Question
Find the sum of all natural numbers less than 1000 that
are multiples of 3 but not of 5.Solution
Sum(multiples of 3 below 1000) − Sum(multiples of 15 below 1000). Multiples of 3 less than 1000: 3, 6, …, 999 This is an AP with a = 3, d = 3, last term l = 999 Number of terms: n₁ = 999/3 = 333 Sum S₁ = n₁/2 × (first + last) = 333/2 × (3 + 999) = 333/2 × 1002 = 333 × 501 = 166,833 Multiples of 15 less than 1000: 15, 30, …, 990 a = 15, d = 15, l = 990 n₂ = 990/15 = 66 Sum S₂ = 66/2 × (15 + 990) = 33 × 1005 = 33 × 1000 + 33 × 5 = 33,000 + 165 = 33,165 Required sum = S₁ − S₂ = 166,833 − 33,165 = 133,668.
How many 3 digit numbers are divisible by 3?
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