Finding Common Ground | CBSE Class 7 Maths Notes

This note covers factors and multiples, the highest common factor, primes and prime factorisation, the lowest common multiple, division procedures, conjectures, counterexamples, generalisation, and relationships between the highest common factor and lowest common multiple.

How do common factors help us choose the largest equal size?

A factor of a positive whole number divides it exactly, leaving no remainder, or amount left over after division. A common factor divides every number in a given group exactly. Here, positive whole numbers are the counting numbers beginning with 1. Factors help us divide quantities into equal parts without leftovers.

Definition: The Highest Common Factor (HCF) of two or more numbers is the greatest of their common factors. It is also called the Greatest Common Divisor (GCD). A divisor is a number by which another number is divided.

Why does the room need a common factor?

Sameeksha's room is 12 ft wide and 16 ft long; ft means feet. She wants square tiles of equal size, with a whole number of feet along each side. The tiles must fit both dimensions exactly, and she wants to use as few as possible.

The side of each square must divide both 12 and 16. The largest suitable side is therefore 4 ft. A larger suitable tile covers more of the floor, so fewer tiles are needed. Choosing the greatest common factor answers the size question.

Worked example 1. Find the largest whole-number tile side for Sameeksha's 12 ft by 16 ft room.

Answer: The common factors are 1, 2 and 4. Their greatest value is 4, so the tiles should have side 4 ft.

The rough rectangular outline has 16 ft marked above it and 12 ft beside it. Small square corner marks show right angles, or square corners. The drawing shows the room's dimensions, not a completed tile arrangement.

Reference: NCERT Class 7, page 47, unnumbered figure

How does the same reasoning apply to rice?

Lekhana has 84 kg of rice from one farm and 108 kg from another; kg means kilograms. Every bag must hold the same whole-number weight, and rice from the two farms must remain separate. Using as few bags as possible requires the greatest suitable weight.

Worked example 2. Find the greatest equal bag weight for 84 kg and 108 kg of rice, keeping the farms' rice separate.

Answer: Their common factors are 1, 2, 3, 4, 6 and 12. The HCF is 12, so each bag should contain 12 kg of rice.

The important condition is exact division of both quantities. A weight that divides just one farm's quantity cannot satisfy the whole problem. After finding common factors, choose the greatest because the question asks for the fewest equal bags.

How does prime factorisation break a number into useful parts?

A prime number is a number greater than 1 whose only factors are 1 and itself. A composite number is a whole number greater than 1 with factors besides 1 and itself. A prime factor is a factor that is prime. Composite factors can be broken down further.

A product is the result of multiplication. Prime factorisation expresses a number greater than 1 as a product of primes. In the calculations below, × means multiplication, = means equality, and ÷ means division. Brackets group factors that are considered together.

Can different starting factors give different primes?

Worked example 3. Compare two ways of finding the prime factorisation of 90.

Answer: One factorisation is 90 = 3 × 3 × 2 × 5. Starting with 3 × 30 gives 90 = 3 × 30 = 3 × 2 × 15 = 3 × 2 × 3 × 5. Both contain two 3s, one 2 and one 5.

The resulting prime factors are always the same, with perhaps only a change in their order. Reordering factors does not change their product. The prime factorisation of a prime number is the prime number itself, so it does not need further splitting into smaller factors.

How does the division method work?

The division method repeatedly divides a composite number by a prime factor. The result of a division is called the quotient. In the displayed procedure, one factor is written on the left and the quotient is written below the number being divided.

1. Choose a prime factor of the starting composite number.
2. Divide by that prime and write the quotient underneath.
3. If the quotient is composite, divide it by a prime factor again.
4. When a prime remains at the bottom, multiply it with the prime factors collected on the left.

Worked example 4. Use the division method to factorise 105 and 30.

Answer: For 105, division by 3 gives 35, and division of 35 by 5 gives 7. Thus 105 = 3 × 5 × 7. For 30, division by 2 gives 15, and division by 3 gives 5. Thus 30 = 2 × 3 × 5.

One chain places 105 above 35 and 7, with 3 and 5 to the left. Another places 30 above 15 and 5, with 2 and 3 to the left. Circles highlight the composite numbers in the explanatory drawings.

Reference: NCERT Class 7, pages 49 to 50, unnumbered figures

Each circled number equals the factor on its left multiplied by the number below. While carrying out factorisation, the circles are usually left out. The written division steps preserve the same multiplication relationships even without the circles.

How can prime factors reveal all factors and test a conjecture?

A subpart of a prime factorisation is a selection of its prime factors, respecting how many times each occurs. Multiplying the selected factors gives a factor of the original number. The unselected factors multiply to the number needed to recover the original product.

How do selected and remaining factors fit together?

Worked example 5. Show that 28 is a factor of 840 using 840 = 2 × 2 × 2 × 3 × 5 × 7.

Answer: Select 2 × 2 × 7 = 28. The remaining factors give 2 × 3 × 5 = 30. Therefore 840 = (2 × 2 × 7) × (2 × 3 × 5) = 28 × 30.

This method keeps track of the available copies of each prime. A selection cannot use more copies than the factorisation contains. For instance, the factorisation of 840 has three occurrences of 2 but just one occurrence of 3. An occurrence means one appearance of a factor.

How can the factors of 225 be listed systematically?

The prime factorisation is 225 = 5 × 5 × 3 × 3. Group the possible selections by how many prime factors they contain. Repeated ways of obtaining the same factor need not create repeated entries in the final list.

Include 1 as well. The complete factor list is 1, 3, 5, 9, 15, 25, 45, 75, 225. The method is systematic, but listing every possible subpart can still be tedious. Finding only the HCF will not require listing all factors.

Why does a larger number not necessarily have a longer factorisation?

A conjecture is a mathematical claim made without proof or verification. A counterexample is an example in which that claim is false. Anshu conjectures that a larger number must have a longer prime factorisation.

Compare 96 = 2 × 2 × 2 × 2 × 2 × 3 with 121 = 11 × 11. Although 121 is larger, its prime factorisation has fewer prime factors. This counterexample disproves Anshu's conjecture. Here, length counts all occurrences, including repeated primes.

Testing a claim means checking what it actually asserts. These two numbers compare both size and factorisation length, so they address the claim directly. Finding a pattern in a few examples does not remove the need for reasoning or verification.

How do we find the HCF directly from prime factorisations?

A common factor must be available as a subpart of every given number's prime factorisation. The HCF uses the largest common subpart. It contains shared primes, with no prime used more often than it occurs in any one of the given numbers.

What do the common subparts show?

Worked example 6. Find the common factors and HCF of 45 and 75.

Answer: Factorise 45 = 3 × 3 × 5 and 75 = 3 × 5 × 5. Their common subparts give 3, 5 and 3 × 5 = 15. Including 1, the common factors are 1, 3, 5 and 15. Their HCF is 15.

Worked example 7. Find the HCF of 112 and 84.

Answer: 112 = 2 × 2 × 2 × 2 × 7, while 84 = 2 × 2 × 3 × 7. The largest common subpart contains two 2s and one 7. Thus the HCF is 2 × 2 × 7 = 28.

For 112 and 84, selecting four 2s would fail because 84 supplies only two. Selecting the 3 would fail because it does not occur in 112. These restrictions explain why we compare minimum occurrences, meaning the smallest count across the given factorisations.

How do we carry out the minimum-occurrence rule?

1. Write the prime factorisation of each number completely.
2. Identify the primes present in every factorisation.
3. For each shared prime, select the minimum number of its occurrences.
4. Multiply these selected factors to obtain the HCF; if there are no shared primes, the HCF is 1.

Worked example 8. Find the HCF of 30 and 72.

Answer: 30 = 2 × 3 × 5 and 72 = 2 × 2 × 2 × 3 × 3. Both contain 2 and 3, but 30 supplies only one of each. Their HCF is 2 × 3 = 6.

For 225 and 750, the factorisations are 3 × 3 × 5 × 5 and 2 × 3 × 5 × 5 × 5. Select one 3 and two 5s. Their HCF is 3 × 5 × 5 = 75.

What if there are no shared primes?

The factorisations 96 = 2 × 2 × 2 × 2 × 2 × 3 and 275 = 5 × 5 × 11 have no shared prime. Their only common factor is 1, so their HCF is 1. Such numbers are called co-prime numbers.

The same minimum-occurrence method applies to more than two numbers. A prime must then occur in every factorisation to contribute to the common subpart. Comparing incomplete composite factors can hide common primes, so finish the prime factorisations before drawing a conclusion.

How do common multiples identify the first shared length or event?

A multiple of a positive whole number is obtained by multiplying it by a positive whole number. A common multiple is a multiple of every given number. Multiples are useful when complete units repeat and we want to find a shared total.

Definition: The Lowest Common Multiple (LCM) of two or more given numbers is the smallest of their common multiples. It is also called the least common multiple. Here, the lists begin with positive multiples.

How do the torans produce lists of multiples?

A toran here is a decoration made by placing cloth strips one next to another. Anshu uses strips of length 6 cm and Guna uses strips of length 8 cm; cm means centimetres. Each total length must be made from complete strips.

Worked example 9. Find the shortest equal toran length possible using 6 cm strips for Anshu and 8 cm strips for Guna.

Answer: The two lists first meet at 24. Although 48 is another common multiple, 24 is smaller. The shortest equal torans are therefore 24 cm long.

The problem asks for a common multiple because the total is built by repeating the strips. It asks for the lowest one because the torans must be as short as possible. The room problem instead divided fixed lengths into equal parts.

How can repeating days be matched?

Kabamai visits a sweet shop once every 10 days. The shop gives free gajak to school children on Mondays. Gajak is a sweet made from sesame seeds, jaggery and ghee. Starting on a Monday when she receives it, compare the two schedules.

Worked example 10. If Kabamai receives free gajak today, a Monday, and returns every 10 days, when will a visit next coincide with a Monday?

Answer: Mondays recur after 7, 14, 21, 28, 35, 42, 49, 56, 63 and 70 days. Visits recur after 10, 20, 30, 40, 50, 60 and 70 days. The first shared value is 70 days.

Listing multiples makes the meaning of LCM visible. However, for larger numbers, the lists may become tedious before a common entry appears. Prime factorisation provides a more direct method while keeping the same requirement: the answer must be a multiple of each given number.

How does prime factorisation give the least common multiple?

A common multiple must contain each original number's prime factorisation as a subpart. To find the LCM, build the smallest product that meets all those requirements. This reverses the HCF comparison: the HCF fits inside every number, while each number must fit inside the LCM.

What does a multiple contain?

For example, 36 = 2 × 2 × 3 × 3. Its multiple 648 can be written as 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3). The factorisation of 36 is present as a complete subpart.

Worked example 11. Find the LCM of 14 and 35.

Answer: 14 = 2 × 7 and 35 = 5 × 7. The product 2 × 5 × 7 contains both required subparts. Removing any factor would lose one of those requirements. Therefore the LCM is 2 × 5 × 7 = 70.

One occurrence of 7 serves both requirements in this example. Including a second 7 would still give a common multiple, but it would not give the lowest one. The aim is to include enough copies of every prime, without unnecessary copies.

How do we compare repeated primes?

Worked example 12. Find the LCM of 96 and 360.

Answer: 96 = 2 × 2 × 2 × 2 × 2 × 3 and 360 = 2 × 2 × 2 × 3 × 3 × 5. Select five 2s, two 3s and one 5. The LCM is 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440.

Use the maximum number of occurrences, meaning the greatest count needed in either factorisation. Include 5 even though it occurs in only one number. Otherwise, the result would not contain the factorisation of 360 and could not be its multiple.

1. Factorise all the given numbers into primes.
2. Collect every prime appearing anywhere in these factorisations.
3. For each prime, select its greatest number of occurrences in any one factorisation.
4. Multiply the selected factors to obtain the LCM.

This procedure extends to more than two numbers. Compare the counts across the whole group and keep the greatest requirement for each prime. Do not add the counts from all the factorisations, because shared factors can satisfy more than one number's requirement.

What properties help us reason about HCF and LCM?

A general statement describes a pattern or property holding in all possible cases under its stated conditions. Forming such a statement is called generalisation. The conditions matter: a pattern for a number and its multiple does not automatically describe every pair of numbers.

Property: A number and its multiple have the number as their HCF

The HCF of 6 and 18 is 6. This happens because 6 is a factor of 18. More generally, let n represent a positive whole number. The expression 5n means 5 × n, so it represents a multiple of n.

The HCF of n and 5n is n. The number n divides both, and no factor of n can be greater than n. Writing the relationship with a letter expresses the same reasoning for different choices of the positive whole number.

Property: Doubling both numbers doubles their HCF

The factorisations 270 = 2 × 3 × 3 × 3 × 5 and 50 = 2 × 5 × 5 give HCF = 2 × 5 = 10. Doubling both numbers gives 540 and 100.

Now 540 = 2 × 2 × 3 × 3 × 3 × 5 and 100 = 2 × 2 × 5 × 5. Their HCF is 2 × 2 × 5 = 20. Each factorisation has gained an extra 2, which joins the largest common subpart.

A common multiplier is a factor used to multiply both numbers. It need not be their HCF. For 14 × 6 and 14 × 9, the multiplier 14 is shared, but the remaining factors 6 and 9 also share 3. The HCF is 14 × 3 = 42.

Property: The HCF and LCM are related to the product

For 105 and 95, the prime factorisations are 105 = 3 × 5 × 7 and 95 = 5 × 19. Their LCM is 3 × 5 × 7 × 19. Their product contains another 5: 105 × 95 = 3 × 5 × 5 × 7 × 19.

Thus 105 × 95 = LCM × 5, and 5 is their HCF. Our observations seem to suggest the relationship HCF × LCM = product of the two numbers. To investigate why, compare the common and non-common prime factors in the two sides.

Note: The relationship above concerns two numbers. Whether it holds for three numbers is a separate question to explore. Keep the number of quantities in view when applying a property.

The product of two numbers is itself a common multiple. Therefore their LCM is never greater than that product. This provides a useful reasonableness check: a proposed least common multiple should not exceed a common multiple already available from the two numbers.

How can a division procedure find both HCF and LCM?

Writing complete prime factorisations is useful, but common factors can also be removed from both numbers together. This common-division procedure records the factors shared at each step. It stops when the two remaining numbers have no common prime factors.

How are common factors removed?

For 84 and 180, divide both by 2 to obtain 42 and 90. Divide both by 2 again to obtain 21 and 45. Dividing both by 3 leaves 7 and 15, which have no common prime factor.

The common factors removed were 2, 2 and 3. Their product gives the HCF. This procedure keeps the shared part separate from what remains, so it can also show which factors are needed to build a common multiple.

Worked example 13. Find both HCF and LCM of 300 and 150 by dividing by common factors.

Answer: Divide by 2, then 5, then 5, then 3. The successive pairs are 150 and 75; 30 and 15; 6 and 3; 2 and 1. The HCF is 2 × 5 × 5 × 3 = 150. Including the final 2 and 1 gives LCM = 300.

The final pair means the two numbers left after all common factors have been removed. For the HCF, multiply the removed common factors. For the LCM, multiply those factors together with both remaining numbers. At this point the remaining numbers have no shared prime factor.

Can we divide by composite common factors?

Yes. Guna divides 300 and 150 by 50 first, giving 6 and 3, then divides both by 3, leaving 2 and 1. The HCF is 50 × 3 and the LCM is 50 × 3 × 2 × 1.

Worked example 14. Find the HCF and LCM of 630 and 770 using larger common factors.

Answer: Divide both by 10 to obtain 63 and 77. Divide both by 7 to obtain 9 and 11. Their HCF is 10 × 7 = 70. Their LCM is 10 × 7 × 9 × 11 = 6930.

We need not restrict the procedure to removing one prime factor at a time. Any identified common factor can be used, provided it divides both current numbers exactly. Continue checking the remaining pair so that a further common factor is not overlooked.

The top pair is 630, 770 with 10 on the left. The next pair is 63, 77 with 7 on the left. The bottom pair is 9, 11. The layout records both shared divisors and remaining numbers.

Reference: NCERT Class 7, page 62, unnumbered figure

For each method, the calculation should answer the original question. A factor divides the given quantities; a multiple contains complete repetitions of them. Interpreting that distinction comes before choosing whether to calculate an HCF or an LCM.

Glossary

• Factor — A number that divides a given whole number exactly, with no remainder.
• Common factor — A factor shared by every number in the group being considered.
• Highest Common Factor — The greatest of the common factors of two or more given numbers.
• Prime number — A number greater than 1 whose only factors are 1 and itself.
• Composite number — A whole number greater than 1 having factors besides 1 and itself.
• Prime factorisation — An expression of a number greater than 1 as a product of primes.
• Quotient — The result obtained when one number is divided by another number.
• Multiple — A number obtained by multiplying a given number by a positive whole number.
• Common multiple — A number that is a multiple of each of the given numbers.
• Lowest Common Multiple — The smallest positive common multiple of two or more given numbers.
• Co-prime numbers — Numbers whose only common factor is 1, so their HCF is 1.
• Conjecture — A mathematical statement or claim made without proof or verification.
• Counterexample — An example in which a conjecture fails, showing that the conjecture is false.
• Generalisation — The process of forming a statement describing a pattern holding in all relevant cases.

Common errors and misconceptions

• Misconception: The largest equal tile side depends only on the room's length. Correct: The side must divide both length and breadth exactly. For the 12 ft by 16 ft room, it is 4 ft.
• Misconception: A larger number has a longer prime factorisation. Correct: The factorisations of 96 and 121 disprove this: 96 has six prime occurrences, while the larger number 121 has two.
• Misconception: The HCF includes every copy of each shared prime. Correct: It uses the minimum count across the factorisations. For 30 and 72, only one 2 and one 3 can be included.
• Misconception: Numbers without a common prime have no common factor. Correct: They still share the factor 1. The numbers 96 and 275 have no shared prime, but their HCF is 1.
• Misconception: The LCM uses only primes present in both numbers. Correct: It needs every prime appearing in either number, with the maximum required count. The LCM of 96 and 360 includes 5.
• Misconception: A common multiplier must be the HCF. Correct: The numbers 14 × 6 and 14 × 9 have HCF 42, because the remaining factors 6 and 9 share another factor, 3.
• Misconception: The common-division procedure must use primes at every step. Correct: Composite common factors can also be removed. Dividing 630 and 770 by 10, then 7, gives their HCF and helps find their LCM.

Exam-style questions with model answers

Q1. Define a prime number and write the prime factorisation of 90. [2 marks]

1. A prime number is greater than 1 and has only 1 and itself as factors.
2. The prime factorisation is 90 = 2 × 3 × 3 × 5; every factor shown in this product is prime.

Q2. A room is 12 ft wide and 16 ft long. Equal square tiles must fit both dimensions exactly, have whole-number side lengths in feet, and be as few as possible. Find the required tile side, showing your reasoning. [3 marks]

1. The factors of 12 are 1, 2, 3, 4, 6 and 12; the factors of 16 are 1, 2, 4, 8 and 16.
2. A suitable square side must divide both room dimensions, so the available whole-number sides are their common factors: 1 ft, 2 ft and 4 ft.
3. Choose the greatest common factor, 4. Tiles with side 4 ft are the largest suitable ones and therefore minimise the number used.

Q3. Find the HCF of 225 and 750 by prime factorisation. Explain how many copies of each common prime are used. [4 marks]

1. The factorisations are 225 = 3 × 3 × 5 × 5 and 750 = 2 × 3 × 5 × 5 × 5.
2. The shared primes are 3 and 5. The factor 2 is excluded because it is not present in the factorisation of 225.
3. Select one 3 because 750 contains only one. Select two 5s because 225 contains only two, the smaller available count.
4. Multiply the largest common subpart: HCF = 3 × 5 × 5 = 75. These counts fit inside both prime factorisations.

Q4. Find the LCM of 96 and 360 using prime factorisation. Justify the number of occurrences chosen for every prime. [5 marks]

1. Factorise both numbers: 96 = 2 × 2 × 2 × 2 × 2 × 3, and 360 = 2 × 2 × 2 × 3 × 3 × 5.
2. Choose five occurrences of 2. This meets the requirement of 96 as well as the smaller requirement of three occurrences in 360.
3. Choose two occurrences of 3, because 360 needs two even though 96 needs only one. A single 3 would not be sufficient.
4. Choose one occurrence of 5. It is required by 360, so it must be included even though it is absent from 96.
5. Multiply the selected factors: LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440. These are the maximum required counts.

Q5. A sweet shop gives free gajak to school children every Monday. Kabamai receives it today, a Monday, and visits once every 10 days. After how many days will a visit next coincide with free gajak? Explain. [3 marks]

1. Mondays recur every 7 days, so the free-gajak days after today include 7, 14, 21, 28, 35, 42, 49, 56, 63 and 70.
2. Kabamai's later visits occur after 10, 20, 30, 40, 50, 60 and 70 days. A suitable day must occur in both lists.
3. The first shared entry is 70, the LCM of 7 and 10. She will therefore next receive free gajak after 70 days.

Q6. Use common division to find both the HCF and LCM of 630 and 770. Begin by dividing by 10, then check for another common factor. [5 marks]

1. Both starting numbers are divisible by 10. Dividing gives 630 ÷ 10 = 63 and 770 ÷ 10 = 77.
2. The remaining numbers 63 and 77 share 7. Divide both by 7 to obtain the next pair, 9 and 11.
3. The final numbers 9 and 11 have no common prime factor, so there is no further shared factor to remove.
4. Multiply the common factors removed from both numbers: HCF = 10 × 7 = 70. These factors form their complete common part.
5. For the LCM, also include both remaining numbers: LCM = 10 × 7 × 9 × 11 = 6930. This includes the shared part and both remaining parts.

Q7. Anshu claims that a larger number must have a longer prime factorisation. Use 96 = 2 × 2 × 2 × 2 × 2 × 3 and 121 = 11 × 11 to test the claim. Name the kind of evidence used. [3 marks]

1. The given factorisation of 96 has six prime occurrences, counting every repeated 2. The factorisation of 121 has just two prime occurrences.
2. Although 121 is greater than 96, it has the shorter prime factorisation. Thus the comparison does not behave as Anshu's claim predicts.
3. This is a counterexample, meaning an example where the conjecture is false. It disproves the claim that larger numbers must have longer prime factorisations.

Q8. Given 105 = 3 × 5 × 7 and 95 = 5 × 19, identify their HCF, write their LCM as a product of primes, and compare HCF × LCM with 105 × 95. [3 marks]

1. The only common prime is 5, occurring once in each factorisation. Therefore the HCF of 105 and 95 is 5.
2. The LCM needs each prime appearing in either number once. Its prime factorisation is therefore 3 × 5 × 7 × 19.
3. The product 105 × 95 is 3 × 5 × 5 × 7 × 19. Multiplying the LCM by the HCF supplies the second 5, giving that same product.

Key takeaways

• The HCF is the greatest common factor, useful when dividing given quantities into the largest possible equal parts.
• Prime factorisation breaks a number into primes; reordering those prime factors does not change their product.
• A factor can be formed from a subpart of the prime factorisation, respecting each prime's available number of occurrences.
• Find the HCF by choosing shared primes with their minimum numbers of occurrences across all the given factorisations.
• Find the LCM by choosing every needed prime with its maximum number of occurrences in any given factorisation.
• The common-division procedure can find both HCF and LCM, and identified composite common factors can shorten the calculation.
• A counterexample disproves a conjecture; the numbers 96 and 121 show that larger numbers need not have longer prime factorisations.
• For two numbers, observations seem to suggest that their HCF multiplied by their LCM equals their product.

Test yourself

What is another name for the Highest Common Factor?

It is also called the Greatest Common Divisor, abbreviated as GCD.

Which factors remain after selecting 2 × 2 × 7 from 840 = 2 × 2 × 2 × 3 × 5 × 7?

The remaining factors are 2 × 3 × 5, whose product is 30. Thus 840 = 28 × 30.

What are all the factors of 225?

The complete list is 1, 3, 5, 9, 15, 25, 45, 75 and 225.

Why is the HCF of 96 and 275 equal to 1?

Their prime factorisations share no prime factor. Therefore their only common factor, and hence their HCF, is 1.

Why does the LCM of 14 and 35 contain just one 7?

Both numbers need one 7. A single occurrence, together with 2 and 5, contains both complete prime factorisations.

What happens to the HCF when both 270 and 50 are doubled?

The HCF doubles from 10 to 20 because both prime factorisations gain one additional factor of 2.

Why is 14 not the HCF of 14 × 6 and 14 × 9?

The remaining factors 6 and 9 share 3, so the HCF is 14 × 3 = 42.

Why can 50 be used as the first divisor for 300 and 150 in common division?

It divides both numbers exactly, giving 6 and 3. The procedure can remove composite common factors as well as primes.