Find the sum of all the multiples of 3 and 5 below 10000

First of all, stop thinking on the number 1000 and turn your attention to the number 990 instead. If you solve the problem for 990 you just have to add 993,995,996 & 999 to it for the final answer. This sum is (a)=3983

Count all the #s divisible by 3: From 3... to 990 there are 330 terms. The sum is 330(990+3)/2, so (b)=163845

Count all the #s divisible by 5: From 5... to 990 there are 198 terms. The sum is 198(990+5)/2, so (c)=98505

Now, the GCD (greatest common divisor) of 3 & 5 is 1, so the LCM (least common multiple) should be 3×5=15.

This means every number that divides by 15 was counted twice, and it should be done only once. Because of this, you have an extra set of numbers started with 15 all the way to 990 that has to be removed from (b)&(c).

Then, from 15... to 990 there are 66 terms and their sum is 66(990+15)/2, so (d)=33165

The answer for the problem is: (a)+(b)+(c)−(d)=233168

Simple but very fun problem.

a

step-by-step explanation:

i need the pts ignore this answer

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.