Number System for SSC CGL: The Short Answer
The number system chapter in SSC CGL covers types of numbers, divisibility rules, factors, remainders, unit digits, trailing zeros and sums of series. It is the most-asked arithmetic base topic: SSC CGL Tier 1 usually has 2 to 3 questions from it, and Tier 2 has more. The fastest wins come from five tools: divisibility rules (2 to 13), prime factorisation for counting factors, the remainder cycle method, the 5-division rule for trailing zeros, and the n(n + 1)/2 family of sum formulas.
Key takeaway for SSC: Learn the divisibility table and the factor formula by heart. Most "find the missing digit", "how many factors" and "find the remainder" questions then take under 45 seconds. Every number on this page was checked in Python.
Unit digits and LCM–HCF are separate notes, so see our unit digit tricks for SSC CGL and LCM and HCF tricks for SSC CGL for those parts of the chapter.
Types of Numbers in Number System (Table)
| Type | Meaning | Examples |
|---|---|---|
| Natural numbers | Counting numbers | 1, 2, 3 … |
| Whole numbers | Natural numbers and 0 | 0, 1, 2 … |
| Integers | Whole numbers and negatives | … −2, −1, 0, 1, 2 … |
| Rational numbers | Can be written as p/q, q ≠ 0 | 3/4, −5, 0.75, 0.333… |
| Irrational numbers | Cannot be written as p/q; non-terminating, non-repeating decimals | √2, √3, π |
| Real numbers | All rational and irrational numbers | — |
| Even / odd | Divisible by 2 / not divisible by 2 | 8 / 9 |
| Prime numbers | Exactly two factors: 1 and itself | 2, 3, 5, 7, 11 |
| Composite numbers | More than two factors | 4, 6, 8, 9 |
| Co-prime numbers | HCF = 1 | 8 and 15 |
Facts SSC tests often:
- 1 is neither prime nor composite.
- 2 is the only even prime.
- There are 25 prime numbers from 1 to 100.
- 0 is a whole number but not a natural number.
- A terminating or repeating decimal is rational; 0.101001000100001… (no repeating block) is irrational.
Divisibility Rules for SSC CGL (2 to 13)
| Divisible by | Rule | Example |
|---|---|---|
| 2 | Last digit even | 3,574 |
| 3 | Sum of digits divisible by 3 | 8,412 → 15 |
| 4 | Last two digits divisible by 4 | 7,316 → 16 |
| 5 | Last digit 0 or 5 | 9,435 |
| 6 | Divisible by both 2 and 3 | 5,412 |
| 7 | Double the last digit and subtract from the rest; repeat | 3,192 → 319 − 4 = 315 → 31 − 10 = 21 ✔ |
| 8 | Last three digits divisible by 8 | 7,316,152 → 152 |
| 9 | Sum of digits divisible by 9 | 675 → 18 |
| 10 | Last digit 0 | 4,560 |
| 11 | (Sum of odd-place digits) − (sum of even-place digits) = 0 or a multiple of 11 | 918,082 → 25 − 3 = 22 ✔ |
| 12 | Divisible by both 3 and 4 | 1,428 |
| 13 | Multiply last digit by 4 and add to the rest; repeat | 1,001 → 100 + 4 = 104 → 10 + 16 = 26 ✔ |
For composite divisors, use co-prime factors: divisible by 15 → check 3 and 5; by 24 → check 3 and 8 (not 4 and 6, which are not co-prime); by 72 → check 8 and 9.
Example 1: Find the digit x if 6x5 is divisible by 9.
6 + x + 5 = 11 + x must be a multiple of 9 → x = 7 (675).
Example 2: Find x if 7x9 is divisible by 11.
Odd places 7 + 9 = 16, even place x. 16 − x = 0 or 11 → x = 5 (759 = 11 × 69).
Example 3: Find the smallest digit x for which 72x6 is divisible by 8.
Last three digits 2x6 must be divisible by 8: 216 works (x = 1). Answer x = 1 (other values: 5 and 9).
Factors of a Number Formula (Number and Sum of Factors)
Write the number as N = aᵖ × bᵠ × cʳ (a, b, c prime).
| Find | Formula |
|---|---|
| Number of factors | (p + 1)(q + 1)(r + 1) |
| Sum of factors | (a⁰ + a¹ + … + aᵖ)(b⁰ + … + bᵠ)(c⁰ + … + cʳ) |
| Number of even factors | Total factors − odd factors (odd factors ignore the power of 2) |
Example 4: How many factors does 360 have, and what is their sum?
360 = 2³ × 3² × 5.
- Number = (3 + 1)(2 + 1)(1 + 1) = 24
- Sum = (1 + 2 + 4 + 8)(1 + 3 + 9)(1 + 5) = 15 × 13 × 6 = 1,170
- Odd factors = (2 + 1)(1 + 1) = 6, so even factors = 24 − 6 = 18.
Example 5: How many factors does 720 have?
720 = 2⁴ × 3² × 5 → (5)(3)(2) = 30.
Remainder Theorem Tricks for SSC CGL
Basic rule: the remainder of a product equals the product of the remainders (then take the remainder again).
Example 6: Find the remainder when 2³¹ is divided by 5.
Powers of 2 mod 5 repeat every 4: 2, 4, 3, 1. 31 = 4 × 7 + 3, so the remainder is the 3rd value: 3.
Example 7: Remainder when 17²³ is divided by 16.
17 leaves remainder 1 when divided by 16, and 1²³ = 1. Remainder 1.
Example 8: Remainder when 7¹⁰⁰ is divided by 6.
7 ≡ 1 (mod 6), so 7¹⁰⁰ leaves 1.
Shortcut rules:
| Form | Remainder |
|---|---|
| (a + 1)ⁿ ÷ a | 1 |
| (a − 1)ⁿ ÷ a, n even | 1 |
| (a − 1)ⁿ ÷ a, n odd | a − 1 |
| (xⁿ + aⁿ) ÷ (x + a), n odd | 0 (divisible) |
| (xⁿ − aⁿ) ÷ (x + a), n even | 0 (divisible) |
| (xⁿ − aⁿ) ÷ (x − a), any n | 0 (divisible) |
Example 9: Remainder when 3⁵⁰ is divided by 8.
3² = 9 leaves 1 when divided by 8. 3⁵⁰ = (3²)²⁵, so the remainder is 1.
Trailing Zeros in Factorial (Number of Zeros Trick)
The number of zeros at the end of n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
| n! | Zeros |
|---|---|
| 60! | 12 + 2 = 14 |
| 100! | 20 + 4 = 24 |
| 250! | 50 + 10 + 2 = 62 |
Zeros come from pairs of 2 and 5, and 5s are always fewer, so we count 5s.
Sum of Series Formulas for SSC CGL
| Series | Formula | Example |
|---|---|---|
| 1 + 2 + … + n | n(n + 1)/2 | 1 to 50 = 1,275 |
| 1² + 2² + … + n² | n(n + 1)(2n + 1)/6 | 1² to 10² = 385 |
| 1³ + 2³ + … + n³ | [n(n + 1)/2]² | 1³ to 10³ = 3,025 |
| First n odd numbers | n² | first 20 odd = 400 |
| First n even numbers | n(n + 1) | first 10 even = 110 |
Example 10: How many numbers from 1 to 200 are divisible by 3 or 5?
Divisible by 3: 66. By 5: 40. By both (15): 13. Answer = 66 + 40 − 13 = 93.
Example 11: Find the largest 3-digit number divisible by 7.
999 ÷ 7 leaves remainder 5, so 999 − 5 = 994.
Number System Tricks: Common Mistakes
- Calling 1 a prime number. It has only one factor.
- Checking 24 with 4 and 6. Use co-prime pairs (3 and 8).
- Forgetting 25, 125 in trailing zeros. 100! has 24 zeros, not 20.
- Counting factors without prime powers. Always factorise fully first.
- Missing the remainder cycle length. Find the cycle (2, 4, 3, 1 for 2 mod 5) before dividing the power.
Number System Questions for SSC CGL (Practice Set with Answers)
- How many factors does 540 have? Answer: 24. 540 = 2² × 3³ × 5 → 3 × 4 × 2.
- How many trailing zeros are there in 150!? Answer: 37. 30 + 6 + 1.
- Find the digit x if 3x54 is divisible by 9. Answer: 6. 3 + x + 5 + 4 = 12 + x → 18.
- Find the remainder when 2¹⁰⁰ is divided by 7. Answer: 2. 2³ = 8 leaves 1; 2¹⁰⁰ = (2³)³³ × 2 → 2.
- Find the remainder when 5⁹⁹ is divided by 4. Answer: 1. 5 leaves 1 when divided by 4.
- Is 1,001 a prime number? Answer: No. 1,001 = 7 × 11 × 13.
- Find the sum of all odd numbers from 1 to 99. Answer: 2,500. There are 50 odd numbers → 50².
- Which of these is a prime number: 91, 97, 119, 133? Answer: 97. 91 = 7 × 13, 119 = 7 × 17, 133 = 7 × 19.
- Find the digit x if 84x2 is divisible by 3, and x is the largest possible digit. Answer: 7. 8 + 4 + x + 2 = 14 + x → 15, 18 or 21 → x = 1, 4 or 7.
- Find the sum of 1² + 2² + … + 10². Answer: 385. 10 × 11 × 21 ÷ 6.
For timed practice, take the SSC CGL Number System Mock Test 01 and the SSC GD Number System Mock Test 01.
Number System FAQs
What topics come under number system in SSC CGL?
Types of numbers, divisibility rules, factors and their sum, remainders, unit digits, trailing zeros, HCF and LCM, and sums of series. SSC CGL Tier 1 usually has 2 to 3 questions from these.
What is the divisibility rule of 7?
Double the last digit and subtract it from the number formed by the remaining digits. If the result is 0 or divisible by 7, the original number is divisible by 7. Repeat for large numbers.
What is the divisibility rule of 11?
Find the difference between the sum of digits in odd places and the sum in even places. If it is 0 or a multiple of 11, the number is divisible by 11.
How do you find the number of factors of a number?
Write the number as a product of prime powers, add 1 to each power, and multiply. For example, 360 = 2³ × 3² × 5 has 4 × 3 × 2 = 24 factors.
How many prime numbers are there between 1 and 100?
There are 25 prime numbers from 1 to 100, starting with 2 and ending with 97.
Is 1 a prime number?
No. A prime number has exactly two factors. 1 has only one factor, so it is neither prime nor composite.
Practice Next
- Mock tests: SSC CGL Number System Mock Test 01, SSC CGL Number System Mock Test 02, SSC CGL Number System Mock Test 03, SSC CGL Number System Mock Test 04 and SSC GD Number System Mock Test 01.
- Related notes: unit digit tricks for SSC CGL, LCM and HCF tricks for SSC CGL, surds and indices for SSC CGL and simplification tricks for bank exams.
- Exam details: SSC CGL 2026 and SSC CHSL 2026.
- Browse all study notes.