In this guide
The number system sits at the start of the CDS maths syllabus for a reason. It is the arithmetic that everything else stands on, and it produces a steady set of short questions in most papers: is this number divisible by 11, what is the unit digit of this power, how many factors does this number have, what remainder does this leave.
These are some of the fastest marks in the paper once you know the rules, and some of the most dangerous if you half-remember them. So this guide explains each rule and why it works. A rule you understand is a rule you will apply correctly under pressure.
Types of numbers
| Type | Meaning | Examples |
|---|---|---|
| Natural numbers | Counting numbers | 1, 2, 3, … |
| Whole numbers | Natural numbers and 0 | 0, 1, 2, … |
| Integers | Whole numbers and their negatives | …, −2, −1, 0, 1, 2, … |
| Rational numbers | Can be written as p/q with q not zero | 3/4, −5, 0.333… |
| Irrational numbers | Cannot be written as p/q; decimal never ends or repeats | √2, √3, π |
| Prime numbers | Exactly two factors, 1 and itself | 2, 3, 5, 7, 11 |
| Composite numbers | More than two factors | 4, 6, 9, 91 |
| Co-primes | Two numbers whose HCF is 1 | 8 and 15 |
Two facts UPSC likes to test: 1 is neither prime nor composite, and 2 is the only even prime.
Divisibility rules, and why they work
| Divisor | Rule | Why it works |
|---|---|---|
| 2, 5, 10 | Check the last digit | 10 is divisible by 2 and 5, so only the units digit matters |
| 4 | Last two digits divisible by 4 | 100 is divisible by 4, so the hundreds and above never matter |
| 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, so a number leaves the same remainder as its digit sum |
| 11 | Difference between the sums of alternate digits is 0 or a multiple of 11 | 10 leaves remainder −1 on division by 11, 100 leaves +1, and so on, so the digits count with alternating signs |
| 6, 12, 72 | Split into co-prime parts: 6 = 2 × 3, 12 = 3 × 4, 72 = 8 × 9 | A number divisible by two co-prime numbers is divisible by their product |
Unit digits of powers
Only the units digits of the numbers being multiplied affect the units digit of the product. So the unit digit of 1,237⁵ depends only on 7⁵. Powers of any digit then repeat their last digits in a cycle of at most four:
| Base ends in | Cycle of unit digits | Cycle length |
|---|---|---|
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 4 | 4, 6 (odd power 4, even power 6) | 2 |
| 9 | 9, 1 (odd power 9, even power 1) | 2 |
| 0, 1, 5, 6 | Always the same digit | 1 |
Method: divide the power by 4 and look at the remainder. Remainder 1, 2 or 3 gives the first, second or third digit of the cycle. Remainder 0 gives the fourth, not the first. That last point is where most errors happen.
Counting factors
If N = a^p × b^q × c^r, where a, b and c are different primes, then N has (p + 1)(q + 1)(r + 1) factors.
Why: every factor of N is built by choosing how many of each prime to use. You can use the prime a anywhere from 0 to p times (p + 1 choices), the prime b from 0 to q times (q + 1 choices), and so on. Multiply the choices.
The same idea counts even factors: an even factor must use 2 at least once, so the power of 2 has p choices instead of p + 1.
Remainders
- The remainder of a sum or product equals the sum or product of the remainders, reduced again. So 17 × 23 divided by 5 gives remainders 2 × 3 = 6, and 6 leaves 1.
- If a leaves remainder 1 when divided by d, then every power of a also leaves remainder 1.
- If a leaves remainder d − 1 (that is, −1), then even powers of a leave 1 and odd powers leave d − 1.
These two facts turn most "huge power" remainder questions into two lines.
Primes, sums and zeros
- Testing a prime: check divisibility only by primes up to the square root. For 391, √391 is just under 20, so test 2, 3, 5, 7, 11, 13, 17, 19. It fails at 17: 391 = 17 × 23.
- Sum of the first n natural numbers: n(n + 1)/2.
- Sum of the first n odd numbers: n².
- Sum of squares of the first n natural numbers: n(n + 1)(2n + 1)/6.
- Trailing zeros of n!: each zero needs a 2 and a 5, and 5s are scarcer. Count them: n ÷ 5 + n ÷ 25 + n ÷ 125 + …, taking whole-number parts.
Worked questions
Question 1: Find the unit digit of 7⁹⁵ − 3⁵⁸.
- 95 ÷ 4 leaves 3, so 7⁹⁵ ends like 7³: 3.
- 58 ÷ 4 leaves 2, so 3⁵⁸ ends like 3²: 9.
- The first number is larger, so subtract with a borrow: 13 − 9 = 4. The unit digit is 4.
Question 2: What is the largest digit k for which the number 7k416 is divisible by 72?
- 72 = 8 × 9, which are co-prime.
- By 8: the last three digits 416 = 8 × 52. Always true.
- By 9: the digit sum 7 + k + 4 + 1 + 6 = 18 + k must be a multiple of 9, so k = 0 or 9.
- The largest is k = 9. Check: 79,416 = 72 × 1,103.
Question 3: How many factors does 360 have, and how many of them are even?
- 360 = 2³ × 3² × 5.
- All factors: (3 + 1)(2 + 1)(1 + 1) = 24.
- Even factors: the power of 2 must be 1, 2 or 3, so 3 × 3 × 2 = 18. (That leaves 6 odd factors: 1, 3, 5, 9, 15, 45.)
Question 4: Find the remainder when 3¹⁰⁰ is divided by 8.
- 3² = 9 leaves remainder 1 on division by 8.
- 3¹⁰⁰ = (3²)⁵⁰, so it leaves 1⁵⁰ = 1.
Question 5: Find the sum 51 + 52 + 53 + … + 100.
- Sum from 1 to 100 = 100 × 101 ÷ 2 = 5,050.
- Sum from 1 to 50 = 50 × 51 ÷ 2 = 1,275.
- Required sum = 5,050 − 1,275 = 3,775.
Question 6: How many zeros are there at the end of 100!?
- Multiples of 5 up to 100: 100 ÷ 5 = 20.
- Multiples of 25 add an extra 5 each: 100 ÷ 25 = 4.
- 125 is beyond 100, so stop. Zeros = 20 + 4 = 24.
Practice set
- Find the unit digit of 3²² × 7¹⁵.
- How many factors does 540 have?
- Is 2,35,796 divisible by 11?
- Find the remainder when 17 × 23 × 29 is divided by 5.
- How many zeros are there at the end of 50!?
- Find the sum of all odd numbers from 1 to 49.
- Is 221 a prime number?
- Find the remainder when 5⁴⁰ is divided by 13.
Answers
- 7. 22 ÷ 4 leaves 2, so 3²² ends in 9. 15 ÷ 4 leaves 3, so 7¹⁵ ends in 3. 9 × 3 = 27 ends in 7.
- 24. 540 = 2² × 3³ × 5, so (2 + 1)(3 + 1)(1 + 1) = 24.
- Yes. From the right, alternate digits 6 + 7 + 3 = 16 and 9 + 5 + 2 = 16. The difference is 0. (2,35,796 = 11 × 21,436.)
- 4. Remainders 2, 3 and 4 multiply to 24, which leaves 4.
- 12. 50 ÷ 5 = 10 and 50 ÷ 25 = 2, so 10 + 2 = 12.
- 625. There are 25 odd numbers from 1 to 49, and their sum is 25² = 625.
- No. √221 is below 15; testing primes up to 13 gives 221 = 13 × 17.
- 1. 5² = 25 leaves 12, which is −1 on division by 13. So 5⁴ leaves (−1)² = 1, and 5⁴⁰ = (5⁴)¹⁰ leaves 1.
What to do next
- Write the divisibility table and the unit-digit cycles on your formula sheet
- Solve all number system questions from the last five CDS papers, timed at a minute each
- Mark every error as a rule error or a slip, and redo the rule errors after two days
- Move on to HCF and LCM, which uses the same prime factorisation
- Then practise fractions and roots to finish the arithmetic base
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 Union Public Service Commission website .
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