Solution :

#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.

#

#3(1+2+3+....+333)+5(1+2+3+.....+199)-15(1+2+3+...+66)

#puts 3*333*334/2+5*199*200/2-15*66*67/2

#

def sum_of_n_consecutive_numbers (n)

n*(n+1)/2

end

def sum_of_multiples_of_m(max,m)

m*sum_of_n_consecutive_numbers(max/m)

end

puts sum_of_multiples_of_m(999,3)+sum_of_multiples_of_m(999,5)-sum_of_multiples_of_m(999,15)

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