SSC CGL Maths · Number System

Factorisation, HCF and LCM

Write a number as a product of prime powers, and its divisors, its common factors with another number and its least common multiple can all be read off directly.

All Number System notes

Why this matters

Eleven questions in the bank, eight EASY and three MODERATE. Every one starts the same way, by writing a number or a sum as a product of primes, and the answer is then one line away: a divisor count, a factor of a sum of powers, a tile size or the least square multiple.

Concept 1 of 3: Prime factorisation and counting divisors

Every divisor of

2^3 \times 3^2

is built by choosing how many 2s (none to three) and how many 3s (none to two) to take. So counting divisors is counting choices, and adding them up is multiplying the sums of each prime's choices.

Definition

Write $N = p^a \times q^b \times r^c$ , with $p, q, r$ different primes. Divide out the small primes 2, 3, 5, 7 in turn.
Number of divisors $= (a + 1)(b + 1)(c + 1)$ . It counts 1 and $N$ itself.
Sum of divisors $= (1 + p + \cdots + p^a)(1 + q + \cdots + q^b)(1 + r + \cdots + r^c)$ .
Divisors that are perfect squares use only even powers of each prime: for $2^4$ the choices are $2^0, 2^2, 2^4$ .
$d$ divides $N$ exactly when every prime power in $d$ also appears in $N$ , at least as high.
For the sum of the first $n$ natural numbers, find it with $\frac{n(n + 1)}{2}$ , then factorise to see what divides it.

Divisors from the prime factorisation

N = p^a q^b r^c: \quad \text{number of divisors} = (a + 1)(b + 1)(c + 1) \qquad \text{sum of divisors} = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b)(1 + r + \cdots + r^c) \qquad 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}

p, q, rthe different prime factors of N
a, b, chow many times each prime appears

Worked example

How many divisors does 540 have, and what is their sum?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 20 Jul 2023, Shift 1 · Q60Moderate

Example 1 · Number System · Factorisation, HCF and LCM

What is the sum of the divisors of 484 that are perfect squares?

Multiplying the powers instead of the powers plus one

2^3 \times 3^2

has

(3 + 1)(2 + 1) = 12

divisors, not

3 \times 2 = 6

. The plus one is the choice of taking none of that prime.

Drill 1 more on prime factorisation and counting divisors

Concept 2 of 3: Taking a common power out of a sum of powers

A sum like

3^{10} + 3^{11} + 3^{12}

is too big to work out, but every term holds

3^{10}

. Take it out, and what is left in the bracket is a small number you can factorise in your head.

Definition

Take out the smallest power: $a^m + a^{m+1} + \cdots + a^{m+k} = a^m(1 + a + \cdots + a^k)$ .
The sum is divisible by every factor of the bracket and by every factor of $a^m$ .
$a^n - b^n$ is always divisible by $a - b$ . So $a^n - 1$ is divisible by $a - 1$ .
$a^n - b^n$ is also divisible by $a + b$ when $n$ is even, and $a^n + b^n$ is divisible by $a + b$ when $n$ is odd.
Fast route: rewrite to a bigger base first. With $m$ even, $8^m - 1 = 64^{m/2} - 1$ , divisible by 63.
For a 'least value of n' question, try the options from the smallest.

Sums and differences of powers

a^m + a^{m+1} + \cdots + a^{m+k} = a^m(1 + a + \cdots + a^k) \qquad a^n - b^n \text{ is divisible by } a - b \qquad a^n - b^n \text{ is divisible by } a + b \ (n \text{ even}) \qquad a^n + b^n \text{ is divisible by } a + b \ (n \text{ odd})

Worked example

Which of 11, 13 and 31 divides

5^{20} + 5^{21} + 5^{22}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 26 Sep 2024, Shift 3 · Q67Easy

Example 2 · Number System · Factorisation, HCF and LCM

7^{15} + 7^{16} + 7^{17}

is divisible by which of the following given numbers?

Forgetting the factors of the power you took out

3^8 + 3^9 = 3^8 \times 4

is divisible by 4, but also by 3, by 81 and by 12. When more than one option divides the sum, check the factors of

3^8

as well as the bracket.

Using a + b on an odd power of a difference

a^n - b^n

is divisible by

a + b

only for even

n

8^3 - 1 = 511 = 7 \times 73

is not divisible by

8 + 1 = 9

Drill 4 more on taking a common power out of a sum of powers

Concept 3 of 3: HCF, LCM and the least square multiple

The HCF is the biggest measure that fits a whole number of times into each number; the LCM is the smallest number each one fits into. In prime powers, the HCF takes the lower power of each common prime and the LCM takes the higher power of every prime.

Definition

HCF: the lowest power of each prime common to all the numbers. LCM: the highest power of every prime that appears.
Co-prime numbers (you met them on the tests page) have HCF 1, and their LCM is their product.
For two numbers: HCF $\times$ LCM $=$ the product of the numbers. Their mean proportional $\sqrt{ab}$ is therefore $\sqrt{\text{HCF} \times \text{LCM}}$ .
The largest square tile for a floor $L \times B$ has side HCF $(L, B)$ , in the same units; the number of tiles is $\frac{L}{h} \times \frac{B}{h}$ .
The least number divisible by several numbers is their LCM.
The least perfect square divisible by several numbers: find the LCM, then multiply by each prime that appears to an odd power, once. A square has every prime to an even power.

HCF, LCM, tiles and the mean proportional

\text{HCF} \times \text{LCM} = a \times b \qquad \sqrt{ab} = \sqrt{\text{HCF} \times \text{LCM}} \qquad \text{tiles} = \frac{L}{h} \times \frac{B}{h}, \ h = \text{HCF}(L, B)

a, bthe two numbers
L, Bthe length and breadth of the floor, in one unit
hthe side of the largest square tile

Worked example

Find the least perfect square that is divisible by 4, 6 and 14.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 1 · Q52Moderate

Example 3 · Number System · Factorisation, HCF and LCM

What is the least square number that is divisible by 8, 12, and 20?

Using HCF times LCM for three numbers

HCF

\times

LCM equals the product only for two numbers. For 2, 4 and 8 the HCF is 2 and the LCM 8, but the product is 64.

Squaring the LCM to get a square multiple

The least square multiple of

12 = 2^2 \times 3

12 \times 3 = 36

, not

12^2 = 144

. Multiply only by the primes whose power is odd.

Drill 3 more on hcf, lcm and the least square multiple

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Prime factorisation and counting divisors
Divisors from the prime factorisation
$N = p^a q^b r^c: \quad \text{number of divisors} = (a + 1)(b + 1)(c + 1) \qquad \text{sum of divisors} = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b)(1 + r + \cdots + r^c) \qquad 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$
Taking a common power out of a sum of powers
Sums and differences of powers
$a^m + a^{m+1} + \cdots + a^{m+k} = a^m(1 + a + \cdots + a^k) \qquad a^n - b^n \text{ is divisible by } a - b \qquad a^n - b^n \text{ is divisible by } a + b \ (n \text{ even}) \qquad a^n + b^n \text{ is divisible by } a + b \ (n \text{ odd})$
HCF, LCM and the least square multiple
HCF, LCM, tiles and the mean proportional
$\text{HCF} \times \text{LCM} = a \times b \qquad \sqrt{ab} = \sqrt{\text{HCF} \times \text{LCM}} \qquad \text{tiles} = \frac{L}{h} \times \frac{B}{h}, \ h = \text{HCF}(L, B)$

Watch out for (5)

Multiplying the powers instead of the powers plus one
→ Prime factorisation and counting divisors
Forgetting the factors of the power you took out
→ Taking a common power out of a sum of powers
Using a + b on an odd power of a difference
→ Taking a common power out of a sum of powers
Using HCF times LCM for three numbers
→ HCF, LCM and the least square multiple
Squaring the LCM to get a square multiple
→ HCF, LCM and the least square multiple

Test yourself on Number System

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Number System chapter test Drill all 11 questions