program that counts from 1 through 100, and prints “D3” if the number is evenly divisible by 3

Question

Write a C++ program that counts from 1 through 100, and prints “D3” if the number is evenly divisible by 3, “D5” if it’s evenly divisible by 5, “D3&5” if it’s even divisible by three and five, or the number for all other cases.

Answer 1

You will be using the modulus operator '%' which will divide one number by another number and return the remainder. If a number mod 3 has a remainder of 0 than it is divisible by 3, if that same number mod 5 has a remainder of 0 then it is also divisible by 5.

We will use conditional statements to check if a number is divisible by 3, 5, both, or none.

1) divisible by 3: (x mod 3 equals 0) and (x mod 5 does not equal 0)

2) divisible by 5: (x mod 5 equals 0) and (x mod 3 does not equal 0)

3) divisible by both: (x mod 3 equals 0) and (x mod 5 also equals 0)

4) any case that is not the 3 listed above

The code will look like this:

for (int i = 1; i < 100; i++) {

if (i % 3 == 0 && i % 5 != 0) {

cout << 'D3';

} else if (i % 5 == 0 && i % 3 != 0) {

cout << 'D5';

} else if (i % 5 == 0 && i % 3 == 0) {

cout << 'D3&5';

} else {

cout << i;

}

}