ITPEC FE Morning April 2022 Question 2

Bitwise equivalence of modulo and addition — convert (A mod 32) + 64 to bitwise form.

Key rules:
- A mod 2^n = A AND (2^n − 1) — extracts the lowest n bits
- + 2^k = OR 2^k — when the bit being set doesn't overlap with existing bits

Step-by-step:
- A mod 32 → 32 = 2⁵ → A AND 31 (31 = 0b11111, extracts bits 0–4)
- + 64 → 64 = 2⁶ = 0b1000000 → bit 6 doesn't overlap with bits 0–4 → OR 64
- Result: (A AND 31) OR 64 → (a)

Why not others:
- (b) A AND 32 extracts only bit 5, not lower 5 bits; OR 32 sets bit 5, not bit 6
- (c) A OR 31 sets lower bits to 1 instead of extracting; AND 64 zeroes everything except bit 6
- (d) A OR 64 sets bit 6 but AND 32 keeps only bit 5, losing all other information

Key rule: mod 2^n ↔ AND (2^n − 1); non-overlapping + 2^k ↔ OR 2^k.

AI-generated — may contain errors