What is the unit digit of 7^53?

The cyclicity of 7 is 4 (7,9,3,1). 53 ÷ 4 gives remainder 1, so the unit digit is 7.

How many divisors does 120 have?

120 = 2³ × 3 × 5. Number of divisors = (3+1)(1+1)(1+1) = 16.

Is 0 a natural number?

No. Natural numbers start from 1. Whole numbers include 0. Integers include negative numbers too.

What is the divisibility rule for 8?

A number is divisible by 8 if its last three digits form a number divisible by 8.

How do you find unit digit of a large power like 2^100?

Cyclicity of 2 is 4 (2,4,8,6). 100 ÷ 4 = 25 remainder 0. When remainder is 0, take the last digit in the cycle, which is 6.

Number System — Formulas, Shortcuts, and Worked Examples | TNPSC, SSC, Bank

Quick Concept Summary

The Number System is the foundation of all quantitative aptitude. Numbers are broadly classified as Natural Numbers (1, 2, 3, ...), Whole Numbers (0, 1, 2, 3, ...), Integers (..., -2, -1, 0, 1, 2, ...), Rational Numbers (p/q form where q ≠ 0), Irrational Numbers (like √2, π), and Real Numbers (all rational + irrational). For TNPSC exams, the focus lies on properties of integers, divisibility, and patterns.

Divisibility rules allow you to test whether a number divides another without performing actual division — a massive time-saver in exams. For instance, to check if 4,872 is divisible by 8, you only need to check the last three digits (872 ÷ 8 = 109 exactly, so yes). These rules are derived from the base-10 positional system and are completely reliable.

Unit digit cyclicity is another high-yield pattern. When you raise a number to a large power, the unit digit of the result follows a cycle of length 1, 2, or 4. Memorising cycles for digits 0–9 lets you answer "unit digit of 7^99" questions in under 10 seconds. Combined with divisibility rules and knowledge of number types, this topic typically contributes 3–5 marks per TNPSC Group II/IIA paper.

Key Formulas

Formula	Description	Notes
Sum of 1 to n = n(n+1)/2	Sum of first n natural numbers	e.g., 1+2+...+100 = 5050
Sum of squares = n(n+1)(2n+1)/6	Sum of 1² + 2² + ... + n²	e.g., n=10 → 385
Sum of cubes = [n(n+1)/2]²	Sum of 1³ + 2³ + ... + n³	Always equals (sum of naturals)²
Sum of first n odd numbers = n²	1+3+5+...+(2n-1) = n²	e.g., first 5 odds → 25
Sum of first n even numbers = n(n+1)	2+4+6+...+2n = n(n+1)	e.g., first 5 evens → 30
LCM × HCF = Product of two numbers	Only valid for exactly two numbers	Do NOT apply to three numbers
Number of divisors = (a+1)(b+1)(c+1)	If n = p^a × q^b × r^c	Includes 1 and the number itself
Divisibility by 3	Sum of digits divisible by 3	e.g., 123 → 1+2+3=6 ✓
Divisibility by 11	(Odd position sum) - (Even position sum) = 0 or 11k	e.g., 121 → (1+1)−2 = 0 ✓
Divisibility by 8	Last 3 digits divisible by 8	e.g., 5,872 → 872 ÷ 8 = 109 ✓

Worked Examples

Example 1: Finding the unit digit of a large power

Problem: Find the unit digit of 7^85.

Step 1: Identify the cyclicity of 7.
        7^1 = 7   → unit digit 7
        7^2 = 49  → unit digit 9
        7^3 = 343 → unit digit 3
        7^4 = 2401→ unit digit 1
        7^5 = ...  → unit digit 7 (cycle repeats)
        Cyclicity of 7 = 4 (the pattern 7, 9, 3, 1 repeats)

Step 2: Divide the power by the cyclicity.
        85 ÷ 4 = 21 remainder 1

Step 3: The remainder is 1, so the unit digit is the 1st digit in the cycle.
        Cycle: 7 (position 1), 9 (position 2), 3 (position 3), 1 (position 4)
        Position 1 → unit digit = 7

Answer: Unit digit of 7^85 = 7

Example 2: Applying divisibility rule for 11

Problem: Is 7,69,428 divisible by 11?

Step 1: Label digit positions from the right.
        Digits (right to left):  8, 2, 4, 9, 6, 7
        Position:                1, 2, 3, 4, 5, 6

Step 2: Sum of digits at odd positions (1st, 3rd, 5th from right):
        8 + 4 + 6 = 18

Step 3: Sum of digits at even positions (2nd, 4th, 6th from right):
        2 + 9 + 7 = 18

Step 4: Difference = 18 - 18 = 0

Step 5: Since the difference is 0 (a multiple of 11), the number IS divisible by 11.

Answer: Yes, 7,69,428 is divisible by 11.

Example 3: Number of divisors

Problem: How many divisors does 360 have?

Step 1: Prime factorisation of 360.
        360 = 2 × 180
            = 2 × 2 × 90
            = 2 × 2 × 2 × 45
            = 2 × 2 × 2 × 3 × 15
            = 2 × 2 × 2 × 3 × 3 × 5
        360 = 2³ × 3² × 5¹

Step 2: Use the divisor formula.
        Number of divisors = (3+1) × (2+1) × (1+1)
                           = 4 × 3 × 2
                           = 24

Answer: 360 has 24 divisors.

Verification check: Some of the 24 divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12,
15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360. ✓

Example 4: Sum formula application

Problem: Find the sum of all odd numbers from 1 to 99.

Step 1: Identify the series.
        Odd numbers: 1, 3, 5, ..., 99
        These are the first n odd numbers.

Step 2: Find n (how many odd numbers are there from 1 to 99).
        The k-th odd number = 2k - 1
        Set 2k - 1 = 99 → 2k = 100 → k = 50
        So there are 50 odd numbers from 1 to 99.

Step 3: Apply the formula: Sum of first n odd numbers = n²
        Sum = 50² = 2500

Answer: Sum of odd numbers from 1 to 99 = 2500

Cross-check: Sum of all numbers 1 to 99 = 99×100/2 = 4950
             Sum of even numbers 2 to 98 = 2+4+...+98 = 49×50 = 2450 (n=49 even numbers... wait)
             Actually: even numbers 2,4,...,98 → n=49, sum = 49×50 = 2450
             Odd sum = 4950 - 2450 = 2500 ✓

Example 5: Unit digit with cyclicity = 1 or 2

Problem: Find the unit digit of (i) 6^200 (ii) 5^37 (iii) 2^46

Part (i): 6^200
        6^1 = 6, 6^2 = 36, 6^3 = 216
        Unit digit of any power of 6 is always 6.
        Cyclicity of 6 = 1.
        Answer: Unit digit = 6

Part (ii): 5^37
        5^1 = 5, 5^2 = 25, 5^3 = 125
        Unit digit of any power of 5 is always 5.
        Cyclicity of 5 = 1.
        Answer: Unit digit = 5

Part (iii): 2^46
        Cycle of 2: 2^1=2, 2^2=4, 2^3=8, 2^4=16 → pattern is 2, 4, 8, 6
        Cyclicity = 4
        46 ÷ 4 = 11 remainder 2
        Remainder 2 → position 2 in cycle → unit digit = 4
        Answer: Unit digit = 4

Shortcuts and Tricks

Cyclicity Table (Memorise this)

Unit Digit	Cycle	Cyclicity
0	0	1
1	1	1
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
5	5	1
6	6	1
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2

Rule when remainder is 0: Take the last digit in the cycle (e.g., 2^20 → 20÷4=5 rem 0 → last of cycle {2,4,8,6} = 6).

Divisibility Quick Reference

Divisor	Rule
2	Last digit even
3	Sum of digits divisible by 3
4	Last 2 digits divisible by 4
5	Last digit 0 or 5
6	Divisible by both 2 and 3
7	Double the last digit, subtract from rest; repeat if needed
8	Last 3 digits divisible by 8
9	Sum of digits divisible by 9
10	Last digit is 0
11	Alternating sum = 0 or multiple of 11
12	Divisible by both 3 and 4

Number of Zeros at the End of n!

Count factors of 5 in n! using: ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

Example: Zeros in 100! = ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24 zeros

Shortcut for Consecutive Number Sums

If asked "sum of integers from 15 to 30":

Total terms = (30 - 15) + 1 = 16
Sum = (first + last) × terms / 2 = (15 + 30) × 16 / 2 = 45 × 8 = 360

TNPSC Exam Pattern

Frequency: 3–5 questions per paper in TNPSC Group II and Group IV. Number system questions appear in both the Aptitude section and the General Studies (Maths) section depending on the paper.

Common Question Types: 1. Unit digit of a^b (always 1 question, easy 1 mark) 2. Divisibility-based questions ("which of the following is divisible by 9?") 3. Sum of series questions (first n naturals, first n odd/even) 4. Number of divisors or factor-count questions 5. Finding the largest/smallest number satisfying a divisibility condition

Common Traps:

"LCM × HCF = product" is only valid for two numbers. Exam setters sometimes use three numbers to trap candidates.
When remainder is 0 in cyclicity problems, students often write 0 as the unit digit instead of taking the last element of the cycle.
Divisibility by 4 rule: candidates confuse it with divisibility by 8 (last 2 vs last 3 digits).
"Natural numbers" vs "whole numbers" — exam may ask "how many natural numbers from 1 to 50 are divisible by both 3 and 5?" — the word "natural" matters for starting point.

Marks Weightage: In TNPSC Group II Preliminary, Aptitude carries 25 marks total. Number System typically accounts for 3–5 of those marks. Every mark counts.

Practice Questions

Try these questions before checking the answers.

Q1. What is the unit digit of 3^71 × 6^52?

Q2. How many numbers from 1 to 200 are divisible by both 4 and 6?

Q3. Find the sum of all natural numbers from 1 to 50 that are divisible by 3.

Q4. How many divisors does 540 have?

Q5. A number when divided by 3 leaves remainder 2, and when divided by 5 leaves remainder 1. What is the smallest such positive number?

Q6. What is the unit digit of (1234)^321 + (5678)^456?

Q7. Is 5,84,663 divisible by 11?

Q8. Find the number of trailing zeros in 50!

Q9. The sum of first 20 odd natural numbers is:

Q10. What is the largest 4-digit number divisible by 13?

Answers

Q	Answer	Key Working
Q1	8	3^71: cycle 4, 71÷4=17 rem 3 → unit digit 7. 6^52: always 6. 7×6=42 → unit digit 8
Q2	16	LCM(4,6)=12. Numbers: 12,24,...,192,204>200. Count = ⌊200/12⌋ = 16
Q3	408	Series: 3,6,9,...,48 (last multiple of 3 ≤ 50 is 48). n=16. Sum = 16×17/2 × 3 = 408
Q4	24	540 = 2² × 3³ × 5. Divisors = (2+1)(3+1)(1+1) = 3×4×2 = 24
Q5	11	Numbers with rem 1 when ÷5: 6,11,16,21... Check ÷3: 6→rem 0, 11→rem 2 ✓. Answer = 11
Q6	5	4^321: cycle 2, odd power → unit 4. 8^456: cycle 4, 456÷4=114 rem 0 → last cycle digit = 6. 4+6=10 → unit 0... Wait: unit of 1234 is 4, unit of 5678 is 8. 4^321: rem 1 (odd) → 4. 8^456: rem 0 → 6. 4+6=10 → unit digit = 0
Q7	No	Odd pos (R to L): 3+6+8 = 17. Even pos: 6+4+5 = 15. Diff = 17-15 = 2 ≠ 0 or 11. Not divisible.
Q8	12	⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12
Q9	400	Sum of first n odd naturals = n². n=20. 20² = 400
Q10	9997	9999 ÷ 13 = 769 rem 2. So 9999 - 2 = 9997