I am stuck on this concept.

How many flip-flops, in a shift register, would be required to hold an initial count of 16?

 

Flip-flops can have a value of 1 or 0, a.k.a. binary, or base 2. To change from base 10 (our normal counting base) to base 2 (or 3 or 4 or 8 or 16, or any other base) by hand, you will need to do some division (r stands for remainder):
So to convert 11 to base 2:

11 / 2 = 5 r 1
then
5 / 2 = 2 r 1
and
2 / 2 = 1 r 0
finally
1 / 2 = 0 r 1

Next, gather the remainders in reverse order: 1, 0, 1, 1. This will give you 11 in base 2. As a proof, convert 1101 back to decimal:

1 * 2E4 (==8) = 8
0 * 2E2 (==4) = 0
1 * 2E1 (==2) = 2
1 * 2E0 (==1) = 1

Add 'em up, and you get 8+2+1 == 11.

HTH,
--Rich

 

great! thanks, this makes perfect sense!