Is there a command for checking whole numbers?

[NinjaFish]

I'm working on some exercises, and not entirely sure how you check for fractional numerics? By this I mean, I need a piece of code or a command that I can use in my IF statement that will let me check if a number i'm dividing is divisible by two numbers. However I do not know how to make it "check" if a decimal is left over.

Pseudo-coding this. I basically am just wanting

if 100 divisable by 3
output
else
output

if 100 divisble by 4
output
else
output


etc etc etc. It just uses a loop to go up to 101, then 102 etc. All that I can do. But I need the part that will tell me if it is or not. Not sure if there is a built in code or something my mathematically retarded brain doesn't know how to figure out

(For reference. I don't need the entire thing written out. I just need a small line or two of code that will let me check and determine if it's divisble or not).

Thanks in advance

[Mozza314]

In C/C++ you can do "integer arithmetic", so when you divide there is no fractional part, e.g. 7 / 2 = 3. So if you're working with integers, it's not appropriate to test for a fractional part.

Regarding your specific problem, divisibility is generally tested using the '%' operator, which gives you the remainder of a division. x is divisible by y iff there's no remainder when x is divided by y, so you can do x % y == 0 to test whether x is divisible by y.

[NinjaFish]

In C/C++ you can do "integer arithmetic", so when you divide there is no fractional part, e.g. 7 / 2 = 3. So if you're working with integers, it's not appropriate to test for a fractional part.

Regarding your specific problem, divisibility is generally tested using the '%' operator, which gives you the remainder of a division. x is divisible by y iff there's no remainder when x is divided by y, so you can do x % y == 0 to test whether x is divisible by y.

Thanks a tonne. So just to make sure I could do

100 % 3 == 0 would give me 'false' and skip my IF and go to else?

[Mozza314]

Thanks a tonne. So just to make sure I could do

100 % 3 == 0 would give me 'false' and skip my IF and go to else?

Yep. 100 % 3 is 1 (i.e. 100 divided by 3 is 33 remainder 1).

[NinjaFish]

Yep. 100 % 3 is 1 (i.e. 100 divided by 3 is 33 remainder 1).

Thank you kindly sir. Worked like a charm and my program works flawlessly. Was very simple to figure out how to implement as well