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(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.
