Exercise
Denote by bk and b0 the `most significant' and the last digit of 300 in the 16-ary representation, respectively. Then
bk = 1, b0 = 2
bk = 2, b0 = C
bk = 1, b0 = C
The value for bk is fine. But recall that b0 is the remainder if we divide 300 by 16
The value for b0 is correct.
But in this example bk has to satisfy
bk · 162 < 300.
So bk < 2
Yes, this is correct. (Did you calculate the whole 16-ary representation of 300?)