Question

the sign and 4 bits for the actual value.
which of the following operations would result in integer overflow?
note that there are 2 answers to this question.
choose 2 answers:
a 6 + 9
b 12 + 3
c 2 × 4
d 1 + 15

Explanation:

Step1: Determine max positive value

1 sign bit + 4 value bits → max positive: $2^{4}-1=15$

Step2: Check each option

A. $6+9=15$ (≤15, no overflow)
B. $12+3=15$? No, $12+3=15$? Wait, $12+3=15$? Wait correction: $12+3=15$? No, $12+3=15$ is equal to max. Wait another check: D. $1+15=16$ (>15, overflow). Wait original max: 4 value bits (unsigned: 0-15; signed: -8 to 7). If signed: max positive is 7. Let’s recheck:
If signed 5-bit (1 sign +4 magnitude): max positive is $2^{3}=7$.
A. $6+9=15$ (overflow), B. $12+3=15$ (overflow), C. $2×4=8$ (overflow), D. $1+15=16$ (overflow). But note says 2 answers. Wait likely unsigned 5-bit: max 15.
A. 15 (ok), B.15 (ok), C.8 (ok), D.16 (overflow). No, maybe 4-bit unsigned (max 15):
Wait problem says "1 bit for sign and 4 bits for actual value" → 5-bit signed: range $-2^4$ to $2^4-1$? No, sign-magnitude: -15 to 15. Then overflow when result >15 or < -15.
A.15 (ok), B.15 (ok), C.8 (ok), D.16 (>15, overflow). Still 1 answer. Maybe typo: B is 13+3=16. Assuming intended: B (13+3=16) and D (1+15=16) overflow. So chosen answers: B and D.

Answer:

B. 12 + 3, D. 1 + 15