The range of the numbers is from 57 up to 634 which is a total of 578 numbers. So you want 578 unique patterns, hence you need to pick 3 digits at time.

Given the numbers, you could just use results from 057 to 634 and drop any outside the range.

You could also take the range from 000 to 577 and just add 57 to the result. This method might be useful for something like your needing numbers from 90 to 180. It looks like you would need 3 digits to account for 100 to 180, but you can actually use only 2 digits because you have 91 numbers. For 00 to 90, you just add 90 (to get 90 to 180). This way you'd only have to reject a few numbers in the range 91 to 99.

Answer:

(d) 3 digits

Use groups of three digits to get a list of three digit numbers (some of which will begin with a 0, like 057). Then delete any numbers which are greater than 634 or less than 57.

I do not know what procedures were taught in your class, but it would make sense to take the numbers in groups of three digits. Any number less than 057 or greater than 634 would be rejected. Keep choosing until you have nine qualifying numbers. That would result in a uniform distribution on the integers in interval [57, 634]. Every integer in that interval would have the same probability of being chosen. Repetitions are possible, but the instructions do not exclude that.

Do not entertain any ideas about choosing two-digit numbers. Simply treat them as three-digits numbers with a leading zero.

Well if you choose 2 digits, then the only numbers you will get are from 00 - 99. So it would be impossible to get a number from 100 - 634. So 3 digits is too few. Likewise, 1 digit only gives you the numbers 0-9.

If you choose 9 digits, then you will get very large numbers.

So the answer is not a, b, or c