HCF and LCM for CUET General Aptitude

Q: How do you find two numbers when their HCF is given?

If the HCF is h, write the numbers as hp and hq, where p and q have no common factor other than 1 (they are co-prime). Then turn whatever else the question tells you into a condition on p and q.

About 15 min to learn · 63 questions

CUET asks HCF and LCM both directly and hidden inside word problems about bells, tiles and remainders. The arithmetic is short. What takes skill is seeing which of the two a question wants. This lesson covers both ideas, the two methods, the HCF × LCM rule, fractions and decimals, and every common word-problem pattern.

Syllabus area: General Mental Ability, Numerical Ability

On this page · 9 sections

01The core idea
02Finding HCF and LCM
03How HCF and LCM are related
04Fractions and decimals
05Word problems
06Exam shortcuts
07Common mistakes
08Quick revision
09Check your understanding

The core idea

Most HCF and LCM questions reduce to one decision. Does the problem want the biggest number that fits into all the given numbers, or the smallest number that all of them fit into? Once you have decided, the calculation takes seconds.

What is HCF?

The HCF (highest common factor) of two or more numbers is the largest number that divides each of them exactly. It is also called the GCD, or greatest common divisor.

The factors of 20 are 1, 2, 4, 5, 10 and 20. The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. They share 1, 2, 5 and 10, and the largest of these is 10:

\operatorname{HCF}(20, 30) = 10

What is LCM?

The LCM (least common multiple) of two or more numbers is the smallest number that each of them divides exactly.

The multiples of 20 are 20, 40, 60, 80 and so on. The multiples of 30 are 30, 60, 90 and so on. The first one they share is 60:

\operatorname{LCM}(20, 30) = 60

What is the difference between HCF and LCM?

The HCF is the largest number that divides all the given numbers. The LCM is the smallest number that all the given numbers divide. So the HCF can never be larger than the smallest number, and the LCM can never be smaller than the largest number.

	HCF	LCM
Meaning	Largest common factor	Smallest common multiple
Size	At most the smallest number	At least the largest number
For 20 and 30	10	60
By prime factors	Common primes, lowest powers	All primes, highest powers
In word problems	Cutting into the largest equal parts	Waiting for things to line up

A five-second check

Before you mark an answer, check two things. The HCF must divide every given number, and every given number must divide the LCM. An "HCF" larger than one of the numbers, or an "LCM" smaller than one of them, is wrong.

Finding HCF and LCM

There are two methods worth knowing. Prime factorisation gives you the HCF and the LCM together, and suits most CUET numbers. The division method finds only the HCF, but it is much faster when the numbers are large or hard to factorise. For very small numbers, listing factors or multiples as above is fine.

How do you find HCF and LCM by prime factorisation?

Write each number as a product of prime powers. For the HCF, take only the primes that appear in every number, each at its lowest power. For the LCM, take every prime that appears in any number, each at its highest power.

Example 1 · Prime factorisation

Find the HCF and LCM of 84, 126 and 210.

Write each number as a product of primes.

$\begin{aligned} 84 &= 2^2 \times 3 \times 7 \\ 126 &= 2 \times 3^2 \times 7 \\ 210 &= 2 \times 3 \times 5 \times 7 \end{aligned}$
For the HCF, the primes found in all three numbers are 2, 3 and 7. Take the lowest power of each: $\operatorname{HCF} = 2 \times 3 \times 7 = 42$ .
For the LCM, take every prime at its highest power: $\operatorname{LCM} = 2^2 \times 3^2 \times 5 \times 7 = 1260$ .

Answer

$\operatorname{HCF} = 42$ and $\operatorname{LCM} = 1260$

Watch out

The prime 5 appears in only one of the numbers. It is left out of the HCF, but it must go into the LCM.

How does the division method find the HCF?

Divide the larger number by the smaller one. Then divide the previous divisor by the remainder, and keep going until the remainder is 0. The last divisor you used is the HCF. This is Euclid's method.

Example 2 · Division method

Find the HCF of 1183 and 1729.

Divide 1729 by 1183: $1729 = 1 \times 1183 + 546$ .
Divide the old divisor, 1183, by the remainder 546: $1183 = 2 \times 546 + 91$ .
Divide 546 by 91: $546 = 6 \times 91 + 0$ . The remainder is 0, so stop.

Answer

\operatorname{HCF}(1183, 1729) = 91

To check: $1183 = 91 \times 13$ and $1729 = 91 \times 19$ .

Exam shortcut

For three numbers, find the HCF of two of them, then the HCF of that result and the third number.

Why this works

Suppose $a = qb + r$ . Any number that divides both $a$ and $b$ also divides $r = a - qb$ . Likewise, any number that divides both $b$ and $r$ also divides $a$ . So $\operatorname{HCF}(a, b) = \operatorname{HCF}(b, r)$ . Each step keeps the HCF the same while the numbers get smaller. When the remainder reaches 0, the last divisor divides the number before it exactly, so it is the HCF.

What is the relationship between HCF and LCM?

For any two numbers, the HCF multiplied by the LCM equals the product of the two numbers. If you know any three of these four quantities, you can find the fourth.

The product rule, for two numbers

\operatorname{HCF}(a, b) \times \operatorname{LCM}(a, b) = a \times b

Example 3 · Finding the missing number

The HCF of two numbers is 14, their LCM is 420, and one of the numbers is 70. Find the other number.

Put the known values into the rule: $14 \times 420 = 70 \times \text{other}$ .
Solve: $\text{other} = \dfrac{14 \times 420}{70} = \dfrac{5880}{70} = 84$ .

Answer

The other number is 84.

To check: $\operatorname{HCF}(70, 84) = 14$ and $\operatorname{LCM}(70, 84) = 420$ .

Two useful facts follow from this rule:

The HCF always divides the LCM. Both numbers are multiples of the HCF, so every common multiple is too. So "HCF 12, LCM 100" is impossible, because 100 is not a multiple of 12.
For co-prime numbers (HCF = 1), the LCM is the product. For example, $\operatorname{LCM}(8, 15) = 120$ .

Watch out

The rule works for exactly two numbers. For three numbers, HCF × LCM is usually not equal to their product.

Why this works

Let $h$ be the HCF, and write the numbers as $a = hp$ and $b = hq$ , where $p$ and $q$ are co-prime. Then $\operatorname{LCM} = hpq$ , so $\operatorname{HCF} \times \operatorname{LCM} = h \times hpq = hp \times hq = a \times b$ .

How do you find two numbers when their HCF is given?

If the HCF is $h$ , write the numbers as $hp$ and $hq$ , where $p$ and $q$ have no common factor other than 1 (they are co-prime). Then turn whatever else the question tells you into a condition on $p$ and $q$ .

Example 4 · Counting pairs

The HCF of two numbers is 9 and their sum is 108. How many such pairs are there?

Write the numbers as $9p$ and $9q$ . Then $9p + 9q = 108$ , so $p + q = 12$ .
List the pairs that add to 12: (1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6).
Keep only the co-prime pairs: (1, 11) and (5, 7).

Answer

2 pairs: (9, 99) and (45, 63).

Watch out

(18, 90) also adds up to 108, but its HCF is 18, not 9. That is why $p$ and $q$ must be co-prime.

Common variation: numbers in a ratio

If two numbers are in the ratio 4 : 9 and their HCF is 6, the numbers are $6 \times 4 = 24$ and $6 \times 9 = 54$ , and $\operatorname{LCM} = 6 \times 4 \times 9 = 216$ . This shortcut works because 4 and 9 are co-prime.

With three terms, check the terms in pairs. For the ratio 2 : 3 : 4 with HCF $h$ , the terms 2 and 4 share a factor. So the LCM is $h \times \operatorname{LCM}(2, 3, 4) = 12h$ , not $24h$ .

Fractions and decimals

How do you find the HCF and LCM of fractions?

Reduce each fraction to its lowest terms first. The HCF of fractions is the HCF of the numerators divided by the LCM of the denominators. The LCM of fractions is the LCM of the numerators divided by the HCF of the denominators.

Fractions in lowest terms

\operatorname{HCF} = \frac{\operatorname{HCF}\text{ of numerators}}{\operatorname{LCM}\text{ of denominators}}

\operatorname{LCM} = \frac{\operatorname{LCM}\text{ of numerators}}{\operatorname{HCF}\text{ of denominators}}

Example 5 · Fractions

Find the HCF and LCM of $\tfrac{2}{5}$ , $\tfrac{6}{7}$ and $\tfrac{8}{15}$ .

The numerators are 2, 6 and 8. Their HCF is 2 and their LCM is 24.
The denominators are 5, 7 and 15. Their HCF is 1 and their LCM is 105.
Use the formulas: $\operatorname{HCF} = \dfrac{2}{105}$ and $\operatorname{LCM} = \dfrac{24}{1} = 24$ .

Answer

$\operatorname{HCF} = \tfrac{2}{105}$ and $\operatorname{LCM} = 24$

To sense-check: $24 \div \tfrac{6}{7} = 28$ , a whole number, so 24 really is a multiple of $\tfrac{6}{7}$ . The HCF of fractions should come out no larger than the smallest fraction, and the LCM no smaller than the largest.

Watch out

The two formulas are mirror images, and swapping the denominator parts is the most common error. The HCF uses the LCM of the denominators. The LCM uses the HCF of the denominators.

How do you find the HCF and LCM of decimals?

Give every number the same number of decimal places and remove the decimal point. Then find the HCF or LCM of the whole numbers, and put the same number of decimal places back.

Example 6 · Decimals

Find the HCF and LCM of 0.8, 1.2 and 0.28.

Write every number with two decimal places: 0.80, 1.20 and 0.28. Then work with 80, 120 and 28.
$\operatorname{HCF}(80, 120, 28) = 4$ , so the HCF is 0.04.
$\operatorname{LCM}(80, 120, 28) = 1680$ , so the LCM is 16.80, which is 16.8.

Answer

HCF = 0.04 and LCM = 16.8

Watch out

If you skip the step of giving every number the same number of decimal places, you end up working with 8, 12 and 28, which gives the wrong answer.

Word problems

How do I know whether a word problem needs HCF or LCM?

Ask what the problem is doing. If it is splitting things into the largest possible equal parts, you need the HCF. If it is waiting for repeating things to line up, or asking for the smallest amount that every size divides, you need the LCM.

Clue in the question	Use	Why
Largest tile, container, rope piece or group size	HCF	The piece must fit exactly into every quantity
Fewest tiles, boxes or rows	HCF, then divide	The largest piece gives the fewest pieces
Bells, lights or runners coming together again	LCM	Every cycle has to line up
Smallest length, amount or group that splits exactly	LCM	Every size must divide it
Greatest number that divides, with remainders	HCF	Remove the remainders first
Least number that is divided, with remainders	LCM	Then add or adjust for the remainder

Example 7 · Bells ringing together

Three sirens sound every 20, 30 and 45 minutes. They sound together at 10:00 am. When do they next sound together?

They coincide again after $\operatorname{LCM}(20, 30, 45) = 180$ minutes.
180 minutes is 3 hours after 10:00 am.

Answer

At 1:00 pm.

Example 8 · Fewest tiles

A room measures 5.4 m by 4.2 m. What is the fewest number of identical square tiles that will cover the floor exactly?

Convert to centimetres: 540 cm by 420 cm.
The fewest tiles means the largest tile: $\operatorname{HCF}(540, 420) = 60$ cm.
Count them: $\dfrac{540}{60} \times \dfrac{420}{60} = 9 \times 7 = 63$ .

Answer

63 tiles, each 60 cm × 60 cm.

Exam shortcut

For runners on a circular track, first work out each runner's lap time (track length ÷ speed). The LCM of the lap times is when they are next back at the starting point together.

How do you find the least number that leaves a given remainder?

Start from the LCM of the divisors. If every division leaves the same remainder, add it. If the gap between each divisor and its remainder is the same, subtract that gap.

Situation	Answer	Example
Same remainder $r$ every time (answer greater than $r$ )	LCM $+\ r$	Divided by 10, 14 and 35, leaves 3: $70 + 3 = 73$
Divisor − remainder is the same $k$ every time	LCM $-\ k$	Divided by 6, 10 and 14, leaves 4, 8 and 12: $210 - 2 = 208$
Same remainder, plus another condition	$\text{LCM} \times n + r$ , trying $n = 1, 2, 3, \ldots$	Example 9 below

Example 9 · Remainder with an extra condition

Find the least number that leaves remainder 2 when divided by 4, 5 and 6, and is also divisible by 7.

$\operatorname{LCM}(4, 5, 6) = 60$ , so the number has the form $60n + 2$ .
Try values of $n$ . For $n = 1$ you get 62, and for $n = 2$ you get 122. Neither divides by 7.
For $n = 3$ you get $182 = 7 \times 26$ , which works.

Answer

182

A faster way to find n

60 leaves remainder 4 when divided by 7. So $60n + 2$ is divisible by 7 exactly when $4n + 2$ is. The smallest $n$ that makes $4n + 2$ a multiple of 7 is $n = 3$ .

Common variation: the largest or smallest n-digit number

Divide the extreme number (9999 for "largest four-digit", 1000 for "smallest four-digit") by the LCM, and take the multiple of the LCM just below it. Add the remainder if the question gives one. If the result now has the wrong number of digits, move one LCM up or down.

For example, the smallest four-digit number that leaves 5 when divided by 12: the multiple of 12 just below 1000 is 996, and $996 + 5 = 1001$ , which already has four digits. Stepping up to 1008 first would give 1013, which is too big.

How do you find the greatest number that divides with given remainders?

Subtract each remainder from its number, then take the HCF of the results. If the question only says the remainder is the same each time, without saying what it is, take the HCF of the differences between the numbers.

Example 10 · Different remainders

Find the greatest number that divides 1657 and 2037, leaving remainders 6 and 5 respectively.

Remove the remainders: $1657 - 6 = 1651$ and $2037 - 5 = 2032$ .
Find the HCF: $1651 = 127 \times 13$ and $2032 = 127 \times 16$ . Since 13 and 16 are co-prime, $\operatorname{HCF} = 127$ .
Check that the divisor is larger than both remainders. 127 is larger than 6 and 5, so the answer is valid.

Answer

127

Watch out

A remainder is always smaller than the divisor. If the HCF comes out no larger than one of the remainders, no number fits: every valid divisor divides that HCF, so none can be larger than it.

Example 11 · Same, unknown remainder

Find the greatest number that divides 158, 318 and 578, leaving the same remainder each time.

Take the differences between the numbers: $318 - 158 = 160$ , $578 - 318 = 260$ and $578 - 158 = 420$ .
Find the HCF: $\operatorname{HCF}(160, 260, 420) = 20$ .

Answer

To check: each of the three numbers leaves remainder 18 when divided by 20.

Exam shortcuts

Test the options first. In an HCF question, cross out every option that does not divide all the numbers. In an LCM question, cross out every option that is not divisible by all of them. Often only one option is left.
The HCF divides the difference. The HCF of two numbers also divides their difference. For example, $\operatorname{HCF}(391, 408)$ must divide $408 - 391 = 17$ . Since 17 is prime and $391 = 17 \times 23$ , the answer is 17.
Consecutive numbers are co-prime. So their LCM is simply their product: $\operatorname{LCM}(15, 16) = 240$ .
Counting common multiples. The count of numbers up to $N$ that are divisible by both $a$ and $b$ is $N \div \operatorname{LCM}(a, b)$ , rounded down.
Least perfect square that is a multiple. Find the LCM, then multiply by each prime that appears to an odd power in its factorisation.

Common mistakes

Using the product instead of the LCM. $\operatorname{LCM}(12, 18)$ is 36, not 216. The product counts the shared factor twice.
Stopping at a common factor that is not the largest. 12 divides both 48 and 72, but the HCF is 24.
Getting the remainder the wrong way round. "Divided by … leaves 4" means you add 4 to the LCM. "Increased by 5 becomes divisible" means you subtract 5.
Mixing units. Convert everything to the same unit first. For example, 6 m 24 cm is 624 cm.
Counting coincidences off by one. Lights that change together every 4 minutes coincide 15 times in the next 60 minutes, or 16 times if you count the start. Read whether the start counts.
Counting calendar days carelessly. 60 days after 1 March is not 1 May. Count the actual days in each month.
Pairs that are not co-prime. When you write the numbers as $hp$ and $hq$ , check that $p$ and $q$ have no common factor other than 1.
Using HCF × LCM = product for three numbers. It only works for two.

Quick revision

Definitions

HCF: the largest number that divides all the given numbers. It is at most the smallest number.
LCM: the smallest number that all the given numbers divide. It is at least the largest number.

Methods

Prime factorisation: for the HCF, common primes at their lowest powers. For the LCM, all primes at their highest powers.
Division (Euclid): divide by the remainder until the remainder is 0. The last divisor is the HCF.

Formulas

Two numbers: $\operatorname{HCF} \times \operatorname{LCM} = a \times b$
HCF of fractions: HCF of numerators ÷ LCM of denominators
LCM of fractions: LCM of numerators ÷ HCF of denominators
Given the HCF: the numbers are $hp$ and $hq$ , with $p$ and $q$ co-prime

HCF or LCM?

Splitting into the largest equal parts, or the fewest pieces: HCF
Things lining up again, or the smallest amount every size divides: LCM

Remainders

Least number, same remainder $r$ : LCM $+\ r$
Least number, when divisor − remainder is the same $k$ each time: LCM $-\ k$
Greatest divisor: subtract the remainders, then take the HCF. The answer must be larger than every remainder.
Greatest divisor, same unknown remainder: the HCF of the differences

Traps

LCM is not the product, unless the numbers are co-prime
Put decimals on the same number of places, and quantities in the same unit
HCF × LCM = product works for two numbers only

This topic connects closely to number system (factors, primes and divisibility) and ratio and proportion.

Next · 08Quick revision

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63 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

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Finding HCF and LCM

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