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5 March, 22:03

Multiples of 3 and 5 Problem 1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

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  1. 6 March, 01:39
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    Answer: 233,168

    Step-by-step explanation:

    Formula: 1 + 2 + 3 + ... + n = n (n+1) / 2

    Sum of all the numbers below 1000 that is divisible by 3:

    3 + 6 + 9 + ... + 999 = 3 (1 + 2 + 3 + ... + 333) = 3 x 333 x 334 / 2 = 166,833

    Sum of all the numbers below 1000 that is divisible by 5:

    5 + 10 + 15 + ... + 995 = 5 (1 + 2 + 3 + ... + 199) = 5 x 199 x 200 / 2 = 99,500

    As we add up 166,833 and 99,500, the numbers that are divisible by 3*5 = 15 would be counted double. Therefore, subtract the result for numbers divisible by 15 just once:

    Sum of all numbers below 1000 that is divisible by 15:

    15 + 30 + 45 + ... + 990 = 15 (1 + 2 + 3 + ... + 66) = 15 x 66 x 67 / 2 = 33165

    Therefore, [ 166,833 + 99,500 ] - 33,165 = 233,168
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