Multiples of 3 and 5

[jasen 4]

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.

It is easy to find the answer using computer, but it's more interesting if we can solve it without coding.

[Thalia 64]

Oops. 1 extra multiple of 5 then. So we subtract 1000 and get 233,168.

[Rob_G 1]

So...

The multiples of 3 are all of the form 3n where n is a natural number. The largest value of n is floor(1000/3) or 333. So we get 3(1), 3(2), 3(3), ..., 3(333). So if we sum them...sum[n=1...333](3n) we get 3* (333*334)/2 or 166,833.

The same follows for 5 1000/5=200. So sum[n=1...200](5n) or 5*(200*201)/2 or 100,500.

Finally we add them together to get 267,333.

[jasen 4]

So...

Hidden Content

good start but wrong, your way to find out sum of multiple 3 and 5 are still wrong.

[Thalia 64]

Adding on to Rob_G

Every 15 (3*5), you get a duplicate number. 66 multiples of 15 in 1000. So if we follow the same method used for 3 and 5, we get 15(66*67)/2=33,165. Subtracting this from 267,333, we get 234,168.

[jasen 4]

Adding on to Rob_G

Hidden Content

very close! just wrong 1 digit.

the question is asking for sum bellow 1000.