Sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5
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Including 100 and 300 ...
multiples of 4 --- there are 51
their sum = (51/2)(300 +100) = 10,200
multiples of 5 --- there are 41
their sum = (41/2)(300 + 100) = 8,200
multiples of 20 have been counted twice --- there are 11
their sum = (11/2)(300+100) = 2,200
total = 10,200 + 8,200 - 2,200 = 16,200
multiples of 4 --- there are 51
their sum = (51/2)(300 +100) = 10,200
multiples of 5 --- there are 41
their sum = (41/2)(300 + 100) = 8,200
multiples of 20 have been counted twice --- there are 11
their sum = (11/2)(300+100) = 2,200
total = 10,200 + 8,200 - 2,200 = 16,200
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