A| + |B| - |A ∩ B|
Given:
- Range: 1 to 11
- Divisors: 3 and 5
Step 1 - Count numbers divisible by 3:
Numbers divisible by 3 = ⌊11/3⌋ = 3
These are: 3, 6, 9, ..., 9
Step 2 - Count numbers divisible by 5:
Numbers divisible by 5 = ⌊11/5⌋ = 2
These are: 5, 10, 15, ..., 10
Step 3 - Count numbers divisible by BOTH 3 AND 5:
Numbers divisible by LCM(3,5) = 15
Count = ⌊11/15⌋ = 0
(These are counted twice in steps 1 and 2)
Step 4 - Apply Inclusion-Exclusion:
Total = (divisible by 3) + (divisible by 5) - (divisible by both)
= 3 + 2 - 0
= 5
Visualization (Venn Diagram concept):
- Circle A: divisible by 3 (3 numbers)
- Circle B: divisible by 5 (2 numbers)
- Intersection: divisible by both (0 numbers)
- Union: 5 numbers
Key Principle: Subtract the intersection to avoid double counting.
Extension to Three Sets:
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|
Common Application: Finding numbers NOT divisible by either = 11 - 5 = 6
Concept: Inclusion-Exclusion Principle - when counting elements in the union of sets, add individual counts and subtract overcounted intersections.
Formula: |A ∪ B| = |A| + |B| - |A ∩ B|
Given:
- Range: 1 to 11
- Divisors: 3 and 5
Step 1 - Count numbers divisible by 3:
Numbers divisible by 3 = ⌊11/3⌋ = 3
These are: 3, 6, 9, ..., 9
Step 2 - Count numbers divisible by 5:
Numbers divisible by 5 = ⌊11/5⌋ = 2
These are: 5, 10, 15, ..., 10
Step 3 - Count numbers divisible by BOTH 3 AND 5:
Numbers divisible by LCM(3,5) = 15
Count = ⌊11/15⌋ = 0
(These are counted twice in steps 1 and 2)
Step 4 - Apply Inclusion-Exclusion:
Total = (divisible by 3) + (divisible by 5) - (divisible by both)
= 3 + 2 - 0
= 5
Visualization (Venn Diagram concept):
- Circle A: divisible by 3 (3 numbers)
- Circle B: divisible by 5 (2 numbers)
- Intersection: divisible by both (0 numbers)
- Union: 5 numbers
Key Principle: Subtract the intersection to avoid double counting.
Extension to Three Sets:
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∩ B ∩ C|
Common Application: Finding numbers NOT divisible by either = 11 - 5 = 6