How do you check whether a number is divisible by another number?

You do this using the modulus operator, %

n % k == 0

evaluates true if and only if n is an exact multiple of k. In elementary maths this is known as the remainder from a division.

In your current approach you perform a division and the result will be either

always an integer if you use integer division, or

always a float if you use floating point division.

It's just the wrong way to go about testing divisibility.

You can simply use % Modulus operator to check divisibility.
For example: n % 2 == 0 means n is exactly divisible by 2 and n % 2 != 0 means n is not exactly divisible by 2.

You can use % operator to check divisiblity of a given number

The code to check whether given no. is divisible by 3 or 5 when no. less than 1000 is given below:

n=0
while n<1000:
    if n%3==0 or n%5==0:
        print n,'is multiple of 3 or 5'
    n=n+1

This code appears to do what you are asking for.

for value in range(1,1000):
    if value % 3 == 0 or value % 5 == 0:
        print(value)

Or something like

for value in range(1,1000):
    if value % 3 == 0 or value % 5 == 0:
        some_list.append(value)

Or any number of things.

I had the same approach. Because I didn't understand how to use the module(%) operator.

6 % 3 = 0 *This means if you divide 6 by 3 you will not have a remainder, 3 is a factor of 6.

Now you have to relate it to your given problem.

if n % 3 == 0 *This is saying, if my number(n) is divisible by 3 leaving a 0 remainder.

Add your then(print, return) statement and continue your

a = 1400
a1 = 5
a2 = 3

b= str(a/a1)
b1 = str(a/a2)
c =b[(len(b)-2):len(b)]
c1 =b[(len(b1)-2):len(b1)]
if c == ".0":
    print("yeah for 5!")
if c1 == ".0":
    print("yeah for 3!")

For small numbers n%3 == 0 will be fine. For very large numbers I propose to calculate the cross sum first and then check if the cross sum is a multiple of 3:

def is_divisible_by_3(number):
    if sum(map(int, str(number))) % 3 != 0:
        my_bool = False
    return my_bool

Try this ...

public class Solution {

  public static void main(String[] args) {
    long t = 1000;
    long sum = 0;

    for(int i = 1; i<t; i++){
            if(i%3 == 0 || i%5 == 0){
                sum = sum + i;
            }
        }
        System.out.println(sum);    
  }
}

The simplest way is to test whether a number is an integer is int(x) == x. Otherwise, what David Heffernan said.