Number Systems & Binary Logic

Binary (base-2, digits 0-1), Octal (base-8, 0-7), Decimal (base-10, 0-9), Hexadecimal (base-16, 0-9 then A-F where A=10...F=15). Memory aid: 'BODH' = Binary-Octal-Decimal-Hex with bases 2-8-10-16. Conversion shortcuts: every 3 binary digits = 1 octal digit; every 4 binary digits = 1 hex digit. So Binary→Octal: group bits in 3s; Binary→Hex: group in 4s (from right). Example: 11010110 → group as 1101|0110 = D6 hex; group as 011|010|110 = 326 octal. SBI PO computer section tests base conversion and digit-validity ('which is NOT a valid octal number'). Largest single hex digit F = 15 = 1111 binary.

To convert decimal→binary, repeatedly divide by 2 and read remainders BOTTOM to TOP. Faster exam trick: subtract powers of 2 (128,64,32,16,8,4,2,1). Example: 45 → 32+8+4+1 → mark positions: 128(0)64(0)32(1)16(0)8(1)4(1)2(0)1(1) = 00101101. Reverse (binary→decimal): multiply each bit by its positional power of 2 and sum. Example: 101101 = 32+8+4+1 = 45. Memorize powers of 2 up to 1024: 1,2,4,8,16,32,64,128,256,512,1024. SBI PO speed tip: for a string of n ones, value = 2^n − 1 (e.g., 11111 = 2^5−1 = 31). This saves seconds in conversion-heavy questions.

Basic gates: AND (output 1 only if ALL inputs 1), OR (1 if ANY input 1), NOT (inverter). Derived: NAND (NOT-AND), NOR (NOT-OR), XOR (1 if inputs DIFFER), XNOR (1 if inputs SAME). Memory aid: XOR = 'eXclusive — outputs 1 when inputs are unequal'. NAND and NOR are 'universal gates' — any circuit can be built from them alone. Truth-table quick recall for 2 inputs: AND=0001, OR=0111, XOR=0110 (for input pairs 00,01,10,11). SBI PO asks gate outputs and universal-gate identification. Boolean: A AND 0 = 0; A OR 1 = 1; A AND 1 = A; A OR 0 = A; NOT(NOT A) = A.

Cover the answer, recall, then check. 12 cards on number systems, conversions and gates.

Q1. State the base (radix) of Binary, Octal, Decimal and Hexadecimal.
A1. Binary = 2, Octal = 8, Decimal = 10, Hexadecimal = 16.

Q2. In hexadecimal, what decimal values do A and F represent?
A2. A = 10 and F = 15 (A–F represent 10–15).

Q3. Convert binary 1010 to decimal.
A3. 8 + 0 + 2 + 0 = 10.

Q4. Convert decimal 45 to binary.
A4. 45 = 32+8+4+1 = 101101.

Q5. Binary→Octal and Binary→Hex grouping rule?
A5. Group binary digits in 3s for octal and in 4s for hexadecimal (from the right).

Q6. Truth of AND, OR and NOT gates in one line each.
A6. AND = 1 only if ALL inputs 1; OR = 1 if ANY input 1; NOT = inverts the input.

Q7. Which two gates are called "universal gates" and why?
A7. NAND and NOR — any logic circuit can be built using only NAND or only NOR gates.

Q8. When does an XOR gate output 1?
A8. When its inputs DIFFER (one 0 and one 1); XNOR outputs 1 when inputs are the SAME.

Q9. How do you get the 2's complement of a binary number?
A9. Take the 1's complement (flip all bits) and add 1; used to represent negative numbers.

Q10. Expand ASCII and give its bit size.
A10. American Standard Code for Information Interchange — 7-bit (128 characters); extended ASCII is 8-bit (256).

Q11. What is BCD and what does Unicode add over ASCII?
A11. BCD = Binary Coded Decimal (each decimal digit in 4 bits); Unicode uses 16+ bits to encode almost every language's characters.

Q12. Convert hexadecimal FF to decimal.
A12. F(15)×16 + F(15) = 240 + 15 = 255.

Number systems bridge Computer Aptitude and Reasoning, so SBI PO's Reasoning & Computer Aptitude section loves them — a base conversion or a logic-gate truth question is almost guaranteed. The maths is tiny; the marks come from knowing the four bases and the seven gates cold.

The four number systems

System	Base	Digits used
Binary	2	0, 1
Octal	8	0–7
Decimal	10	0–9
Hexadecimal	16	0–9, A–F (A=10 … F=15)

Conversions: Binary→Decimal = multiply each bit by its power of 2 and add. Decimal→Binary = repeatedly divide by 2, read remainders bottom-to-top (or subtract powers of 2). Group 3 bits = 1 octal digit, 4 bits = 1 hex digit. MSB = most significant (leftmost) bit; LSB = least significant (rightmost).

Logic gates

• Basic: AND (1 if all inputs 1), OR (1 if any input 1), NOT (inverter).
• Derived: NAND (NOT-AND), NOR (NOT-OR), XOR (1 if inputs differ), XNOR (1 if inputs same).
• NAND and NOR are universal gates — any circuit can be built from either alone.

Negative numbers use 2's complement = 1's complement (flip bits) + 1. Character codes: ASCII = 7-bit/128 chars, Unicode = 16-bit+, BCD codes each decimal digit in 4 bits.

Exam Tricks & Tips

• 🎯 Base order mnemonic "BODH" = Binary-Octal-Decimal-Hex with bases 2-8-10-16.
• 🎯 Binary→Decimal fast: mark position values 128-64-32-16-8-4-2-1 and add only where the bit is 1.
• 🎯 XOR = eXclusive → outputs 1 only when inputs are unequal; XNOR is its opposite.
• 🎯 Universal gates are NAND & NOR — remember "the two N-gates rule them all."
• 🎯 2's complement = "flip and add 1".
• ❌ Common mistake: reading decimal→binary remainders top-to-bottom. Always read them bottom-to-top.

Expected exam pattern

1–2 questions: convert a small binary/decimal/hex value, identify a gate's output, or state which gates are universal. Numbers are kept small for mental calculation.

Quick recap

Bases 2-8-10-16 (BODH); group 3 bits=octal, 4 bits=hex; AND/OR/NOT + NAND/NOR (universal)/XOR/XNOR; 2's complement = flip+1; ASCII 7-bit, Unicode 16-bit.

Worked Example

Problem: Solved computation: Convert the hexadecimal number (2F)₁₆ into decimal and then into binary.

Solution:
Hexadecimal is base-16, where F = 15.
To decimal: (2F)₁₆ = 2 × 16¹ + 15 × 16⁰ = 32 + 15 = 47.
To binary: convert each hex digit into 4 bits.
2 → 0010
F → 1111
So (2F)₁₆ = 0010 1111 = (101111)₂ (dropping leading zeros).
Check: 32 + 8 + 4 + 2 + 1 = 47 ✓.

Answer: (2F)₁₆ = 47₁₀ = (101111)₂.