Is it possible to solve this problem in c++?

Write a program that counts the numbers from 3 to 117. But for multiples of three add 3 instead of 1 and for the multiples of five add 5 instead of 1. For numbers which are multiples of both three and five add 15 instead of 1. Ex: If we are looking at numbers 5 to 15 (inclusive), the program would output 39.

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Jump to PostYou will need a for loop and an understanding of modulus, and a little bit of good old fashioned logic.

Good luck!

Jump to PostWell if

`X % 3`

is zero then you know X is divisible by 3.

Jump to Postwhen something is evenly divisible by a number the modulo is 0.

`15 / 3 = 5 15 % 3 = 0`

Knowing this you can write the following code to see if something is divisble by a number:

`if(someNUmber % 3 == …`

Jump to Post`x<=i<=y`

Is not valid c++. If you need to loop between all number in x and y then you can use:

`for (int i = x; i <= y; i++)`

Wich is translated to: start with

`i`

equall to`x`

…

Jump to PostWhat did you plug in for the missing values and what does your for loop look like now? Remember when you are doing the summation that if the number is divisible by 15 you add 15, if it is divisible by 5 you add 5, if it is divisible by …

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