Chapter 8 — Divisibility

Meaning: Divisibility tells us when one whole number fits exactly into another. If a divides b, then b can be written as b = a × k for some whole number k. In this case a is called a factor of b and b is called a multiple of a.

Factors: A factor (or divisor) of a number is a number that multiplies with another whole number to give the original number. For example, factors of 12 are 1, 2, 3, 4, 6 and 12.

Multiples: Multiples of a number are results of multiplying it by whole numbers. For example, the first few multiples of 4 are 4, 8, 12, 16, 20, … Multiples go on without end, while factors are limited for a fixed number.

Related ideas: Every number has 1 and itself as factors. Numbers with only these two factors are called prime; others are composite. Zero is a special multiple because every number divides zero, but zero has no factors in the usual sense. Negative numbers also have factors and multiples but here we work with positive whole numbers.

How to find factors: To find factors of a number, try dividing by small numbers in pairs: if a × b = n then a and b are factors. Check up to √n to avoid repetition. For a classroom number such as 36, you check 1, 2, 3, 4, 6 and so on.

Why it matters: Knowing factors and multiples helps in reducing fractions, solving division problems and understanding later topics like LCM, HCF and algebraic factoring.

Basic rules and why they work: Divisibility rules for 2, 5 and 10 are simple because they depend only on the last digit (unit digit) of a number. This is because our number system is base ten: every place to the left represents a multiple of 10. Since tens, hundreds, thousands etc. are all multiples of 10, only the units digit can change the remainder when dividing by numbers that are factors of 10.

Divisible by 2: If the last digit is even (0, 2, 4, 6, 8), the whole number is divisible by 2. The tens and higher places contribute multiples of 10 which are themselves divisible by 2, so they do not affect whether the number is even. This test also tells us whether a number is even or odd quickly.

Divisible by 5: If the last digit is 0 or 5, the number is divisible by 5. That is because 10 is divisible by 5 and so every tens block contributes a multiple of 5; only the units digit decides whether the whole is a multiple of 5.

Divisible by 10: If the last digit is 0, the number is divisible by 10. In practice this means numbers ending with 0 are multiples of 10. Recognising a trailing zero helps when simplifying fractions, rounding or checking calculations.

Combined observations: If a number ends with 0 it is divisible by 2, 5 and 10 at the same time. If it ends with 5 it is divisible by 5 but not by 2 or 10. Using these three rules together speeds up many mental calculations — for example when multiplying, checking parity of products, or when quickly simplifying fractions by removing factors of 2 or 5.

Classroom practice: Ask students to look only at the last digit and call out which of the three divisibilities apply. Practice with large numbers to build habit: e.g., 12,340 (divisible by 2,5,10), 7,695 (divisible by 5 only), 8,246 (divisible by 2 only).

The rule: A number is divisible by 3 if the sum of its digits is divisible by 3. To use the test, add all the digits of the number; if the total is a multiple of 3 (for example 3, 6, 9, 12, 15 ...), then the original number is divisible by 3.

Why the rule works: Consider any number written in base ten: its value is a sum of digits times powers of 10. Since 10 leaves remainder 1 when divided by 3 (10 = 3×3 + 1), each power of 10 also leaves remainder 1 modulo 3. That means 10k ≡ 1 (mod 3) for any k. So the whole number is congruent to the sum of its digits modulo 3. If the digit sum is divisible by 3, so is the whole number.

How to apply for big numbers: For large numbers, after summing digits you may get a large sum. It is enough to continue reducing by adding digits of the sum until you reach a small number between 0 and 9; this final value is the digital root. If the digital root is 3, 6 or 9 (or the reduced sum is 0 when using exact multiples), the number is divisible by 3.

Practical examples and checks: For 123,456 add digits 1+2+3+4+5+6 = 21, then 2+1=3, hence divisible by 3. For 5,678: 5+6+7+8 = 26, 2+6=8, not divisible by 3. This method is fast and useful for checking sums, multiplication answers and simplifying fractions when a factor of 3 is suspected.

Teaching tip: Give pairs of numbers and ask students to quickly decide divisibility by 3 using digit sums. Show why a number divisible by 9 is also divisible by 3, but not vice versa, and connect this to the next test (divisibility by 9).

The rule: A number is divisible by 9 if the sum of its digits is divisible by 9. You add the digits; if the total is 9, 18, 27, 36 ... then the number is divisible by 9. If the sum is still large, repeat the digit-sum process until you reach a small number; if the final single-digit result is 9 (or zero in full-sum terms) the number is divisible by 9.

Reasoning: This rule follows from the same idea used for divisibility by 3. Since 10 ≡ 1 (mod 9), every 10k contributes remainder 1 when dividing by 9. So the number is congruent to the sum of its digits modulo 9. If the digit-sum is a multiple of 9, the whole number is a multiple of 9. This is why the digital root test works well: repeated digit-sum gives the remainder modulo 9.

Difference from divisibility by 3: While both 3 and 9 use digit-sum, the 9-test is stronger. Every number divisible by 9 is also divisible by 3, but a number divisible by 3 may not be divisible by 9. For example, 18 is divisible by 9, but 12 is divisible by 3 only.

Uses and checks: The 9-test is useful for checking arithmetic: when adding several numbers, the sum of their digit-sums modulo 9 should match the digit-sum of the result; if not, an error occurred. It also helps to spot mistakes after multiplication. For long numbers, reduce the digit-sum step by step to reach a small number quickly.

Practice idea: Give pupils many numbers and ask them to decide divisibility by 9 using the digit-sum method. Also practise combining rules: for numbers ending in 0 or 5, check 5-rule first then 9-rule if needed. Use examples like 7,839 or 54,321 to reinforce the steps and the reason why the rule works.

Divisibility by 4 — the last-two-digit test: The test for 4 says: a number is divisible by 4 if the number formed by its last two digits is divisible by 4. This works because 100 is a multiple of 4, so any hundreds, thousands and higher place values contribute multiples of 100 which are divisible by 4. Therefore, only the last two-digit block (tens and units) can affect the remainder modulo 4.

How to apply: Look at the last two digits; if that two-digit number (or single digit treated as two-digit with leading zero) is divisible by 4 then the whole number is divisible by 4. For instance, 3,124 → check 24 → 24 ÷ 4 = 6 so the whole number is divisible by 4. For 308 → check 08 → 8 ÷ 4 = 2 so divisible.

Divisibility by 25 — specific endings: A number is divisible by 25 if its last two digits are 00, 25, 50 or 75. This stems from the fact that 100 is divisible by 25; hence the remainder modulo 25 depends solely on the last two digits. The sequence of two-digit endings that give multiples of 25 repeats every 100.

Understanding with examples: For 7,225 the last two digits are 25 so it is divisible by 25; dividing 7225 ÷ 25 = 289. For 2,500 the ending 00 shows divisibility by both 4 and 25 (and by 100). A number ending with 75 like 1,275 is divisible by 25 but not by 4 (since 75 ÷ 4 leaves remainder).

Class exercises and links: Give students lists of numbers and ask them to check divisibility by both 4 and 25 using last-two-digits only. Also show connection: if last two digits are 00, the number is divisible by 4, 25 and 100. Practice helps students quickly identify divisibility without full division and prepares them for LCM/HCF work where factors 2 and 5 often appear.

Divisibility by 6 — use 2 and 3: The number 6 factors as 2×3 and because 2 and 3 are coprime, a number is divisible by 6 exactly when it is divisible by both 2 and 3. This gives a simple two-step test: check that the last digit is even (divisible by 2), and check that the sum of digits is divisible by 3. If both conditions hold, the number is divisible by 6.

Why it is correct: If a number is divisible by 2 and by 3 then it has both 2 and 3 among its prime factors, hence it must be divisible by 6. Conversely, if it is divisible by 6 then it leaves no remainder when divided by 2 or 3 individually, so both tests must pass.

Divisibility by 8 — last-three-digit test: The test for 8 requires checking the last three digits of the number. A number is divisible by 8 if the integer formed by its last three digits is divisible by 8. This is because 1000 = 125×8, so thousands and higher place values are multiples of 8 and do not change the remainder. For numbers with fewer than three digits, check the full number.

How to use the last-three-digit test: For large numbers, write down the last three digits and divide that three-digit number by 8. If it divides exactly, the whole number is divisible by 8. For instance, for 12,456 check 456; since 456 ÷ 8 = 57 exactly, 12,456 is divisible by 8. For 2,016 check 016 → 16 ÷ 8 = 2 so yes.

Class practice and tips: Combine the 6-test with the 2 and 3 tests in quick mental checks: if a number ends with an even digit and its digit-sum is multiple of 3, it is divisible by 6. For 8, practise dividing three-digit endings by 8 mentally or with short division. Use examples like 1,344 for 6 (even + digit-sum 12) and 5,432 for 8 (check 432). These tests make many divisibility checks fast without full division.

Divisibility by 7 — subtract twice last digit: While rules for 7 are less direct than for 2 or 5, there is a useful test: remove the last digit of the number, double it and subtract that from the remaining truncated number. If the result is divisible by 7 (including zero), then the original number is divisible by 7. Repeat the process until a small number is reached which you can check easily.

Why this method works (short idea): The test rests on the relation 10 ≡ 3 (mod 7). Algebraically, writing the number as 10×(rest) + last digit and adjusting with twice the last digit creates an equivalent remainder modulo 7, allowing reduction of large numbers to smaller equivalents.

Procedure and examples: For 203: remove last digit 3, double → 6, subtract from 20 gives 14 which is divisible by 7, so 203 is divisible by 7. For 1,092: last digit 2 → double 4 → 109 − 4 = 105 → divisible by 7.

Divisibility by 11 — alternating sum: For 11 the test is simpler to describe: compute the alternating sum of the digits (add digits in odd positions and subtract digits in even positions, or vice versa). If the result is 0 or a multiple of 11 (like 11, −11, 22, etc.), the number is divisible by 11. This works because 10 ≡ −1 (mod 11), so powers of 10 alternate signs when reduced modulo 11.

Examples for 11: For 2,431: (2 − 4 + 3 − 1) = 0 so divisible by 11. For 1,452: (1 − 4 + 5 − 2) = 0 so divisible. Show students with 2-, 3- and 4-digit numbers until they see the pattern. These tests are handy for medium-size numbers and for factoring problems where 7 or 11 might be a divisor.

Prime numbers — definition and examples: A prime number greater than 1 has exactly two positive factors: 1 and itself. Examples are 2, 3, 5, 7, 11, 13, 17, 19 and so on. Notice that 2 is the only even prime because every other even number is divisible by 2 and hence has at least three factors (1, 2 and itself).

Composite numbers — definition and examples: A composite number is a number greater than 1 that has more than two positive factors. For example, 12 is composite because its factors are 1, 2, 3, 4, 6 and 12. Composite numbers can be built by multiplying primes together.

Number 1 and 0: The number 1 is neither prime nor composite by definition because it has exactly one positive factor. Zero is not considered a prime or composite in this context: zero is a multiple of every number, but it has no meaningful prime factorisation.

How to test if a number is prime: To test whether n is prime, try dividing it by prime numbers up to √n. If none of these primes divide n exactly, then n must be prime. For class-6 numbers this method is straightforward: for 29, check divisibility by 2, 3 and 5 (primes ≤ √29 ≈ 5.4); none divide 29 so it is prime.

Prime factorisation: Any composite number can be written as a product of prime factors. Use factor trees to find prime factors: for 60, split to 6×10, then 6 → 2×3 and 10 → 2×5, so 60 = 2×2×3×5. Recording prime factors helps find HCF and LCM later.

Why primes matter: Primes are the building blocks of numbers: every integer greater than 1 can be written uniquely (up to order) as a product of primes. Practise listing primes up to 50 and building factor trees to become confident in recognising primes and composites.

What is HCF? The Highest Common Factor (HCF), also called GCD (greatest common divisor), of two or more numbers is the largest number that exactly divides each of them. For example, HCF of 18 and 24 is 6 because 6 is the greatest number that divides both without remainder.

Method 1 — List factors: For small numbers list all factors of each number and choose the greatest common one. For 36 and 60, factors of 36 are 1,2,3,4,6,9,12,18,36 and of 60 are 1,2,3,4,5,6,10,12,15,20,30,60. The greatest common factor is 12.

Method 2 — Prime factorisation: A clearer method uses prime factors. Express each number as a product of primes, then take the common primes with the smallest exponent. Example: 18 = 2×3×3 = 2×3^2 and 24 = 2×2×2×3 = 2^3×3. Common primes are one 2 and one 3, so HCF = 2×3 = 6. This method extends easily to three or more numbers.

Method 3 — Repeated division: Use repeated division by common primes: divide both numbers by a common prime until no common prime divides both. Multiply the common divisors to get HCF. This is practical for classroom use when factor trees are not fully written.

Applications: HCF is used to simplify fractions, to divide objects into the largest equal groups with no leftovers, and it links to LCM by the formula a×b = HCF(a,b) × LCM(a,b). Practise with problems like splitting sweets or simplifying fractions to see HCF in action.

What is LCM? The Least Common Multiple (LCM) of two or more numbers is the smallest positive number that is a multiple of each of them. For example, the LCM of 6 and 8 is 24 because 24 is the smallest number divisible by both 6 and 8.

Method 1 — List multiples: For small numbers we can list multiples until a common one appears. For 6: 6,12,18,24,... and for 8: 8,16,24,... the first common multiple is 24. This method is simple but slow for larger numbers.

Method 2 — Prime factorisation (best): Write each number as product of primes and take each prime with the highest power that appears in any factorisation. For example, 6 = 2×3 and 8 = 2×2×2 = 2^3. The highest power of 2 between them is 2^3, and include 3 once, so LCM = 2^3×3 = 24. Use this method for three or more numbers as well: find highest power of each prime across all numbers and multiply them.

Relation to HCF: For two numbers a and b we have a×b = HCF(a,b)×LCM(a,b). Thus if you know HCF and the product a×b, you can compute LCM by dividing product by HCF. This relation is handy for quick calculations.

Applications: LCM is used for adding fractions with different denominators, scheduling repeating events (when two cycles meet), and solving problems where two or more periodic actions must coincide. Practise examples like LCM of 12 and 15 or of 4,6 and 10 to gain confidence.

Goal and idea: Simplifying a fraction means writing it in lowest terms so numerator and denominator have no common factor other than 1. Divisibility tests are useful to find common small factors quickly so we can reduce the fraction step by step. For many classroom fractions, checking divisibility by 2, 3 and 5 will simplify most cases.

Step-by-step method: First, check easy small primes: if both numerator and denominator end with even digits, divide both by 2. If both end with 0 or 5, try dividing by 5. Use digit-sum to see if both are divisible by 3. Continue dividing by common primes until no further small common factor exists. Alternatively, find HCF by prime factorisation and divide both by it in one step.

Worked reduction: For example 48/180: both even → divide by 2 → 24/90; again even → divide by 2 → 12/45. Now check 3: digit sums 1+2=3 and 4+5=9, both divisible by 3 → divide by 3 → 4/15. Now 4 and 15 have no common factors, so the fraction is in lowest terms. Using HCF method: prime factors 48=2^4×3 and 180=2^2×3^2×5, common = 2^2×3 = 12, dividing gives 4/15 directly.

Why this helps: Simplified fractions are easier to compare, add and use in calculations. Quick divisibility checks save time in exams and reduce errors. Teach students to try 2, 3, 5 first; these cover many cases. For practice, give mixed problems including ones needing larger HCFs so students learn both stepwise and prime-factor methods.

Types of word problems: Many real-life questions use divisibility: sharing items equally among people, forming groups with equal numbers, arranging objects in rows with no leftover, and scheduling repeating events. Translate the words into arithmetic and then use divisibility or HCF/LCM to find the solution.

Using HCF for equal grouping: When asked to divide several collections into the largest possible equal groups so that each group has the same composition, use HCF. Example: to split 48 apples and 72 oranges into equal fruit packs where each pack has same number of apples and same number of oranges, compute HCF(48,72) = 24. This gives 24 packs, each containing 2 apples and 3 oranges.

Using LCM for repeating events: When events repeat at different intervals and you want to know when they occur together, use LCM. Example: a bus every 15 minutes and a train every 20 minutes will come together after LCM(15,20)=60 minutes. Draw timelines for visual understanding.

Sharing equally: If asked how many each person gets when items are shared equally, divide total by number of people and check divisibility. If there is a remainder, some items will be left over. Use divisibility tests to see quickly whether equal sharing is possible without performing full division.

Problem-solving steps and tips: Read the problem carefully, identify numbers involved, decide whether the question asks about factors (use HCF), multiples (use LCM) or simple division, apply divisibility tests to speed checks, and show work. Encourage drawing boxes, rows or timelines to visualise grouping or cycles before calculating.

Why checking matters: Divisibility tests give quick checks to spot mistakes in arithmetic. Before writing final answers in exams, students should perform a fast divisibility check — it often reveals a misplaced digit or calculation error immediately.

Common shortcuts: Use last digit rules for 2, 5 and 10 to check parity and endings. Use digit-sum rules for 3 and 9 to check sums and products: the digit-sum of a product modulo 9 should match the product of the digit-sums modulo 9. For multiplication, checking the last digit of the product (units digit) is a fast sanity check: it must equal the product of the unit digits of the factors modulo 10.

Examples of checks: If you multiply 234×56 and get 13,104, check the last digit: 4×6 ends with 4 so product should end with 4 — it does. For addition, compute digit-sums of all addends and reduce modulo 9; the digit-sum of the total should match. For fractions, after simplification check that numerator and denominator are coprime (no common small prime factors remain).

Using divisibility to pick shortcuts: When simplifying or cancelling in fractions, try removing factors of 2 or 5 first if numbers end with even digits or 5/0. Use the relation HCF×LCM = product to find missing values quickly when two of the three are known. For repeated arithmetic tasks, build small mental tables of divisibility to speed work.

Teaching habit: Encourage students to write beside their answer which divisibility test or check they used. Practise problems where a correct result must also pass a given divisibility test, so checking becomes a natural part of problem solving.