Factors Divisibility and Prime Factorisation

What you will learn

Recognise the structure of a positive integer before choosing a calculation. You will find factors and multiples, distinguish prime and composite numbers, use divisibility tests with reasons, and write a complete prime factorisation. You will also explain when divisibility by two numbers guarantees divisibility by their product.

This lesson develops the number-system foundations used by HCF, LCM, fractions and square roots later in this module. It does not predict topic-wise exam marks. Unless stated otherwise, all factors, multiples, divisors and numbers being tested here are positive integers. This avoids silently mixing in zero or negative-factor conventions.

1 Retrieve three useful ideas

1. 42 ÷ 6 = 7 because 6 × 7 = 42. This division has no remainder.
2. 3² = 3 × 3 = 9. An exponent counts repeated factors.
3. 43 = 6 × 7 + 1. Here the quotient is 7 and the remainder is 1, so 6 does not divide 43 exactly.

For positive integer division, the divisor is positive and the quotient is a nonnegative integer. Write dividend = divisor × quotient + remainder, with the remainder an integer at least 0 and less than the divisor. You do not need a fraction or decimal to decide whether an integer divides another exactly. If this is unfamiliar, check 29 ÷ 5 by writing 29 = 5 × 5 + 4, and 30 ÷ 5 by writing 30 = 5 × 6 + 0.

2 A factor divides and a multiple is built by multiplication

If n = d × k for positive integers d and k, then d and k are factors of n, and n is a multiple of both. The same statement can be read in either direction, but the words are not interchangeable: 6 is a factor of 42; 42 is a multiple of 6.

Every positive integer has 1 and itself as factors; for 1 these are the same factor. All its positive factors lie between 1 and the number, so there are finitely many. Its positive multiples are obtained by multiplying by 1, 2, 3, …, so the list never ends. Zero is a multiple in a wider integer convention, but it is not a positive multiple and is excluded from our lists.

Worked example 1 Build a complete factor list

List all positive factors of 48 and its first four positive multiples.

1. Form factor pairs: 1 × 48, 2 × 24, 3 × 16, 4 × 12 and 6 × 8.
2. The trial factor 5 fails, since 48 = 5 × 9 + 3. Once the smaller trial factor would exceed its partner, subsequent pairs repeat in reverse. In this case 7 × 7 > 48, so there cannot be a pair with both factors at least 7.
3. Collect both members of each pair: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
4. The first four positive multiples are 48 × 1, 48 × 2, 48 × 3, 48 × 4: 48, 96, 144, 192.

Error check: writing only one member of each factor pair misses valid factors. If a pair repeats the same factor, such as 5 × 5 = 25, list that factor only once.

3 Prime is a property of one number and co-prime is a relationship

A prime number has exactly two distinct positive factors, 1 and itself. A composite number is an integer greater than 1 with more than two positive factors. The number 1 is neither: it has only one positive factor. The number 2 is the only even prime; every larger even number has the additional factor 2. An odd number need not be prime.

Two positive integers are co-prime when their only common positive factor is 1. This does not require either number to be prime. Common factors are simply factors that appear in both factor lists. The next lesson develops the highest of the common factors into the HCF method.

Worked example 2 Test the definition rather than the appearance

Classify 1, 2, 39 and 47, then decide whether 8 and 15 are co-prime.

• 1 has one positive factor, so it is neither prime nor composite.
• 2 has factors 1 and 2, so it is prime.
• 39 = 3 × 13, so it is composite despite being odd.
• To test 47, observe that 7 × 7 > 47. A nontrivial factor pair would have a factor from 2 through 6. It is enough to test the primes 2, 3 and 5: 4 and 6 contain 2 or 3. None divides 47, so 47 is prime.
• Factors of 8 are 1, 2, 4, 8; factors of 15 are 1, 3, 5, 15. Their only common factor is 1, so 8 and 15 are co-prime, though both are composite.

Error check: do not confuse “both prime” with “co-prime”. Also, being not divisible by 2 is only one test; it does not prove primality.

4 Divisibility tests save calculation because of place value

A divisibility test decides whether the remainder is zero. It does not usually give the quotient. The following tests apply to positive integers written in decimal digits:

• 2: last digit is 0, 2, 4, 6 or 8
• 5: last digit is 0 or 5
• 10: last digit is 0
• 4: the number formed by the last two digits is divisible by 4
• 8: the number formed by the last three digits is divisible by 8
• 3: the sum of all digits is divisible by 3
• 9: the sum of all digits is divisible by 9
• 6: the number is divisible by both 2 and 3
• 11: the difference between the sums of alternate digits is 0 or a positive or negative multiple of 11

For a number with fewer than two or three digits, test the whole number for 4 or 8. A suffix such as 00 or 000 has value 0 and passes the relevant divisibility test.

Why the last digits work

Every complete group of 10 is divisible by 2 and 5, leaving only the last digit to check. For divisibility by 10, the last digit must be 0. Similarly, 100 is divisible by 4 and 1000 is divisible by 8. The higher-place part is already divisible, so only the last two or three digits can leave a remainder.

Worked example 3 Test 5312 by its endings

Test divisibility by 2, 4, 5, 8 and 10.

1. The last digit is 2: divisible by 2, but not by 5 or 10.
2. The last two digits form 12 = 4 × 3. Since 5312 = 5300 + 12 and 5300 is divisible by 4, the number is divisible by 4.
3. The last three digits form 312 = 8 × 39. Since 5312 = 5000 + 312 and 5000 is divisible by 8, it is divisible by 8.

Error check: an even last digit alone does not prove divisibility by 4 or 8. Use the required number of ending digits.

Why the digit sum works

Write 7389 as 7 × 1000 + 3 × 100 + 8 × 10 + 9. Each of 1000, 100 and 10 is one more than a multiple of 9. Replacing them by 1 therefore removes only multiples of 9, leaving 7 + 3 + 8 + 9. The original number and this digit sum have the same remainder on division by 9. The removed quantities are also multiples of 3, so the same reasoning works for 3.

Worked example 4 Combine a digit-sum test with an evenness test

Test 7389 for divisibility by 3, 9 and 6.

1. Its digit sum is 7 + 3 + 8 + 9 = 27.
2. Since 27 is divisible by 3 and 9, 7389 is divisible by both 3 and 9.
3. 7389 is odd, so it is not divisible by 2. Therefore it is not divisible by 6.

Why does 6 require both tests? Since 6 = 2 × 3 and 2 and 3 are distinct primes, the number must contain both factors. A number divisible by 3 alone need not be divisible by 9 or by 6. For instance, 15 is divisible by 3, but not by 9 or 6.

Why alternate digit sums work for 11

Successive powers of 10 are alternately 1 below or 1 above a multiple of 11: 10 = 11 − 1, 100 = 99 + 1, 1000 = 1001 − 1. This turns a place-value sum into an alternating digit sum. You may begin at either end; reversing the signs does not change whether the difference is a multiple of 11.

Worked example 5 Find a missing digit

Which digit d makes the four-digit number 47d3 divisible by 11?

1. A digit is an integer from 0 to 9. The alternating difference is (4 + d) − (7 + 3) = d − 6.
2. As d ranges from 0 to 9, that difference ranges from −6 to 3. The only multiple of 11 in that range is 0.
3. Thus d − 6 = 0, giving d = 6. Verify: 4763 = 11 × 433.

Error check: the difference need not equal 11. Zero, 11, −11, 22 and −22 all pass, when they can arise from the given digits.

5 Prime factorisation reveals the building blocks

A factorisation expresses a number as a product of factors. A prime factorisation goes further: every factor in the product is prime. Keep splitting composite factors until none remain, or divide repeatedly by prime divisors. Repeated copies of a prime can then be written as a power. For every integer greater than 1, the list of prime factors and their multiplicities is unique apart from order.

Worked example 6 Finish and verify the factorisation

Write 840 as a product of powers of primes.

1. Split into manageable factors: 840 = 84 × 10.
2. Continue: 84 = 2 × 2 × 3 × 7, and 10 = 2 × 5.
3. Combine: 840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.
4. Check by multiplying back: 8 × 3 × 5 × 7 = 24 × 35 = 840.

Starting with 840 = 8 × 105 leads to the same final primes. A different first split is allowed; a different final count of a prime means a calculation was missed. Do not stop at 4 × 210: 4 and 210 are composite. Do not add 1 to the list of prime factors.

6 When can two divisibility tests be combined into a product

For co-prime positive integers a and b, a number divisible by both a and b is divisible by a × b. In prime-factor language, the requirements do not share any prime factors, so both complete groups of factors must occur. If a and b are not co-prime, some prime requirements overlap and the product can demand extra copies that were never guaranteed.

Worked example 7 State the guarantee carefully

A number is divisible by both 4 and 9. Must it be divisible by 36? Compare the conditions “divisible by 4 and 6”.

1. 4 = 2² and 9 = 3² have no common prime factor. A number divisible by both must contain two factors 2 and two factors 3, so it is divisible by 2² × 3² = 36.
2. With 4 = 2² and 6 = 2 × 3, the requirements overlap in a factor 2. The number 12 is divisible by both 4 and 6, but not by 24. Thus the product conclusion is not guaranteed.
3. It is not impossible either: 48 is divisible by 4, 6 and their product 24.

Error check: “not guaranteed” is different from “never possible”. State the condition and test the claim with an actual number. The next lesson explains how the LCM handles these overlapping requirements systematically.

7 Independent practice

Choose one option in each question before reading the explanations. Write the factor pair, relevant digit test or counterexample that supports the choice. This is free, untimed learning practice with no negative marking or pass cutoff, not a full RRB CBT simulation.

1. Is 7 a factor of 91?

A. Yes, because 91 = 7 × 13 B. No, because 91 is odd C. Yes, because 9 + 1 = 10 D. No, because 7 is prime

2. Which number appears in both lists: 6, 12, 18, 24, … and 10, 20, 30, 40, …? Choose the least positive common value.

A. 12 B. 20 C. 24 D. 30

3. Why is 1 not a prime number?

A. It is odd B. It has only one positive factor C. It is divisible by every positive integer D. It has no factors

4. Which pair is co-prime?

A. 9 and 15 B. 16 and 24 C. 9 and 16 D. 21 and 28

5. Which statement is correct for 2356?

A. Divisible by 4 but not by 8 B. Divisible by 8 but not by 4 C. Divisible by both 4 and 8 D. Divisible by neither 4 nor 8

6. Which statement is correct for 6243?

A. Divisible by 9 but not by 3 B. Divisible by both 3 and 9 C. Divisible by neither 3 nor 9 D. Divisible by 3 but not by 9

7. Which number is divisible by 6?

A. 4515 B. 4518 C. 4520 D. 4523

8. Using the 11-test on 9185, what conclusion follows?

A. Not divisible because the digit sum is 23 B. Not divisible because the alternating difference is not 0 C. Divisible because (9 + 8) − (1 + 5) = 11 D. Divisible because its last digit is 5

9. Which is the prime factorisation of 1260?

A. 2² × 3² × 5 × 7 B. 2 × 3² × 5 × 7 C. 2² × 3 × 5 × 7 D. 4 × 9 × 5 × 7

10. What is the smallest digit d that makes the four-digit number 52d4 divisible by 9?

A. 2 B. 5 C. 6 D. 7

11. A list claims to show all positive factors of 40: 1, 2, 4, 5, 10, 20, 40. Which factor is missing?

A. 6 B. 8 C. 12 D. 16

12. A positive integer is divisible by both 6 and 9. Which statement is always justified?

A. It must be divisible by 54 B. It cannot be divisible by 54 C. Divisibility by 54 is not guaranteed; 18 is a counterexample D. It must be a prime number

8 Explained answer key

1. A. 91 = 7 × 13, so division by 7 leaves no remainder. Oddness and primality do not prevent divisibility; the digit sum shown is not a test for 7.

2. D. Continue the lists: 30 = 6 × 5 = 10 × 3. Earlier positive multiples of 10 are 10 and 20, neither a multiple of 6.

3. B. A prime has exactly two distinct positive factors. The only positive factor of 1 is 1. It is neither prime nor composite.

4. C. 9 has positive factors 1, 3, 9; 16 has 1, 2, 4, 8, 16. Only 1 is shared. The other pairs share 3, 8 and 7, respectively.

5. A. The last two digits form 56 = 4 × 14. The last three form 356 = 8 × 44 + 4, so the 8-test fails.

6. D. The digit sum is 6 + 2 + 4 + 3 = 15. It is divisible by 3, but not by 9.

7. B. 4518 is even and its digit sum is 18, so both the 2 and 3 tests pass. 4515 is odd; 4520 has digit sum 11; 4523 is odd and has digit sum 14.

8. C. The alternating difference is 11, a multiple of 11. A difference of 0 is allowed but not required. Check: 9185 = 11 × 835.

9. A. 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260 and all the bases 2, 3, 5, 7 are prime. D gives the right product but still contains composite factors 4 and 9. B and C give 630 and 420.

10. D. The digit sum is 11 + d. For d from 0 to 9 it ranges from 11 to 20, whose only multiple of 9 is 18. Thus d = 7. Check: 5274 = 9 × 586.

11. B. The pair 5 × 8 = 40 reveals the missing 8. The complete list is 1, 2, 4, 5, 8, 10, 20, 40.

12. C. 18 is divisible by 6 and 9, but not by 54, so A fails. Yet 54 itself is divisible by all three, so B also fails. The shared factor 3 means the product rule is not guaranteed.

Sources and next step

NCERT Mathematics Class VI, Playing with Numbers, via IIT Kanpur SATHEE, §§3.2–3.6, supports factors, primes, co-primality, divisibility and prime factorisation; the summary states the coprime-product rule. The web edition is cited by section rather than an unverified printed page. NIOS Mathematics 211 Chapter 2, §2.2, printed pp.44–45, supports prime-power notation. The explanations, chosen worked numbers and practice here are original and independently calculated.

Before continuing, explain one last-digit test, one digit-sum test and the difference between prime and co-prime. If a mistake remains, revisit that section and retry its question; there is no pass gate. Next, HCF LCM and Choosing the Right Model uses this factor structure to solve grouping and repetition problems.

Analogy

Packets and labels

Imagine 35 counters that must be placed in equal packets with none left over. Packets of 5 work because 5 × 7 = 35; packets of 6 leave 5 counters after filling five packets. The allowed packet sizes are factors of 35. If you instead keep adding full packets of 5, the totals 5, 10, 15, … are multiples of 5.

This distinguishes a size that divides a fixed total from totals built by repeating a fixed size. A divisibility test is a quick check of whether a packet size fits exactly; it does not count the packets for you. Prime factorisation is a numerical decomposition, not a claim that physical packets must be prime-sized.

Quick reference

Definitions and domain

This lesson uses positive integers and positive factors/multiples. A factor divides exactly; a multiple is obtained by multiplication. A prime has exactly two distinct positive factors. A composite is greater than 1 and has more than two. 1 is neither prime nor composite; 2 is the only even prime. Co-prime numbers have only 1 as a common positive factor and may both be composite.

Divisibility checklist

• 2: last digit even
• 5: last digit 0 or 5
• 10: last digit 0
• 4: last two-digit suffix divisible by 4
• 8: last three-digit suffix divisible by 8
• 3 or 9: digit sum divisible by 3 or 9, respectively
• 6: both the 2 and 3 tests pass
• 11: alternate digit-sum difference is 0 or a multiple of 11, allowing negative differences

For a short number, use all its digits. Suffix 00 or 000 has value 0.

Verification

Collect both factors in each pair, avoiding duplicates. A complete prime factorisation contains only primes; multiply back to check the number and count repeated factors before writing exponents. A number divisible by two co-prime positive integers is divisible by their product. Without co-primality the implication is not guaranteed, rather than impossible.