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State Board · Class 7 · Mathematics

Lesson 3: HCF and LCM

5 concepts · 40 practice questions with answers · Free to read

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Co-prime and twin prime numbers

From daily life

You have 8 bananas and 15 guavas. You want to make equal packets so that every packet holds both fruits and nothing is left over. The number of packets must divide both 8 and 15 exactly. 8 is divisible by 1, 2, 4, 8 and 15 by 1, 3, 5, 15. Only 1 appears in both lists, so only one packet can be made. Such numbers, 8 and 15, are called co-prime numbers.

Simple definition

Two numbers are co-prime if they have no common divisor (factor) other than 1.

Prime number: A number that has exactly two divisors, 1 and itself, is a prime number: 2, 3, 5, 7, 11, 13 ... 1 is not prime because it has only one divisor. A number with more than two divisors is a composite number (e.g. 12).

Twin primes: Two prime numbers that differ by 2, e.g. 11 and 13.

Divisors of 121234612Divisors of 3515735

In the figure: The divisors of 12 and 35 are shown. Only 1 is in both rows (gold), so 12 and 35 are co-prime.

Rule 1: Checking whether two numbers are co-prime — Write all the divisors of both numbers. If only 1 is common, the numbers are co-prime.

  • Easy: 8 and 15 → divisors of 8 are 1, 2, 4, 8 and of 15 are 1, 3, 5, 15; only 1 is common, so they are co-prime.
  • Medium: 14 and 21 → divisors of 14 are 1, 2, 7, 14 and of 21 are 1, 3, 7, 21; 1 and 7 are common, so they are not co-prime.
  • Exam type: 26 and 45 → divisors of 26 are 1, 2, 13, 26 and of 45 are 1, 3, 5, 9, 15, 45; only 1 is common, so they are co-prime (even though neither number is prime).

Rule 2: Consecutive numbers and two different primes — Two consecutive numbers are always co-prime. Two different prime numbers are also always co-prime. There is no need to search for common divisors.

  • Easy: 17 and 18 are consecutive, so they are co-prime.
  • Medium: 19 and 23 are both prime and different, so they are co-prime (divisors of 19 are 1, 19 and of 23 are 1, 23).
  • Exam type: 99 and 100 are consecutive, so they are co-prime (99 = 9 × 11 and 100 = 4 × 25; no factor is the same).
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Co-prime and twin prime numbers

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Rule 3: Twin primes — Both numbers must be prime and their difference must be 2. Twin primes are always co-prime.

  • Easy: 41 and 43 are both prime and the difference is 43 − 41 = 2, so they are twin primes.
  • Medium: 23 and 25 differ by 2, but 25 = 5 × 5 is not prime, so they are not twin primes.
  • Exam type: The primes from 100 to 110 are 101, 103, 107, 109. 103 − 101 = 2 and 109 − 107 = 2, so the twin prime pairs are 101 and 103, and 107 and 109.
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Co-prime and twin prime numbers

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Remember

Co-prime numbers need not themselves be prime: 8 and 15 are not prime, yet they are co-prime. Twin primes are always co-prime, but co-prime numbers are not always twin primes.

Common mistake

Taking 1 to be a prime number. A prime needs two different divisors (1 and itself). 1 has only one divisor, so 1 is neither prime nor composite.

Shortcut

Two even numbers are never co-prime, because both are divisible by 2 (e.g. 14 and 22). Consecutive numbers, however, are always co-prime.

Spot the difference

PairCo-prime?Twin primes?
8 and 15YesNo (not prime)
11 and 13YesYes
14 and 21No (7 is common)No
19 and 23YesNo (difference 4)

Practice questions

Easy
  1. 1Are 9 and 10 co-prime?
  2. 2Are 12 and 18 co-prime? Write their common divisors.
  3. 3Write all the twin prime pairs from 3 to 20.
Medium
  1. 4Write all the numbers below 10 that are co-prime to 16.
  2. 5Which pair of twin primes lies between 60 and 80?
  3. 6Which of the pairs 21 and 28, 25 and 36, 27 and 39 is co-prime?
Challenging
  1. 7The product of two consecutive numbers is 156. Which numbers are they? Are they co-prime?
  2. 842 laddoos and 55 karanjis are to be shared in equal packets (each packet must hold both items and nothing may be left over). Can more than 1 packet be made? Why?
Show answers (try them yourself first)
  1. 1Yes, because 9 and 10 are consecutive. 2) No; the common divisors are 1, 2, 3, 6. 3) 3 and 5, 5 and 7, 11 and 13, 17 and 19.
  2. 4The divisors of 16 are 1, 2, 4, 8, 16; so 1, 3, 5, 7, 9 (2, 4, 6, 8 and 16 share 2). 5) The primes are 61, 67, 71, 73, 79; only 71 and 73 differ by 2. 6) 25 and 36, because 21 and 28 share 7 and 27 and 39 share 3.
  3. 712 × 13 = 156; the two numbers are consecutive, so 12 and 13 are co-prime.
  4. 8The divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42 and of 55 are 1, 5, 11, 55; only 1 is common, so 42 and 55 are co-prime. Hence only 1 packet can be made, not more than 1.

Prime factorisation

From daily life

A wall is built from bricks and a brick cannot be broken down further. In the same way every composite number is built from prime "bricks". If 30 chairs are placed in 5 rows, each row has 6, and 6 chairs can be placed as 2 × 3. So 30 = 5 × 6 = 5 × 2 × 3. Now 2, 3 and 5 cannot be broken down any more; these are the prime factors of 30.

Simple definition

Writing a number as a product of prime numbers is called prime factorisation. Every number has only one set of prime factors, whatever the order.

6023021535

In the figure: 60 is split into two factors: 60 = 2 × 30, then 30 = 2 × 15 and 15 = 3 × 5. The gold circles are the prime factors: 60 = 2 × 2 × 3 × 5.

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Prime factorisation

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Rule 1: The division ladder — Divide the number by the smallest prime (2, 3, 5, 7 ...). Keep dividing each quotient again until you reach 1. The primes on the left are the prime factors.

  • Easy: 45 ÷ 3 = 15; 15 ÷ 3 = 5; 5 ÷ 5 = 1. So 45 = 3 × 3 × 5.
  • Medium: 72 ÷ 2 = 36; 36 ÷ 2 = 18; 18 ÷ 2 = 9; 9 ÷ 3 = 3; 3 ÷ 3 = 1. So 72 = 2 × 2 × 2 × 3 × 3.
  • Exam type: 462 ÷ 2 = 231; 231 ÷ 3 = 77; 77 ÷ 7 = 11; 11 ÷ 11 = 1. So 462 = 2 × 3 × 7 × 11.

Rule 2: The factor tree — Split the number into any two factors. Then keep splitting each factor until every factor is prime. Whichever way you split, you end up with the same prime factors.

  • Easy: 36 = 4 × 9; 4 = 2 × 2 and 9 = 3 × 3. So 36 = 2 × 2 × 3 × 3.
  • Medium: 90 = 9 × 10 = (3 × 3) × (2 × 5), so 90 = 2 × 3 × 3 × 5. The other way, 90 = 6 × 15 = (2 × 3) × (3 × 5) = 2 × 3 × 3 × 5; the same factors.
  • Exam type: 140 = 14 × 10 = (2 × 7) × (2 × 5) = 2 × 2 × 5 × 7. The other way, 140 = 4 × 35 = (2 × 2) × (5 × 7); the same prime factors.

Rule 3: Knowing which prime to divide by — If the units digit is 0, 2, 4, 6 or 8, the number is divisible by 2. If the sum of the digits is divisible by 3, the number is divisible by 3. If the units digit is 0 or 5, the number is divisible by 5.

  • Easy: 130 has 0 in the units place, so 130 is divisible by 2 and by 5.
  • Medium: In 261, 2 + 6 + 1 = 9 and 9 is divisible by 3, so 261 ÷ 3 = 87. In 87, 8 + 7 = 15, divide by 3 again: 87 ÷ 3 = 29, and 29 is prime. So 261 = 3 × 3 × 29.
  • Exam type: In 273, 2 + 7 + 3 = 12, so divide by 3: 273 ÷ 3 = 91. 91 is not divisible by 2, 3 or 5, but it is divisible by 7: 91 ÷ 7 = 13. So 273 = 3 × 7 × 13.
Remember

In the end every factor must be a prime number. Do not write 1 as a prime factor. If the number itself is prime, it cannot be split (e.g. 29).

Common mistake

Stopping at 36 = 4 × 9. 4 and 9 are not prime; split them further: 4 = 2 × 2 and 9 = 3 × 3. Only then 36 = 2 × 2 × 3 × 3 is the prime factorisation.

Shortcut

Divide by the small primes in order: 2, 3, 5, 7, 11. As long as a prime divides the number, keep dividing by it, then move to the next prime.

Spot the difference

NumberTypePrime factors
29PrimeCannot be split
45Composite3 × 3 × 5
72Composite2 × 2 × 2 × 3 × 3
1Neither prime nor compositeNone

Practice questions

Easy
  1. 1Write the prime factors of 28.
  2. 2Write the prime factors of 54.
  3. 3Which of the numbers 17, 21, 29 and 39 are prime?
Medium
  1. 4Find the prime factors of 168 with a division ladder.
  2. 5Write the prime factors of 225.
  3. 6Of which number are 2 × 2 × 3 × 5 × 5 the prime factors?
Challenging
  1. 7Find the prime factors of 360. How many times does 2 occur in them?
  2. 8Draw the factor tree of 120 in two ways (first split 12 × 10 in one and 8 × 15 in the other). Do you get the same prime factors?
Show answers (try them yourself first)
  1. 128 ÷ 2 = 14; 14 ÷ 2 = 7; 7 ÷ 7 = 1; so 28 = 2 × 2 × 7. 2) 54 ÷ 2 = 27; 27 ÷ 3 = 9; 9 ÷ 3 = 3; 3 ÷ 3 = 1; so 54 = 2 × 3 × 3 × 3. 3) 17 and 29 are prime; 21 = 3 × 7 and 39 = 3 × 13 are composite.
  2. 4168 ÷ 2 = 84; 84 ÷ 2 = 42; 42 ÷ 2 = 21; 21 ÷ 3 = 7; 7 ÷ 7 = 1; so 168 = 2 × 2 × 2 × 3 × 7. 5) 225 ÷ 3 = 75; 75 ÷ 3 = 25; 25 ÷ 5 = 5; 5 ÷ 5 = 1; so 225 = 3 × 3 × 5 × 5. 6) 2 × 2 × 3 × 5 × 5 = 300.
  3. 7360 ÷ 2 = 180; 180 ÷ 2 = 90; 90 ÷ 2 = 45; 45 ÷ 3 = 15; 15 ÷ 3 = 5; 5 ÷ 5 = 1; so 360 = 2 × 2 × 2 × 3 × 3 × 5 and 2 occurs three times.
  4. 8120 = 12 × 10 = (2 × 2 × 3) × (2 × 5) = 2 × 2 × 2 × 3 × 5. 120 = 8 × 15 = (2 × 2 × 2) × (3 × 5) = 2 × 2 × 2 × 3 × 5. Yes, the prime factors are the same.

HCF (Highest Common Factor)

From daily life

A shopkeeper has 40 pencils and 56 erasers. He wants to make as many identical packets as possible; every packet will hold both items and nothing will be left over. The number of packets must divide both 40 and 56 exactly and must be the largest possible. That number is 8: 8 packets, each with 5 pencils and 7 erasers. So the HCF of 40 and 56 is 8.

Simple definition

The largest of all the common divisors that divide the given numbers exactly is the HCF (Highest Common Factor) of those numbers.

Rule 1: Listing method — Write all the divisors of each number. The largest of the common divisors is the HCF. (An easy method for small numbers.)

  • Easy: 18 and 27 → divisors of 18 are 1, 2, 3, 6, 9, 18 and of 27 are 1, 3, 9, 27; common 1, 3, 9; HCF = 9.
  • Medium: 30 and 45 → divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 and of 45 are 1, 3, 5, 9, 15, 45; common 1, 3, 5, 15; HCF = 15.
  • Exam type: 16, 24 and 40 → divisors of 16 are 1, 2, 4, 8, 16; of 24 are 1, 2, 3, 4, 6, 8, 12, 24; of 40 are 1, 2, 4, 5, 8, 10, 20, 40; common to all three 1, 2, 4, 8; HCF = 8.

Rule 2: Prime factor method — List the prime factors of every number. Pick the prime factors that all the numbers share and multiply them.

Prime factors of 252Prime factors of 210232375

In the figure: 252 = 2 × 2 × 3 × 3 × 7 and 210 = 2 × 3 × 5 × 7. The middle part holds the common prime factors 2, 3, 7. Their product is 2 × 3 × 7 = 42, so the HCF of 252 and 210 is 42.

  • Easy: 20 = 2 × 2 × 5 and 50 = 2 × 5 × 5. Common 2 and 5; HCF = 2 × 5 = 10.
  • Medium: 72 = 2 × 2 × 2 × 3 × 3 and 108 = 2 × 2 × 3 × 3 × 3. Common 2, 2, 3, 3; HCF = 2 × 2 × 3 × 3 = 36.
  • Exam type: 60 = 2 × 2 × 3 × 5; 90 = 2 × 3 × 3 × 5; 150 = 2 × 3 × 5 × 5. Common to all three 2, 3, 5; HCF = 2 × 3 × 5 = 30.
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HCF (Highest Common Factor)

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Rule 3: Division method — Divide the larger number by the smaller one. If there is a remainder, divide the divisor by that remainder. Continue until the remainder is 0. The divisor when the remainder is 0 is the HCF. Every step: dividend = divisor × quotient + remainder.

DividendDivisorRemainder25210542105422142210

In the figure: 252 and 105 → 252 = 105 × 2 + 42; 105 = 42 × 2 + 21; 42 = 21 × 2 + 0. The divisor when the remainder became 0 is 21, so the HCF = 21.

  • Easy: 54 and 24 → 54 = 24 × 2 + 6; 24 = 6 × 4 + 0. HCF = 6.
  • Medium: 176 and 66 → 176 = 66 × 2 + 44; 66 = 44 × 1 + 22; 44 = 22 × 2 + 0. HCF = 22.
  • Exam type: 391 and 187 → 391 = 187 × 2 + 17; 187 = 17 × 11 + 0. HCF = 17.
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HCF (Highest Common Factor)

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Remember

The HCF divides every given number exactly and is never greater than the smallest given number. The HCF of co-prime numbers is always 1.

Common mistake

Multiplying prime factors that are not common as well. 20 = 2 × 2 × 5 and 50 = 2 × 5 × 5: multiplying them all gives 2 × 2 × 5 × 5 = 100, which is greater than 20. Take only the common 2 and 5: 10.

Shortcut

If one number divides the other exactly, the smaller number is the HCF. E.g. 91 ÷ 7 = 13, so the HCF of 7 and 91 is 7.

Spot the difference

MethodHow to do itWhen it is easy
Listing methodWrite all divisors and take the largest common oneSmall numbers
Prime factor methodMultiply the common prime factorsMedium numbers, more than two numbers
Division methodKeep dividing the divisor by the remainder until the remainder is 0Large numbers

Practice questions

Easy
  1. 1Use the listing method to get the HCF of 12 and 30.
  2. 2Use prime factors to get the HCF of 45 and 75.
  3. 3What is the HCF of 17 and 51?
Medium
  1. 4Find the HCF of 168 and 120.
  2. 5Find the HCF of 66, 99 and 165.
  3. 6Use the division method to get the HCF of 323 and 437.
Challenging
  1. 7The HCF of three numbers is 6. Two of them are 24 and 42. If the third number lies between 31 and 39, which number is it?
  2. 8Two ribbons are 96 cm and 156 cm long. They are to be cut into equal pieces, as long as possible, with nothing left over. How long is each piece and how many pieces are there in all?
Show answers (try them yourself first)
  1. 1The divisors of 12 are 1, 2, 3, 4, 6, 12 and of 30 are 1, 2, 3, 5, 6, 10, 15, 30; common 1, 2, 3, 6; HCF = 6. 2) 45 = 3 × 3 × 5 and 75 = 3 × 5 × 5; common 3, 5; HCF = 3 × 5 = 15. 3) 51 = 17 × 3, so the HCF = 17.
  2. 4168 = 2 × 2 × 2 × 3 × 7 and 120 = 2 × 2 × 2 × 3 × 5; common 2, 2, 2, 3; HCF = 2 × 2 × 2 × 3 = 24. 5) 66 = 2 × 3 × 11; 99 = 3 × 3 × 11; 165 = 3 × 5 × 11; common 3, 11; HCF = 3 × 11 = 33. 6) 437 = 323 × 1 + 114; 323 = 114 × 2 + 95; 114 = 95 × 1 + 19; 95 = 19 × 5 + 0; HCF = 19.
  3. 7The third number must be a multiple of 6. The only multiple of 6 between 31 and 39 is 36. Check: 24 = 2 × 2 × 2 × 3; 42 = 2 × 3 × 7; 36 = 2 × 2 × 3 × 3; common 2, 3; HCF = 6.
  4. 896 = 2 × 2 × 2 × 2 × 2 × 3 and 156 = 2 × 2 × 3 × 13; common 2, 2, 3; HCF = 12. Each piece is 12 cm. 96 ÷ 12 = 8 and 156 ÷ 12 = 13, so 8 + 13 = 21 pieces.

LCM (Lowest Common Multiple)

From daily life

Laddoos come in packs of 6 and karanjis in packs of 4. If you want the same number of both, what is the least you must buy? Multiples of 6: 6, 12, 18 ...; multiples of 4: 4, 8, 12 ... The first common multiple is 12. That means 2 packs of laddoos and 3 packs of karanjis; 12 of each. This 12 is the LCM of 6 and 4.

Simple definition

Of all the common multiples of some numbers, the smallest one is their LCM (Lowest Common Multiple).

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960Multiples of 4Multiples of 10Both

In the figure: In the numbers 1 to 60, the multiples of 4 are blue, the multiples of 10 are orange and the common multiples are gold. The first gold number is 20, so the LCM of 4 and 10 is 20. The next gold numbers, 40 and 60, are also multiples of 20.

Rule 1: Listing multiples — Write the multiples of each number. The first common multiple is the LCM.

  • Easy: 6 and 8 → multiples of 6 are 6, 12, 18, 24, 30 and of 8 are 8, 16, 24, 32; the first common multiple is 24; LCM = 24.
  • Medium: 9 and 15 → multiples of 9 are 9, 18, 27, 36, 45 and of 15 are 15, 30, 45; the first common multiple is 45; LCM = 45.
  • Exam type: 14 and 21 → multiples of 14 are 14, 28, 42 and of 21 are 21, 42; the first common multiple is 42; LCM = 42.

Rule 2: Prime factor method — List the prime factors of every number. Take each prime factor as many times as it occurs at most in any one number, and multiply them all.

Prime factors of 60Prime factors of 8452237

In the figure: 60 = 2 × 2 × 3 × 5 and 84 = 2 × 2 × 3 × 7. Take the common part 2 × 2 × 3 only once. Add the 5 left over from 60 and the 7 left over from 84: 2 × 2 × 3 × 5 × 7 = 420, so the LCM of 60 and 84 is 420.

  • Easy: 18 = 2 × 3 × 3 and 24 = 2 × 2 × 2 × 3. 2 at most three times (in 24), 3 at most twice (in 18); LCM = 2 × 2 × 2 × 3 × 3 = 72.
  • Medium: 24 = 2 × 2 × 2 × 3 and 54 = 2 × 3 × 3 × 3. 2 at most three times, 3 at most three times; LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216.
  • Exam type: 15 = 3 × 5; 20 = 2 × 2 × 5; 24 = 2 × 2 × 2 × 3. 2 at most three times, 3 once, 5 once; LCM = 2 × 2 × 2 × 3 × 5 = 120.
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LCM (Lowest Common Multiple)

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Rule 3: Two special cases — (a) If the numbers are co-prime, the LCM = their product. (b) If one number is a multiple of the other, the LCM = the larger number.

  • Easy: 7 and 9 are co-prime, so the LCM = 7 × 9 = 63.
  • Medium: 40 = 8 × 5, so 40 is a multiple of 8; hence the LCM of 8 and 40 is 40.
  • Exam type: 4, 8 and 24 → 24 = 8 × 3 and 24 = 4 × 6, so 24 is a multiple of both 4 and 8; hence the LCM is 24.

Rule 4: The link between HCF and LCM — For two numbers: HCF × LCM = the product of the two numbers. (This does not work for three or more numbers.)

  • Easy: 12 and 18: HCF 6 and LCM 36; 6 × 36 = 216 = 12 × 18.
  • Medium: Two numbers have a product of 360 and an HCF of 6. LCM = 360 ÷ 6 = 60.
  • Exam type: HCF 4, LCM 120 and one number is 24. The other number = (4 × 120) ÷ 24 = 480 ÷ 24 = 20. Check: the HCF of 24 and 20 is 4 and the LCM is 120.
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LCM (Lowest Common Multiple)

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Remember

The LCM is never smaller than the largest given number, and every given number divides it exactly. The HCF is small, the LCM is large.

Common mistake

Swapping HCF and LCM. The HCF is the product of the common prime factors; the LCM takes every prime factor as many times as it occurs at most. 12 and 18: HCF 6 and LCM 36.

Shortcut

Check the multiples of the larger number and find the first one that the smaller number divides. 6 and 8: 8, 16, 24; 24 is divisible by 6, so the LCM is 24.

Spot the difference

PointHCFLCM
MeaningLargest common divisorSmallest common multiple
Prime factorsOnly the common onesEach one as many times as it occurs at most
Compared with the given numbersSmaller or equalLarger or equal
For 12 and 18636

Practice questions

Easy
  1. 1What is the LCM of 5 and 7?
  2. 2Find the LCM of 10 and 25 by listing multiples.
  3. 3What is the LCM of 6 and 18?
Medium
  1. 4Find the LCM of 42 and 63 by the prime factor method.
  2. 5Find the LCM of 8, 12 and 18.
  3. 6The HCF of two numbers is 5 and their LCM is 150. If one number is 30, what is the other?
Challenging
  1. 7Which is the smallest 3-digit number that is divisible by 6, 8 and 12?
  2. 8Find the LCM of 18, 27 and 45.
Show answers (try them yourself first)
  1. 15 and 7 are co-prime, so the LCM = 5 × 7 = 35. 2) Multiples of 10 are 10, 20, 30, 40, 50 and of 25 are 25, 50; the first common multiple is 50, so the LCM = 50. 3) 18 is a multiple of 6 (6 × 3 = 18), so the LCM is 18.
  2. 442 = 2 × 3 × 7 and 63 = 3 × 3 × 7; LCM = 2 × 3 × 3 × 7 = 126. 5) 8 = 2 × 2 × 2; 12 = 2 × 2 × 3; 18 = 2 × 3 × 3; LCM = 2 × 2 × 2 × 3 × 3 = 72. 6) The other number = (5 × 150) ÷ 30 = 750 ÷ 30 = 25.
  3. 76 = 2 × 3; 8 = 2 × 2 × 2; 12 = 2 × 2 × 3; LCM = 2 × 2 × 2 × 3 = 24. Multiples of 24 are 24, 48, 72, 96, 120; the first 3-digit multiple is 120.
  4. 818 = 2 × 3 × 3; 27 = 3 × 3 × 3; 45 = 3 × 3 × 5; LCM = 2 × 3 × 3 × 3 × 5 = 270.

Uses of HCF and LCM

From daily life

You have three ropes, 30 metres, 45 metres and 75 metres long. You want to cut them into equal pieces, as long as possible, with nothing left over. The length of a piece must divide all three lengths exactly and be as large as possible, that is, the HCF.

30 m45 m75 m

In the figure: The HCF of 30, 45 and 75 is 15. Each piece is 15 metres long; the ropes give 2, 3 and 5 pieces respectively.

Simple definition

To cut, share or arrange in rows and get the "largest equal part", use the HCF. When something repeats and you need "when will they meet again" or "the smallest equal number", use the LCM.

Rule 1: The largest equal piece, group or row → HCF — Find the largest number that divides all the given numbers.

  • Easy: Two ribbons 20 metres and 30 metres long are to be cut into equal pieces, as long as possible. HCF = 10, so each piece is 10 metres; pieces 20 ÷ 10 = 2 and 30 ÷ 10 = 3, 5 in all.
  • Medium: Square tiles, as large as possible, are to be laid on a floor 540 cm long and 420 cm wide without cutting any tile. Side of a tile = HCF = 60 cm. Tiles = (540 ÷ 60) × (420 ÷ 60) = 9 × 7 = 63.
  • Exam type: Three cans hold 60, 84 and 108 litres of oil. The largest measure that can measure all three exactly is the HCF = 12 litres (60 = 2 × 2 × 3 × 5; 84 = 2 × 2 × 3 × 7; 108 = 2 × 2 × 3 × 3 × 3; common 2 × 2 × 3 = 12). Measures: 5 + 7 + 9 = 21.

Rule 2: Meeting again or happening together → LCM — Find the first common multiple of all the given numbers.

0122436486072ABMinutes

In the figure: Watchman A passes the gate every 9 minutes and watchman B every 12 minutes. Both are at the gate at the start. They are at the gate together again after 36 minutes (the LCM of 9 and 12 is 36), and once more after 72 minutes.

  • Easy: One lamp blinks every 8 seconds and another every 12 seconds. If both start together, they will blink together again after 24 seconds, the LCM of 8 and 12.
  • Medium: Three buses leave every 12, 20 and 30 minutes. All three left together at 6 in the morning. LCM = 60 minutes, so they will leave together again at 7.
  • Exam type: Three bells ring every 9, 15 and 18 minutes. All three rang together at 12 noon. 9 = 3 × 3; 15 = 3 × 5; 18 = 2 × 3 × 3; LCM = 2 × 3 × 3 × 5 = 90 minutes. So they will ring together again after 90 minutes (1 hour 30 minutes), that is, at 1:30 in the afternoon.
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Uses of HCF and LCM

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Rule 3: The smallest number that leaves the same remainder every time → LCM + remainder — Work out the LCM of the divisors and add the remainder to it. (The remainder must be smaller than every divisor.)

  • Easy: Dividing by 6 and 9 leaves 1 every time. LCM = 18, so the number = 18 + 1 = 19. Check: 19 = 6 × 3 + 1 and 19 = 9 × 2 + 1.
  • Medium: Dividing by 7, 8 and 14 leaves 4 every time. LCM = 56, so the number = 56 + 4 = 60.
  • Exam type: Dividing by 6 and 10 leaves 2 every time and the number is greater than 40. LCM = 30. 30 + 2 = 32 (less than 40), 60 + 2 = 62 (greater than 40), so the smallest number is 62.

Rule 4: Use in fractions — To add fractions, use the LCM of the denominators as the common denominator. To simplify a fraction, divide the numerator and the denominator by their HCF.

  • Easy: 14 + 16 → the LCM of 4 and 6 is 12; 14 = 312 and 16 = 212; sum = 312 + 212 = 512.
  • Medium: 310 + 215 → LCM 30; 310 = 930 and 215 = 430; sum = 930 + 430 = 1330.
  • Exam type: Simplify 84126 → the HCF of 84 and 126 is 42; 84 ÷ 42 = 2 and 126 ÷ 42 = 3, so 84126 = 23.
Remember

Words like "largest", "cut", "share", "rows" point to the HCF; words like "together again", "at the same time", "smallest number divisible by all" point to the LCM.

Common mistake

Stopping at the LCM in a question that says "remainder". Dividing by 6 and 9 leaves 1: the answer is not 18, it is 18 + 1 = 19.

Shortcut

Check your answer: if the answer should be smaller than or equal to the given numbers, use the HCF; if it should be larger than or equal to them, use the LCM.

Spot the difference

Words in the questionWhat to useExample
Largest equal piecesHCFRibbons of 20 m and 30 m → piece 10 m
Meeting againLCM8 and 12 seconds → 24 seconds
Smallest number leaving the same remainderLCM + remainder6 and 9, remainder 1 → 19
Simplifying a fractionHCF of numerator and denominator84 and 126 → 42

Practice questions

Easy
  1. 1Two ropes 36 metres and 54 metres long are to be cut into equal pieces, as long as possible. How long is each piece?
  2. 2Two lamps blink every 6 and 15 seconds. If they start together, after how many seconds will they blink together again?
  3. 316 + 19 = ?
Medium
  1. 4Simplify 4560.
  2. 5What is the smallest number that leaves remainder 6 when divided by 10, 15 and 25?
  3. 6Three friends go to the library every 4, 6 and 10 days respectively. All three were there together today. After how many days will they all be there together again, and how many visits will each friend have made by then?
Challenging
  1. 7A school has 120 boys and 168 girls. As many groups as possible are to be made, with the same number of boys in every group and the same number of girls in every group. How many groups are there, and how many boys and girls in each group?
  2. 8Two buses leave together at 7 in the morning. One leaves every 14 minutes and the other every 35 minutes. At what time will they leave together again?
Show answers (try them yourself first)
  1. 136 = 2 × 2 × 3 × 3 and 54 = 2 × 3 × 3 × 3; common 2, 3, 3; HCF = 2 × 3 × 3 = 18; each piece is 18 metres. 2) 6 = 2 × 3 and 15 = 3 × 5; LCM = 2 × 3 × 5 = 30 seconds. 3) The LCM of 6 and 9 is 18; 16 = 318 and 19 = 218; sum = 318 + 218 = 518.
  2. 4The HCF of 45 and 60 is 15; 45 ÷ 15 = 3 and 60 ÷ 15 = 4; so 4560 = 34. 5) The LCM of 10, 15 and 25 is 150; 150 + 6 = 156. 6) The LCM of 4, 6 and 10 is 60; so they will all be together after 60 days; visits 60 ÷ 4 = 15, 60 ÷ 6 = 10 and 60 ÷ 10 = 6.
  3. 7120 = 2 × 2 × 2 × 3 × 5 and 168 = 2 × 2 × 2 × 3 × 7; HCF = 2 × 2 × 2 × 3 = 24, so 24 groups. Each group has 120 ÷ 24 = 5 boys and 168 ÷ 24 = 7 girls.
  4. 814 = 2 × 7 and 35 = 5 × 7; LCM = 2 × 5 × 7 = 70 minutes, that is 1 hour 10 minutes. So they will leave together again at 10 minutes past 8.