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Number System Module ✦ Concept 11 / 13

Unit Digits & Last Two Digits

Master power cyclicity (0–9), Binomial Expansion for last 2 digits, Euler's totient application, and 24/76 magic rules for CAT & MBA CET.

The Bodhi Vault / Quant Vault / Unit & Last 2 Digits
DEFINITION

Modular Remainder Model

Finding the Unit Digit is equivalent to finding the remainder mod 10. Finding the Last Two Digits is equivalent to finding the remainder mod 100.

Unit Digit = N (mod 10) | Last 2 Digits = N (mod 100)
CORE INTUITION ⚡

Cyclicity & Binomial Anchors

The unit digit of any product depends ONLY on the unit digits of individual terms. For last 2 digits, we convert bases to end in 01, 24, or 76:

  • • Universal Cyclicity of Unit Digits = 4
  • • Base ending in 1: (...a1)...b → Unit digit = 1, Tens digit = UnitDigit(a × b)
  • • Even Base Anchor: 210 = 1024 → Ends in 24!
Never calculate actual large powers — reduce base and exponent mod 4 or mod 100!
💡 WHY THIS CONCEPT MATTERS IN MBA EXAMS

Unit digits and last two digits are frequent target areas in CAT, NMAT, SNAP, and MBA CET to test speed and modular shortcuts:

CAT Power Series
Cyclicity Table (0-9)
Ending in 1 Binomial Rule
24 & 76 Magic Rules
Factorial Sum Unit Digits
SNAP 30-Sec Elimination
NMAT Speed Calculations
Euler Totient Last 2 Digits
📐 UNIT DIGIT CYCLICITY TABLE (0–9)
Base Unit Digit N¹ N² N³ N⁴ Cyclicity Rule Summary
0, 1, 5, 6 0, 1, 5, 6 0, 1, 5, 6 0, 1, 5, 6 0, 1, 5, 6 1 Always remains unchanged!
4 4 6 4 6 2 Odd power → 4 | Even power → 6
9 9 1 9 1 2 Odd power → 9 | Even power → 1
2, 3, 7, 8 2, 3, 7, 8 4, 9, 9, 4 8, 7, 3, 2 6, 1, 1, 6 4 Universal cyclicity of 4 for all digits

Ending in 1 Rule: (...a1)...b

Unit Digit = 1
Tens Digit = UnitDigit(a × b)

Example: Last 2 digits of 41³7 → Unit digit = 1, Tens digit = 4 × 7 = 28 → 8. Answer: 81.

The 24 & 76 Magic Rules

76ⁿ → Ends in 76 for ANY positive n
24(even) → 76 | 24(odd) → 24

Anchor: 2¹⁰ = 1024 ends in 24. So (2¹⁰)⁵⁴ = 24⁵⁴ (even) → 76.

Euler Totient Mod 100

φ(100) = 40
a⁴⁰ ≡ 01 (mod 100)

For gcd(a, 100) = 1, reduce power E modulo 40 when calculating last 2 digits.

PRO-TIP HACK

Odd Numbers Ending in 3, 7, or 9

Convert base to end in 1:
• For 9: Use 9² = 81 → (...9)E = (...81)E/2
• For 3: Use 3⁴ = 81 → (...3)E = (...81)E/4
• For 7: Use 7⁴ = 2401 (ends in 01) → (...7)E = (...01)E/4

📝 STEP-BY-STEP SOLVED EXAMPLES
LEVEL 1: EASY Unit Digit Cyclicity

Find the unit digit of 7³⁴⁵ × 8¹²² × 3⁸⁷.

• 7³⁴⁵: 345 ÷ 4 = 86 rem 1 → 7¹ = 7

• 8¹²²: 122 ÷ 4 = 30 rem 2 → 8² = 64 → 4

• 3⁸⁷: 87 ÷ 4 = 21 rem 3 → 3³ = 27 → 7

• Product of unit digits = 7 × 4 × 7 = 196 → 6.

LEVEL 2: MEDIUM Last 2 Digits of Odd Base

Find the last two digits of 37¹⁰⁴.

• Step 1: 37² = 1369 (ends in 69).

• Step 2: 37¹⁰⁴ = (37²)⁵² = 69⁵² = (69²)²⁶.

• Step 3: 69² = (70 - 1)² = 4900 - 140 + 1 = 4761 (ends in 61).

• Step 4: Apply (...61)²⁶ rule → Unit digit = 1, Tens digit = 6 × 6 = 36 → 6.

• Answer: Last 2 digits are 61.

LEVEL 3: HARD (CAT PATTERN) Even Base Power

Find the last two digits of 2⁵⁴³.

• Step 1: 2⁵⁴³ = (2¹⁰)⁵⁴ × 2³.

• Step 2: Replace 2¹⁰ with 24 → (24)⁵⁴ × 8.

• Step 3: 24⁵⁴ is 24 raised to an EVEN power → 76.

• Step 4: 76 × 8 = 608 → Last 2 digits are 08.

⚠️ COMMON PITFALLS TO AVOID

Mistake 1: Exponent mod 4 = 0

When exponent E ÷ 4 has remainder 0, do NOT take power 0! Take power 4 (e.g. 2⁴ = 16 → 6).

Mistake 2: Single Digit Output

When calculating last 2 digits and the result is 8 or 5, format as 08 or 05 for two-digit representation!

Mistake 3: Euler Totient Conditions

You can only use a⁴⁰ ≡ 1 (mod 100) if gcd(a, 100) = 1 (i.e. a is not divisible by 2 or 5). For even numbers, use the 2⁴⁰ = 76 shortcut!

🎯 PRACTICE QUESTIONS

Q1: Find the unit digit of 1! + 2! + 3! + ... + 100!.

View Solution
For n ≥ 5, n! ends in 0. So unit digit = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 → 3.

Q2: What are the last two digits of 57⁸²?

View Solution

• 57² = 3249 → ends in 49.

• 57⁸² = (57²)⁴¹ = (49)⁴¹ = (49²)²⁰ × 49.

• 49² = 2401 → ends in 01.

• (01)²⁰ × 49 = 01 × 49 = 49.

Q3: Find the last two digits of 3²⁰²⁵.

View Solution

• 3⁴ = 81 (ends in 1).

• 3²⁰²⁵ = (3⁴)⁵⁰⁶ × 3¹ = (81)⁵⁰⁶ × 3.

• Last 2 digits of (81)⁵⁰⁶: Unit digit = 1, Tens digit = 8 × 6 = 48 → 8. So 81.

• 81 × 3 = 243 → Last 2 digits are 43.

❓ FREQUENTLY ASKED QUESTIONS

What is cyclicity in unit digits?

Cyclicity is the periodic repetition pattern of the unit digit when a number is raised to successive positive powers. Powers of 2 end in 2, 4, 8, 6 repeatedly with a cyclicity of 4.

How do you find the last two digits of a number ending in 1?

For a number ending in 1 like (...a1)...b, the unit digit is always 1, and the tens digit is given by the unit digit of (a × tens/unit digit of exponent b).

How do you find the last two digits of odd numbers ending in 3, 7, or 9?

Convert the base to end in 1 by raising it to suitable powers: for ending in 9, use N² (as 9² = 81); for 3 and 7, use N⁴ (as 3⁴ = 81 and 7⁴ = 2401 ending in 01). Then apply the (...a1)b rule.

How do you find the last two digits of powers of 2 or even numbers?

Use the benchmark 2¹⁰ = 1024 (ends in 24). Note that 24(even power) ends in 76, and 24(odd power) ends in 24. Also 76 raised to any positive integer power always ends in 76!

How does Euler's Totient function apply to finding last two digits?

Since φ(100) = 40, for any number a co-prime to 100, a⁴⁰ ≡ 01 (mod 100). Thus we can reduce large exponents modulo 40 when calculating last two digits.

Explore Related Number System Concepts

Concept 08

Factors & Multiples →

 Concept 09

HCF & LCM →

 Concept 10

Remainder Theorems →