SOLUTION: 322base5÷4base5 Figure the problem using base 5blocks. Figure the problem using standard algorithm. Figure the problem using repeated subtraction.

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Question 1161439: 322base5÷4base5
Figure the problem using base 5blocks.
Figure the problem using standard algorithm.
Figure the problem using repeated subtraction.
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


(1) I don't know what "base 5 blocks" are....

(2) To divide using long division, it is helpful to have a multiplication table for 4 in base 5:

1*4 = 4 (base 10) = 4 (base 5)
2*4 = 8 = 13 (base 5)
3*4 = 12 = 22 (base 5)
4*4 = 16 = 31 (base 5)

Then in the subtraction part of the algorithm, you of course need to do the subtraction in base 5.

          4 1 remainder 3
      --------
    4 ) 3 2 2
        3 1      [4*4 = 31]
       ------
          1 2    [subtract; bring down the next digit]
            4    [4*1 = 4]
          ---
            3    [subtract to get the remainder]

(3) Solving using repeated subtraction is straightforward, but gives us a lot of practice in base 5 arithmetic....
1  322-4 = 313
2  313-4 = 304
3  304-4 = 300
4  300-4 = 241
5  241-4 = 232
6  232-4 = 223
7  223-4 = 214
8  214-4 = 210
9  210-4 = 201
10 201-4 = 142
11 142-4 = 133
12 133-4 = 124
13 124-4 = 120
14 120-4 = 111
15 111-4 = 102
16 102-4 =  43
17  43-4 =  34
18  34-4 =  30
19  30-4 =  21
20  21-4 =  12
21  12-4 =   3

We subtracted 4 21 times, which is 41 in base 5; and we were left with a remainder of 3. So again we got the result 322 divided by 4 in base 5 = 41 remainder 3.