**Objective**: Write Given two integers ‘number’ and ‘divisor’, Write an algorithm to find the remainder if ‘number’ is divided by ‘divisor’.

**Condition**: You are not allowed to use modulo or % operator.

**Example**:

num = 10, divisor = 4 remainder = 2 num = 11, divisor = 2 remainder = 1

This is fun puzzle which is asked in the interview.

**Approach**:

1. This problem will become very trivial if use of modulo or % operator is allowed.

2. Idea is Keep subtracting the divisor from number till number>=divisor.

3. Once the step above is done, remaining value of number will be the remainder.

**Example**:

number = 10, divisor = 4 number = number – divisor => 10 – 4 = 6 number = number – divisor => 6 – 4 = 2 remainder = 2

**Code**:

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public class FindRemainder { | |

public static void remainder(int n, int divisor){ | |

if(divisor==0){ | |

System.out.println("Cannot divide by 0"); | |

return; | |

} | |

//if either number or divisor is negative | |

if(n<0) | |

n *=-1; | |

if(divisor<0) | |

divisor *= –1; | |

int number = n; | |

//subtract divisor from n till n>=divisor | |

while(n>=divisor){ | |

n -= divisor; | |

} | |

System.out.println("Number: " + number + " , divisor: " + divisor + ". remainder: " + n); | |

} | |

public static void main(String[] args) { | |

int n = 10; | |

int divisor = 4; | |

remainder(n, divisor); | |

} | |

} |

**Output**:

Number: 10, divisor: 4. remainder: 2

Continuously subtracting is not such an effective way to go. Instead, we can do it like this:

mul = int(number/divisor);

remainder = number – mul * divisor;