In this guide
Number system questions are some of the quickest marks in the CHSL maths section. They test a small set of rules (divisibility, counting, remainders, unit digits) and rarely need long calculation. The trap is that the rules look so simple that candidates learn them as slogans and then misapply them. If you know why each rule works, you won't.
This topic also runs under everything else. HCF and LCM, simplification, and even some algebra questions lean on divisibility and factors, so time spent here pays off across the paper.
Types of numbers
| Type | Examples | Note |
|---|---|---|
| Natural | 1, 2, 3… | Counting numbers |
| Whole | 0, 1, 2… | Natural numbers and 0 |
| Integers | …, −2, −1, 0, 1, 2… | Whole numbers and their negatives |
| Prime | 2, 3, 5, 7, 11… | Exactly two factors; 2 is the only even prime |
| Composite | 4, 6, 8, 9… | More than two factors |
| Co-prime pair | 8 and 15 | HCF is 1; neither needs to be prime |
| Rational | 2/3, 0.5, 0.333… | Can be written as p/q with q ≠ 0 |
| Irrational | √2, √3, π | Cannot be written as p/q; decimal never ends or repeats |
Is a number prime? Test divisibility only by primes up to its square root. Why: if n = a × b and both a and b were bigger than √n, their product would be bigger than n. So any factor pair has one member at or below √n.
Example: is 221 prime? √221 is just under 15, so test 2, 3, 5, 7, 11 and 13. 221 = 13 × 17, so it is not prime.
Divisibility tests and why they work
| By | Test | Why |
|---|---|---|
| 2, 5, 10 | Last digit | 10 is divisible by 2 and 5, so every digit except the last is a multiple of 10 |
| 4 | Last two digits divisible by 4 | 100 is divisible by 4 |
| 8 | Last three digits divisible by 8 | 1,000 is divisible by 8 |
| 3, 9 | Sum of digits divisible by 3 or 9 | 10, 100, 1,000… each leave remainder 1 when divided by 9 (and 3), so a number leaves the same remainder as its digit sum |
| 11 | (Sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11 | 10 leaves remainder −1 when divided by 11, so place values alternate +1, −1, +1… |
| 6 | Divisible by 2 and 3 | 6 = 2 × 3, and 2 and 3 are co-prime |
| 12 | Divisible by 3 and 4 | 12 = 3 × 4, and 3 and 4 are co-prime |
Counting multiples and factors
Multiples of k from 1 to N = the whole-number part of N ÷ k.
Multiples of k from A to B = (count up to B) − (count up to A − 1).
Number of factors. Write N = pᵃ × qᵇ × rᶜ in primes. Then N has (a + 1)(b + 1)(c + 1) factors.
Why: every factor picks a power of p from 0 to a (that is a + 1 choices), a power of q from 0 to b, and so on.
Example: 72 = 2³ × 3², so it has 4 × 3 = 12 factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Useful sums:
- 1 + 2 + … + n = n(n + 1)/2. So 1 + 2 + … + 50 = 50 × 51/2 = 1,275.
- Sum of the first n odd numbers = n².
- Sum of the first n even numbers = n(n + 1).
Remainders and unit digits
Remainder of a product = remainder of the product of the individual remainders.
Why: write each number as (multiple of the divisor) + remainder; when you multiply out, every term except remainder × remainder is a multiple of the divisor.
Unit digits of powers repeat in cycles of at most 4:
| Base ends in | Cycle of unit digits |
|---|---|
| 0, 1, 5, 6 | Always the same digit |
| 2 | 2, 4, 8, 6 |
| 3 | 3, 9, 7, 1 |
| 4 | 4, 6 |
| 7 | 7, 9, 3, 1 |
| 8 | 8, 4, 2, 6 |
| 9 | 9, 1 |
To find the unit digit of aⁿ, divide n by 4. A remainder of 1, 2 or 3 gives the 1st, 2nd or 3rd digit of the cycle; a remainder of 0 gives the 4th.
Recurring decimals
- Pure recurring: write the repeating block over as many 9s. 0.666… = 6/9 = 2/3; 0.1818… = 18/99 = 2/11.
- Mixed recurring: (whole digits up to the end of the first block − the non-repeating part) ÷ (as many 9s as repeating digits, followed by as many 0s as non-repeating digits). 0.1666… = (16 − 1)/90 = 15/90 = 1/6.
Six worked questions
Q1. How many numbers from 100 to 300 are divisible by 6?
Up to 300: 300 ÷ 6 = 50. Up to 99: 99 ÷ 6 = 16.5, so 16. Answer: 50 − 16 = 34.
Q2. If the five-digit number 73x45 is divisible by 11, find x.
Digits in odd places (1st, 3rd, 5th): 7 + x + 5 = 12 + x. Even places (2nd, 4th): 3 + 4 = 7. Difference: (12 + x) − 7 = 5 + x. This must be 0 or a multiple of 11, and x is a digit, so 5 + x = 11 and x = 6. Check: 73,645 = 11 × 6,695.
Q3. What is the remainder when 2²⁰ is divided by 5?
2⁴ = 16 leaves remainder 1. So 2²⁰ = (2⁴)⁵ leaves 1⁵ = 1.
Q4. What is the remainder when 17 × 23 × 31 is divided by 7?
Remainders: 17 → 3, 23 → 2, 31 → 3. Product of remainders = 3 × 2 × 3 = 18, and 18 leaves 4. Check: 17 × 23 × 31 = 12,121 = 7 × 1,731 + 4.
Q5. Find the unit digit of 13⁴⁷ × 18³².
For 13⁴⁷, the base ends in 3 (cycle 3, 9, 7, 1). 47 ÷ 4 leaves 3, so the unit digit is 7. For 18³², the base ends in 8 (cycle 8, 4, 2, 6). 32 ÷ 4 leaves 0, so the unit digit is 6. The product ends in the unit digit of 7 × 6 = 42, which is 2.
Q6. How many numbers from 1 to 100 are divisible by 3 or 5?
Divisible by 3: 33. By 5: 20. By both, that is by 15: 6. Those 6 were counted twice, so the answer is 33 + 20 − 6 = 47.
Practice
- Is 4,356 divisible by 11?
- How many numbers from 1 to 200 are divisible by 9?
- Find the sum of the first 20 odd numbers.
- What is the unit digit of 7³⁰?
- What is the remainder when 3¹² is divided by 8?
- Which is larger: 4/9 or 5/11?
- How many factors does 180 have?
- How many numbers from 1 to 100 are divisible by neither 2 nor 5?
Answers:
- Yes. Odd places: 4 + 5 = 9; even places: 3 + 6 = 9; difference 0. (4,356 = 11 × 396.)
- 22. 200 ÷ 9 = 22.2…, so 22.
- 400. Sum of the first n odd numbers is n², and 20² = 400.
- 9. Cycle 7, 9, 3, 1; 30 ÷ 4 leaves 2, so the 2nd digit.
- 1. 3² = 9 leaves 1, so 3¹² = (3²)⁶ leaves 1.
- 5/11. Cross-multiply: 4 × 11 = 44 and 9 × 5 = 45, and 45 is bigger.
- 18. 180 = 2² × 3² × 5, so 3 × 3 × 2 = 18.
- 40. Divisible by 2: 50; by 5: 20; by 10: 10. Divisible by 2 or 5 = 50 + 20 − 10 = 60, so 100 − 60 = 40.
What to do next
- Write the divisibility table from memory, including the "why" column.
- Learn the unit-digit cycles until you can recall them instantly.
- Solve 30 mixed number system questions from previous CHSL papers, timed at 40 seconds each.
- Move on to HCF and LCM, which builds directly on prime factors, and keep squares and cubes fresh with the squares, cubes and roots guide.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
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