Three rolls: With each roll assign a 1 for an odd number, and a 0 for an even number.   Each roll contributes one digit of a three digit binary number between 0 and 7.   

Two rolls: No one came up with a method guaranteed to work every time with two rolls.  Nevertheless, consider the following algorithm:  Roll the die twice with each roll contributing one digit of a two digit decimal number.  This provides thirty-six possible numbers, which we can divide into nine sets of size four.  Depending on which set your number falls into, you can then select a number from 1 to 9.  If the two rolls generate a 9, roll the die once more and use the technique for three rolls.  This last roll will only be needed one time in nine, so the expected number of rolls is 2 1/9.  

One roll:  Yes, IT CAN BE DONE with one roll.  This technique was submitted by Ron Welch. Roll the die only once and write down the number that appears on top AND the one on the face most facing you.  (This will only fail in the case when there are two faces equally facing you, which occurs with probability zero.)  Since the top and bottom faces always add to seven, this procedure will provide a two-digit number in the set 

{12,13,14,15,21,23,24,26,31,32,35,36,41,42,45,46,51,53,54,56,62,63,64,65}.

There are twenty-four numbers in the set which you can be divided into eight sets of size three and with which you can select your problem.  Way to go Ron!  

Several people "proved" it couldn't be done with fewer than three rolls.  It is highly unlikely, however, that it can be done with fewer than one roll!

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