Perform Modulo Operations with 17 and 5

The % operator returns what division leaves behind

When you write 17 % 5, the % operator gives you the remainder left after dividing 17 by 5. It does not give you the integer division quotient. The quotient is 3, because three complete groups of 5 fit into 17, but the remainder is 2.

int remainder = 17 % 5;  // remainder is 2
int quotient = 17 / 5;    // quotient is 3

int remainder = 17 % 5;  // remainder is 2
int quotient = 17 / 5;    // quotient is 3

The equation 17 = 3 * 5 + 2 proves the remainder is correct

The division relationship is dividend = quotient * divisor + remainder. For 17 and 5, substitute the quotient 3 and the remainder 2: 17 = 3 * 5 + 2. The right side becomes 15 + 2, which reconstructs the dividend 17 exactly. That reconstruction is why 2 is the remainder produced by 17 % 5.

A positive divisor of 5 restricts the remainder to 0-4

When the divisor is positive, the remainder must be at least 0 and smaller than the divisor. Because the divisor here is 5, the only possible remainder values are 0, 1, 2, 3, and 4. Therefore 5 cannot be the result of 17 % 5, and neither can any value larger than 5. The value 2 fits the required range.

The % operator is evaluated at the same precedence level as * and /

The % operator shares a precedence level with multiplication and division, so it is evaluated before addition. In 17 + 5 % 5, the modulo operation happens first: 5 % 5 is 0, so the expression becomes 17 + 0 and produces 17. Parentheses change that order. In (17 + 5) % 5, the addition happens first, producing 22, and then 22 % 5 produces 2.

int first = 17 + 5 % 5;       // 17 + 0 = 17
int second = (17 + 5) % 5;    // 22 % 5 = 2

int result = 17 + 5 % 5;

A zero right operand makes a remainder expression invalid

The divisor in 17 % 5 must not be replaced by 0. There is no valid remainder from dividing by zero, so the expression 17 % 0 is not an ordinary calculation. The right operand of % must be nonzero.