Program: 73466
Project: 73466
Buggy source:
/** 
 * Divide this Time by the other Time object, and produce a new Time object as a result. If the dividend (this Time structure) is a perfect multiple of the time resolution, then the divideAndRemainder method of BigInteger is used, with the quotient as the new _timeValue. If the method returned a non-zero remainder, the divisor and remainder values are saved in the new Time structure. <p> If the dividend already has a non-null divisor and remainder fields, then we need to represent the final result of the division in a non-null divisor and remainder. The final representation is as follows: let q1 be the quotient, b be the divisor, and r1 be the remainder of this time structure (the dividend time structure). Let c be the divisor, let q2 be q1/c, let r2 be q1%c, then the final result can be represented as: q2 + (r1 + r2*b) / (b*c). In other words, the returned Time structure would have _timeValue = q2 + (r1 + r2*b) / (b*c), while the divisor would be b*c, and remainder would be (r1 + r2*b) % (b*c).
 * @param dividend The dividend of this division.
 * @param resolutionInverse The inverse of the time resolution
 * @param divisor The divisor of this divisor.
 * @return a new Time structure of the division.
 */
private Time _divide(BigInteger resolutionInverse,Time time){
  if (_divisorAndRemainder == null && time._divisorAndRemainder == null) {
    return _bigIntegerDivideByBigInteger(_timeValue,time._timeValue,resolutionInverse);
  }
 else   if (_divisorAndRemainder != null && time._divisorAndRemainder == null) {
    return _timeDivideByBigInteger(this,time._timeValue,resolutionInverse);
  }
 else   if (_divisorAndRemainder == null && time._divisorAndRemainder != null) {
    BigInteger divisor=time._timeValue.multiply(time._divisorAndRemainder[0].divide(resolutionInverse).add(time._divisorAndRemainder[1]));
    return _bigIntegerDivideByBigInteger(_timeValue.multiply(time._divisorAndRemainder[0]),divisor,resolutionInverse);
  }
 else {
    BigInteger divisor=time._timeValue.multiply(time._divisorAndRemainder[0].divide(resolutionInverse).add(time._divisorAndRemainder[1]));
    Time dividend=this.multiply(new Time(_director,time._divisorAndRemainder[0],null,null));
    return _timeDivideByBigInteger(dividend,divisor,resolutionInverse);
  }
}

Fixed version (ground truth):
/** 
 * Divide this Time by the other Time object, and produce a new Time object as a result. If the dividend (this Time structure) is a perfect multiple of the time resolution, then the divideAndRemainder method of BigInteger is used, with the quotient as the new _timeValue. If the method returned a non-zero remainder, the divisor and remainder values are saved in the new Time structure. <p> If the dividend already has a non-null divisor and remainder fields, then we need to represent the final result of the division in a non-null divisor and remainder. The final representation is as follows: let q1 be the quotient, b be the divisor, and r1 be the remainder of this time structure (the dividend time structure). Let c be the divisor, let q2 be q1/c, let r2 be q1%c, then the final result can be represented as: q2 + (r1 + r2*b) / (b*c). In other words, the returned Time structure would have _timeValue = q2 + (r1 + r2*b) / (b*c), while the divisor would be b*c, and remainder would be (r1 + r2*b) % (b*c).
 * @param dividend The dividend of this division.
 * @param resolutionInverse The inverse of the time resolution
 * @param divisor The divisor of this divisor.
 * @return a new Time structure of the division.
 */
private Time _divide(BigInteger resolutionInverse,Time time){
  if (_divisorAndRemainder == null && time._divisorAndRemainder == null) {
    return _bigIntegerDivideByBigInteger(_timeValue,time._timeValue,resolutionInverse);
  }
 else   if (_divisorAndRemainder != null && time._divisorAndRemainder == null) {
    return _timeDivideByBigInteger(this,time._timeValue,resolutionInverse);
  }
 else   if (_divisorAndRemainder == null && time._divisorAndRemainder != null) {
    BigInteger divisor=time._timeValue.multiply(time._divisorAndRemainder[0]).add(time._divisorAndRemainder[1]);
    return _bigIntegerDivideByBigInteger(_timeValue.multiply(time._divisorAndRemainder[0]),divisor,resolutionInverse);
  }
 else {
    BigInteger divisor=time._timeValue.multiply(time._divisorAndRemainder[0]).add(time._divisorAndRemainder[1]);
    Time dividend=this.multiply(time._divisorAndRemainder[0].doubleValue());
    if (dividend._divisorAndRemainder == null) {
      return _bigIntegerDivideByBigInteger(dividend._timeValue,divisor,resolutionInverse);
    }
    return _timeDivideByBigInteger(dividend,divisor,resolutionInverse);
  }
}