Find the sum of all natural numbers less than 1,000 that

Solution

Divisible by both 3 and 5 ⇒ divisible by LCM(3,5) = 15. First list multiples of 15 below 1000, then subtract those divisible by 7 (i.e., by 105). Multiples of 15 less than 1000: 15, 30, …, 990 This is an AP: a = 15, d = 15, l = 990 Number of terms n₁ = 990/15 = 66 Sum S₁ = n₁/2 × (first + last) = 66/2 × (15 + 990) = 33 × 1005 = 33 × (1000 + 5) = 33,000 + 165 = 33,165 Now subtract multiples of 105 (LCM of 15 and 7): Multiples of 105 less than 1000: 105, 210, …, 945 AP: a = 105, d = 105, l = 945 Number of terms n₂ = 945/105 = 9 Sum S₂ = 9/2 × (105 + 945) = 4.5 × 1050 = 4.5 × 1050 = 4.5 × 1000 + 4.5 × 50 = 4500 + 225 = 4,725 Required sum = S₁ − S₂ = 33,165 − 4,725 = 28,440.