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

#1

1-

2-https://ide.codingblocks.com/s/224282

BOTH THESE CODES ARE COORECT. BUT THERE IS A ERROR OF TIME LIMIT. PLEASE EXPLAIN ME A MORE EFFICIENT WAY OF THIS PROBLEM.

#2

this is just O(1) solution that I suggest. First calculate the numbers of numbers below N divisible a, let their count be c1 then the sum of all these numbers should be a * ( 1 + 2 + 3 + c1 terms) . Do the same for b. Then subtract from the answer the same thing for their lcm.

#3

the answer is coming wrong. or i did not understand properly.if i take N=20.

#4

then sum of 3 multiples=63, sum of 5 multiples =30 .total 93. lcm=20. remain 93-20=73. but answer is 78

#5

I think you didn’t understand completely
How do you get the sum of multiples of 5 as 30. It should be 50. and their lcm i.e. 15 has only a single multiple below 20 i.e. 15 so common 15 so ans is 30 + 50 - 15

#6

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