OPERATING SYSTEM(S):
Windows XP

FULL JDK VERSION(S):
java version "1.5.0_05"
Java(TM) 2 Runtime Environment, Standard Edition (build 1.5.0_05-b05)
Java HotSpot(TM) Client VM (build 1.5.0_05-b05, mixed mode)

DESCRIPTION:
The method divideToIntegralValue(BigDecimal, MathContext) does not behave
to spec in certain circumstances.
  From the API: "Throws ArithmeticException if mc.precision > 0 and the
result requires a precision of more than mc.precision digits."
This does not occur with mc.precision=1 and with operands such as: 11 by 1,
22 by 2,  33 by 3 etc...

The following test case should throw an exception, since 33/3 = 11. 11 can
only be represented with a precision of at least 2, but we are setting the
required precision to 1.

import java.math.*;
class d96029 {
             public static void main (String [] args){
                         java.math.BigDecimal jmLHS = new
java.math.BigDecimal("33");
                         java.math.BigDecimal jmRHS = new
java.math.BigDecimal("3");
                         System.out.println("About to divide 33 by 3 to
integral value with max precision of 1.");
                         java.math.BigDecimal result  =
jmLHS.divideToIntegralValue(
                                                 jmRHS, new
java.math.MathContext(1, RoundingMode.UP));
                         System.out.println("Should have thrown execption
since 11 requires precision of 2. Result:");
                         System.out.println(result);
             }
}

Actual output is:
About to divide 33 by 3 to integral value with max precision of 1.
Should have thrown execption since 11 requires precision of 2. Result:
1E+1