How to Solve Week 1 Exam Questions

1. Change base 6 → decimal: (431)₆ = 4×6² + 3×6¹ + 1×6⁰ = 4×36 + 3×6 + 1 = 144 + 18 + 1 = 163
2. Change decimal → hex by dividing by 16: 163 ÷ 16 = 10, remainder 3 10 ÷ 16 = 0, remainder 10 → A
3. Read the remainders bottom → top: A3
4. Check: A3 = 10×16¹ + 3×16⁰ = 160 + 3 = 163 ✔

Trap: don’t write “103”. The remainder 10 is one hex digit: A.

1. Left side: (143)ₐ = $1 \times a^{2} + 4 \times a^{1} + 3 \times a^{0} =$ $a^{2} + 4 a + 3$
2. Right side: (212)ᵦ = $2 \times b^{2} + 1 \times b^{1} + 2 \times b^{0} =$ $2 b^{2} + b + 2$
3. Test the options:

1. Left: (234)ₐ = $2 \times a^{2} + 3 \times a + 4$
2. Right: (143)ᵦ = $1 \times b^{2} + 4 \times b + 3$
3. Test $a = 4 , b = 5$: left = 2×16 + 3×4 + 4 = 32 + 12 + 4 = 48; right = 25 + 20 + 3 = 48 ✔
4. Every other option gives two different numbers (for example $a = 5 , b = 6$ gives 69 vs 63).

Note: strictly, base 4 can only use the digits 0–3, so “234 in base 4” isn’t a valid number. The official answer is still $a = 4 , b = 5$, because it is the only option that makes both sides equal. In the exam, trust the equation.

1. Left: (344)ₐ = $3 \times a^{2} + 4 \times a + 4$
2. Right: (243)ᵦ = $2 \times b^{2} + 4 \times b + 3$
3. Test $a = 5 , b = 6$: left = 3×25 + 4×5 + 4 = 75 + 20 + 4 = 99; right = 2×36 + 4×6 + 3 = 72 + 24 + 3 = 99 ✔
4. Check one wrong option, $a = 4 , b = 5 : 4 8 + 1 6 + 4 = 6 8$ vs 50 + 20 + 3 = 73 ✗

Trap: don’t try to solve this with algebra. Testing 5 options takes about 1 minute.

1. “11” and “21” are written in base b. Expand them: 11ᵦ = $1 \times b^{1} + 1 \times b^{0} =$ $b + 1$ 21ᵦ = $2 \times b^{1} + 1 \times b^{0} =$ $2 b + 1$
2. Roots add up to the middle number: $3 + 5 = b + 1 \to 8 = b + 1 \to$ $b = 7$
3. Check with the product: 3 × 5 = 15 and $2 b + 1 = 2 \times 7 + 1 = 1 5$ ✔

1. 11ᵦ = $1 \times b + 1 =$ $b + 1$; 22ᵦ = $2 \times b + 2 =$ $2 b + 2$
2. Roots add up to the middle number: $3 + 6 = b + 1 \to 9 = b + 1 \to$ $b = 8$
3. Check with the product: 3 × 6 = 18 and 2×8 + 2 = 18 ✔

1. Find the powers of 2 around 128: 2⁷ = 128 and 2⁸ = 256.
2. 2⁷ ≤ 128 < 2⁸, so $n =$ 8.
3. Check by writing it: 128 = 1×2⁷ = 10000000. Count the digits: 8 ✔

Trap: 7 bits only reach 2⁷ − 1 = 127. 128 needs one more bit.

1. 2⁹ = 512 and 2¹⁰ = 1024.
2. 512 ≤ 515 < 1024, so $n =$ 10.
3. Check: 515 = 512 + 2 + 1 = 1×2⁹ + 1×2¹ + 1×2⁰ = 1000000011 → 10 digits ✔

1. Each flip-flop stores 1 bit, so n flip-flops give 2ⁿ states.
2. 4 flip-flops → 2⁴ = 16 states.
3. 16 < 64, so 4 is not enough. (You need 6, because 2⁶ = 64.)

1. 6 flip-flops → 2⁶ = 64 states. 64 ≥ 64 ✔, so it is enough.

1. 6 flip-flops → 2⁶ = 64 states. 64 < 65, so it is one short. You need 7 (2⁷ = 128).

1. 32 ÷ 2 = 16
2. 16 = 2 × 2 × 2 × 2 = 2⁴ → 4 doublings
3. 4 × 2 years = 8 years, not 7 → FALSE

1. same steps as above: 16× = 2⁴ = 4 doublings = 8 years ✔

1. it needs 8 years. In 4 years you get only 2 doublings: 2 → 4 → 8 million, not 32.

1. Moore’s Law is about the number of transistors doubling. Transistors get smaller, which is why more of them fit. One wrong word (“size”) makes the sentence false.