
Shortcut Trick
To find the count of numbers in a range, use integer division by subtracting the quotient of the lower limit from the upper limit.
Three-digit numbers range from 100 to 999.
Divisible by 3: (999 ÷ 3) − (99 ÷ 3) = 333 − 33 = 300
Divisible by 5: (999 ÷ 5) − (99 ÷ 5) = 199 − 19 = 180
Divisible by 15 (LCM of 3 and 5): (999 ÷ 15) − (99 ÷ 15) = 66 − 6 = 60
Total numbers = Divisible by 3 + Divisible by 5 − Divisible by 15 = 300 + 180 − 60 = 420
∴ The correct answer is 420.

Alternate Method
Given:
The set of three-digit numbers starts at 100 and ends at 999.
Formula Used:
Number of terms in an Arithmetic Progression (AP): n = [(Last Term − First Term) ÷ Common Difference] + 1
Inclusion-Exclusion Principle: n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Calculations:
⇒ For numbers divisible by 3: Smallest three-digit number = 102, Largest = 999
⇒ n(3) = [(999 − 102) ÷ 3] + 1 = (897 ÷ 3) + 1 = 299 + 1 = 300
⇒ For numbers divisible by 5: Smallest three-digit number = 100, Largest = 995
⇒ n(5) = [(995 − 100) ÷ 5] + 1 = (895 ÷ 5) + 1 = 179 + 1 = 180
⇒ For numbers divisible by 15 (LCM of 3 and 5): Smallest three-digit number = 105, Largest = 990
⇒ n(15) = [(990 − 105) ÷ 15] + 1 = (885 ÷ 15) + 1 = 59 + 1 = 60
⇒ Total divisible by 3 or 5 = n(3) + n(5) − n(15)
⇒ Total numbers = 300 + 180 − 60 = 420
∴ The correct answer is 420.

Additional Information
Inclusion-Exclusion Principle
When finding the count of elements in the union of two sets, we add the counts of each set and subtract the count of their intersection to prevent double counting: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
Arithmetic Progression Number of Terms
The total number of terms 'n' in an Arithmetic Progression is determined by: n = [(l − a) ÷ d] + 1, where 'l' is the last term, 'a' is the first term, and 'd' is the common difference.
LCM and Common Divisibility
If a number is divisible by two different co-prime numbers, it must also be divisible by their Least Common Multiple (LCM). For 3 and 5, the LCM is 15.