Number Systems: Question 1

Worked solution

Step 1: Understand what a logical right shift does

A logical right shift moves every bit in the register a fixed number of places to the right. Each bit that falls off the right-hand end is discarded, and 0s are filled in on the left-hand end so that the register stays a fixed width (8 bits here). A logical right shift by n places is the same as dividing the original value by 2^n and discarding any remainder.

Step 2: Write out the original register

Register (bit 7 down to bit 0): 0 1 0 1 1 0 1 0

Step 3: Shift every bit 2 places to the right

Drop the rightmost 2 bits (1 0), move the remaining 6 bits two places to the right, and fill the 2 new leftmost positions with 0s.

• Remaining bits after dropping the rightmost 2: 0 1 0 1 1 0
• Fill 2 zeros on the left: 00 + 010110 = 00010110

So after the shift, the register stores 00010110.

Step 4: Check using denary values

The original register 01011010 = 64 + 16 + 8 + 2 = 90 in denary. A logical right shift of 2 places divides by 2^2 = 4, discarding the remainder: 90 / 4 = 22 remainder 2, so the result is 22.

22 in 8-bit binary = 16 + 4 + 2 = 00010110, which matches Step 3, confirming 00010110 is correct.

Why the other options are wrong

• B (01101000, denary 104): this is what you get from shifting left by 2 places instead (90 x 4 = 360, which is truncated to 8 bits as 360 - 256 = 104), the opposite direction to the one asked for.
• C (10010110, denary 150): this is a rotate right by 2 places, where the 2 bits that fall off the right end (1 0) are wrapped around onto the left end instead of being discarded and replaced with 0s.
• D (00101101, denary 45): this is a logical right shift of only 1 place (90 / 2 = 45), not the 2 places asked for in the question.

Final answer

A, the register stores 00010110 after a logical right shift of 2 places.