Overview
This unit introduces students to numbers and basic ideas used when working with whole numbers. It covers how to find factors and multiples, recognise prime and composite numbers, and break numbers into prime factors. You will learn how to find the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two or more numbers and see how they are connected. The unit also teaches simple divisibility tests that help decide quickly whether one number divides another without long division. There are short methods and practical puzzles which develop number sense and speed in mental calculations. These skills matter because they are used in daily life and in higher mathematics: simplifying fractions, solving problems about grouping or sharing, scheduling events, and working with ratios. Practising this chapter builds careful reasoning, accuracy with arithmetic, and confidence in handling larger calculations. By the end of the unit, students will be able to solve many routine school problems, check results quickly using divisibility, and prepare well for topics like fractions, algebra and mensuration.
Learning Objectives
- Recognise and list factors and multiples of a given whole number.
- Identify prime and composite numbers and test primality for small numbers.
- Write the prime factorisation of a number using a factor tree or division method.
- Calculate the HCF and LCM of two or more numbers using prime factorisation and repeated division.
- Apply tests of divisibility for 2, 3, 5, 9 and 10 to check divisibility quickly.
- Use the relation between HCF and LCM to verify calculations and solve simple problems.
- Solve word problems that require finding HCF or LCM for grouping, sharing and scheduling.
- Explain basic properties of whole numbers such as even/odd and the arithmetic laws (commutative, associative, distributive).
Topics in this chapter
14 topics · tap a topic title to jump straight to it.
Natural numbers and whole numbers
What are natural and whole numbers?
Natural numbers are the counting numbers we use every day: 1, 2, 3, 4, and so on. Whole numbers are the natural numbers together with 0, so they form the list 0, 1, 2, 3, 4, ... . These sets form the foundation of arithmetic and most of the problems in this chapter use whole numbers.
Properties and operations
When you add or multiply two whole numbers, the result is always a whole number. This property is called closure under addition and multiplication. Subtraction of whole numbers may not always stay in whole numbers (for example, 3 − 5 = −2, which is not a whole number). Division also often leads outside whole numbers unless the divisor fits exactly into the dividend.
Even and odd numbers
Even numbers are whole numbers divisible by 2 and end with the digits 0, 2, 4, 6 or 8. Odd numbers are those not divisible by 2 and end with 1, 3, 5, 7 or 9. Knowing whether a number is even or odd helps predict the result of sums and products quickly: even + even = even, odd + odd = even, even + odd = odd; product with an even factor gives an even number.
Place value and number line
Whole numbers are shown on a number line starting from 0 at left and increasing to the right. Place value (units, tens, hundreds) tells us the value of each digit in a number. Together these ideas help compare numbers, order them, and perform operations like addition or subtraction using column methods.
Why learn these now?
Clear understanding of natural and whole numbers prepares you for factors and multiples, divisibility tests and for later work with integers, fractions and algebra. Practise by writing lists, ordering numbers, and working with simple operations to build confidence.
- The first five natural numbers are 1, 2, 3, 4, 5.
- Whole numbers up to 6 are 0, 1, 2, 3, 4, 5, 6.
- Natural numbers: N = {1, 2, 3, ...}
- Whole numbers: W = {0, 1, 2, 3, ...}
- Even number: n = 2k where k is an integer
- Odd number: n = 2k + 1 where k is an integer
Factors of a number
What is a factor?
A factor (or divisor) of a whole number is another whole number which divides it exactly leaving no remainder. If a × b = n then a and b are factors of n. Every number has at least two factors: 1 and itself. Factors come in pairs: for example, 12 has factor pairs (1,12), (2,6), (3,4). Finding factors helps when simplifying fractions, finding HCF and solving grouping problems.
How to find all factors
To find all factors of a number n, you can try dividing n by successive integers starting from 1 up to √n (square root of n). When you find a divisor d, include both d and n/d as factors. You only need to test up to √n because larger divisors pair with smaller ones already found. Write the list of factors in ascending order after collecting both members of each pair.
Quick methods and tips
Start testing with small primes 2, 3, 5, 7 because they often divide numbers. Use divisibility tests to reduce work: for example if the last digit is even, 2 is a factor; if digit sum divisible by 3 then 3 is a factor. If a number ends with 0 or 5, then 5 is a factor. Factor trees (breaking a number into two factors repeatedly) and the ladder (division) method are reliable when numbers get larger.
Special cases and notes
1 is a factor of every number, but it does not tell much about the number’s structure. Prime numbers only have two factors: 1 and itself. Composite numbers have more than two. Always check factor lists by multiplying pairs to return the original number; this avoids missing any factor.
Practice
- Find factor pairs and check both sides.
- Use a factor tree to reach prime ends.
- Remember to stop testing at √n to be efficient.
- Factors of 18: divide by 1,2,3,6 giving pairs (1,18), (2,9), (3,6); list: 1,2,3,6,9,18.
- Factors of 17: only (1,17); so 17 is prime.
- If a × b = n then a and b are factors of n
- To test all factors of n, check divisors up to √n
Multiples of a number
Definition of multiples
A multiple of a number n is any number obtained by multiplying n by a whole number (including 0). Thus the multiples of 4 are 0, 4, 8, 12, 16, ... because 4×0, 4×1, 4×2, 4×3, etc. Multiples of a number go on without end; there are infinitely many. Every number is a multiple of itself and 0 is a multiple of every number.
How to list multiples and common multiples
List multiples by repeatedly adding the number or by multiplying by 1,2,3,... For two numbers, common multiples are numbers that appear in both lists. For example multiples of 3 are 3,6,9,12,... and of 4 are 4,8,12,16,... so the common multiples start 12,24,36,... The smallest positive common multiple is the LCM (Least Common Multiple).
Uses of multiples
Multiples are helpful for many problems: scheduling events to occur together (like bells or buses), finding a common denominator for adding fractions, and solving puzzles about equal spacing. When arranging items in rows or columns, multiples tell possible total counts for complete rows.
Tests and shortcuts
To check if a number m is a multiple of n, divide m by n and see if remainder is zero. Using divisibility tests speeds up this check for small divisors: for 2,3,5,9 and 10 use the rules you will learn. For two numbers a and b, quick computation of their LCM (explained later) gives the smallest common multiple without listing many values.
Practice tips
- Write short lists of multiples and look for first common value to find LCM for small numbers.
- Use multiplication tables to remember multiples up to a point.
- Be careful with zero: 0 is a multiple but usually LCM is taken as positive and non-zero.
- Multiples of 5: 0, 5, 10, 15, 20, ...
- Common multiples of 6 and 8: 24, 48, 72, ...; LCM is 24.
- Multiples of n: n × 0, n × 1, n × 2, n × 3, ...
- A number m is a multiple of n if m mod n = 0
Prime and composite numbers
Definitions and quick recognition
A prime number is a whole number greater than 1 that has exactly two distinct factors: 1 and itself. Composite numbers are whole numbers greater than 1 with more than two factors. The number 1 is neither prime nor composite since it has only one factor. Recognising primes and composites is important for factorisation, HCF and LCM.
How to test a number for primality
To test whether a number n is prime, check whether any integer from 2 up to the square root of n divides it. If none divide n, then n is prime. This works because any factor larger than √n pairs with one smaller than √n, so a divisor must exist below or equal to √n if the number is composite. For small numbers, memorising primes up to 50 helps speed tests during exams.
Special notes
2 is the only even prime number; every other even number is composite because 2 divides it. Many small numbers have small prime divisors (2,3,5,7), so test these first using divisibility rules: last digit for 2 and 5, digit sum for 3 and 9. If a number fails these early tests, try larger divisors up to √n.
Why primes matter
Primes are the building blocks of whole numbers— every number greater than 1 can be written as product of primes (prime factorisation). Knowing primes helps simplify fractions, compute HCF and LCM, and reason about number properties. Identifying prime numbers also prepares students for more advanced topics in algebra and number theory.
Practice suggestions
- Memorise primes up to 50 and use quick tests for small divisors.
- Try factor trees to see when a composite number breaks into primes.
- Remember 1 is neither prime nor composite and treat it as a special case.
- 17 is prime because only 1 and 17 divide it.
- 12 is composite because it has factors 1,2,3,4,6,12.
- Prime: exactly two factors {1, p}
- Composite: more than two factors
Prime factorisation and factor tree
What is prime factorisation?
Prime factorisation is expressing a whole number as a product of prime numbers. Every integer greater than 1 can be written in only one way as a product of primes (order of factors does not matter). Prime factorisation reveals the prime building blocks of a number and is very useful in finding HCF, LCM and in simplifying fractions.
Methods: factor tree and division ladder
Factor tree: Start with the number and split it into any two factors. Continue factoring each composite factor until all leaves are prime. The primes at the ends multiplied together give the original number. Division (ladder) method: divide the number repeatedly by the smallest prime that goes into it, writing quotients under each division step, until the final quotient is 1. The primes used are the prime factors.
How to record powers
Often the same prime appears more than once. Write repeated primes using exponents: for example 60 = 2 × 2 × 3 × 5 = 2^2 × 3 × 5. When comparing prime factors for HCF or LCM you will use these exponent counts to choose the lowest or highest powers.
Common mistakes
Avoid stopping too early—ensure all final numbers in a factor tree are primes. If you split into 1 and the number, that does not help; use non-trivial factor pairs. When using division, always use prime divisors in increasing order for clarity (2,3,5,7,...).
Practice
- Try factor trees on several composite numbers and write prime factor form with exponents.
- Use prime factors to check HCF and LCM calculations later.
- Prime factorisation of 84: 84 = 2 × 42 = 2 × 2 × 21 = 2 × 2 × 3 × 7 = 2^2 × 3 × 7.
- Factor tree for 90 gives 90 = 2 × 45 = 2 × 3 × 15 = 2 × 3 × 3 × 5 = 2 × 3^2 × 5.
- Write n as product of primes: n = p1^a × p2^b × ... where p1,p2 are primes
- Read primes from factor tree or repeated division
Highest Common Factor (HCF) — concept and methods
Definition and meaning
The Highest Common Factor (HCF), also called the greatest common divisor (GCD), of two or more positive whole numbers is the largest whole number that divides each of them without leaving a remainder. If you want to split items into the largest equal groups without leftover, you use the HCF.
Methods to find HCF
1. Listing factors: This method suits small numbers. Write all factors of each number and choose the greatest one common to all lists. 2. Prime factorisation: Express each number as product of primes and compare prime powers; for each prime, take the smallest power present in all numbers. Multiply these prime powers to get HCF. 3. Repeated subtraction or division (Euclid’s idea): Repeatedly subtract the smaller number from the larger or use successive division until remainder becomes 0; the last non-zero remainder is the HCF. While Euclid’s algorithm is efficient, prime factor method is often easier to show in school answers.
Worked idea
Take 48 and 180. Prime factors: 48 = 2^4 × 3^1, 180 = 2^2 × 3^2 × 5^1. Common primes are 2 and 3; choose the lowest exponent for each (2^2 and 3^1), then multiply: HCF = 2^2 × 3 = 4 × 3 = 12. This value is the largest number dividing both exactly.
Applications
HCF is used to divide things into the largest possible equal groups (sweets into identical packets), reduce fractions to lowest terms by dividing numerator and denominator by their HCF, and solve problems in geometry where identical pieces are needed.
Tips
- For more than two numbers, find HCF pairwise or use common prime powers across all numbers.
- Check work by dividing original numbers by the HCF to ensure no remainder.
- HCF of 30 and 45: 30 = 2 × 3 × 5, 45 = 3 × 3 × 5; common primes 3 and 5 → HCF = 3 × 5 = 15.
- HCF of 14 and 49: factors of 14 (1,2,7,14) and 49 (1,7,49) → HCF = 7.
- HCF from prime factors: HCF = product of common primes with least powers
- \[If n = ∏ p_i^{a_i} and m = ∏ p_i^{b_i} then HCF = ∏ p_i^{min(a_i,b_i)}\]
Least Common Multiple (LCM) — concept and methods
Definition and uses
The Least Common Multiple (LCM) of two or more positive integers is the smallest positive integer that is a multiple of each of them. LCM is useful when two or more repeating events must be synchronised, or when finding a common denominator to add or compare fractions.
Methods to find LCM
1. Listing multiples: Write the multiples of each number until you find the smallest common one. This is straightforward for small numbers but becomes slow for larger ones. 2. Prime factorisation: Express each number as product of primes; for each distinct prime, take the highest power that appears in any factorisation. Multiply these chosen prime powers to get the LCM. 3. Using HCF: For two numbers a and b, you can use the relation LCM(a,b) = (a × b) / HCF(a,b). This is convenient when HCF is easy to compute.
Example using prime powers
Find LCM of 12 and 18. Factor: 12 = 2^2 × 3^1, 18 = 2^1 × 3^2. Take highest powers: 2^2 and 3^2. LCM = 2^2 × 3^2 = 4 × 9 = 36. If you listed multiples you would find 12's multiples: 12,24,36,... and 18's: 18,36,... so 36 is indeed the smallest common multiple.
Multiple numbers and care
When more than two numbers are involved, prime factor method is best: list prime powers for all numbers and choose the highest power of each prime across the list. Avoid multiplying numbers directly without reducing by common factors; otherwise you may get a larger number than necessary.
Practical tips
- Check by dividing the LCM by each original number to ensure integer quotients.
- Use the HCF–LCM relation to spot mistakes: a × b should equal HCF × LCM for two numbers.
- LCM of 4 and 6: 4 = 2^2, 6 = 2 × 3 → LCM = 2^2 × 3 = 12.
- LCM of 8, 9 and 6: 8 = 2^3, 9 = 3^2, 6 = 2 × 3 → LCM = 2^3 × 3^2 = 8 × 9 = 72.
- LCM from prime factors: LCM = product of all primes each to the highest power
- For two numbers a,b: LCM(a,b) = (a × b) / HCF(a,b)
Relation between HCF and LCM
The product relation
For two positive integers a and b there is a useful and often-used relation: a × b = HCF(a,b) × LCM(a,b). This equation means the product of the two numbers equals the product of their highest common factor and their least common multiple. It provides a way to check calculations and to find one unknown when the other three values are known.
Why it is true (idea)
Think in terms of prime factors. Suppose you write a and b as products of primes with exponents: a = ∏ p_i^{a_i} and b = ∏ p_i^{b_i}. For each prime p_i, the HCF uses min(a_i, b_i) and the LCM uses max(a_i, b_i). Multiplying HCF and LCM therefore gives p_i^{min(a_i,b_i) + max(a_i,b_i)} = p_i^{a_i + b_i}. But a × b is p_i^{a_i} × p_i^{b_i} = p_i^{a_i + b_i}. Hence a × b = HCF × LCM. This prime-exponent idea explains why the formula holds for pairs of numbers.
How to use it
If you know HCF and one number, you can find the LCM from LCM = (a × b) / HCF. It also serves as a quick check: after computing HCF and LCM independently, multiply them and verify the product equals a × b. If not, there is a mistake in one of the computations.
Limitations and extension
This exact product formula is valid for two numbers. For three or more numbers there is no simple single-product rule; you must use prime factor method to compute LCM and HCF among many numbers. In exam problems, the two-number relation often saves time and helps avoid long factor work.
Example and verification
Let a = 12 and b = 18: HCF = 6, LCM = 36. Multiply HCF and LCM: 6 × 36 = 216; a × b = 12 × 18 = 216. The equality holds, confirming our values.
- For a = 12, b = 18: a × b = 216. HCF = 6, LCM = 36 and 6 × 36 = 216.
- If HCF of 14 and 35 is 7 and product is 490 then LCM = 490 ÷ 7 = 70.
- For two positive integers a and b: a × b = HCF(a,b) × LCM(a,b)
- Thus LCM(a,b) = (a × b) / HCF(a,b)
Tests of divisibility
What is a divisibility test?
Divisibility tests are quick rules that tell whether a number is divisible by another without doing full division. They save time when finding factors, checking primes, or simplifying fractions. In this chapter we learn simple tests for 2, 3, 5, 9 and 10.
Common tests
Divisible by 2 if the last digit is 0,2,4,6,8. Divisible by 5 if the last digit is 0 or 5. Divisible by 10 if the last digit is 0. Divisible by 3 if the sum of digits is divisible by 3. Divisible by 9 if the sum of digits is divisible by 9. These rules work because of how our base-10 place-value system distributes powers of 10.
Using tests stepwise
When testing a number, start with easy checks: last digit rules for 2, 5 and 10 are immediate. For 3 and 9 add the digits; if the sum is large you may apply the test again to the sum. Combine tests for compound divisors: if a number passes tests for 3 and 5 it is divisible by 15. For 6 test divisibility by both 2 and 3.
Examples and practice
- Check 4,205: last digit 5 → divisible by 5 but not by 2 or 10.
- Check 3,726: sum 3+7+2+6 = 18 → 18 divisible by 3 and 9? 18 divisible by 3 and by 9? 18 is divisible by 9? No, 18 ÷ 9 = 2 so yes divisible by 9. So 3,726 divisible by both 3 and 9.
Where to be careful
These rules give yes/no answers for divisibility but not the quotient. For divisors like 7, 11 or 13 there are more complex tests taught later. Use these simple tests first to reduce work in larger factorisation tasks.
- Is 4,976 divisible by 2? Last digit 6 → yes, divisible by 2.
- Is 7,239 divisible by 3? Sum = 7+2+3+9=21, 21 divisible by 3 → yes.
- Divisible by 2 ⇔ last digit ∈ {0,2,4,6,8}
- Divisible by 5 ⇔ last digit ∈ {0,5}
- Divisible by 10 ⇔ last digit = 0
- Divisible by 3 ⇔ sum of digits divisible by 3
- Divisible by 9 ⇔ sum of digits divisible by 9
Division algorithm, quotient and remainder
Division statement
When a whole number (dividend) a is divided by a positive whole number (divisor) b, we obtain a quotient q and a remainder r that satisfy the equation a = b×q + r with 0 ≤ r < b. This is called the division algorithm. The quotient tells how many full groups of size b fit into a, and the remainder is what is left over.
Understanding remainder
If the remainder is zero then division is exact and b is a factor of a. If remainder is non-zero it must be less than the divisor. For instance, dividing 25 by 4 gives quotient 6 and remainder 1 because 25 = 4×6 + 1. If you ever find a remainder equal to or larger than the divisor, increase the quotient and reduce the remainder accordingly.
Long division and short division
Long division arranges digits in columns to divide large numbers step by step; short division uses mental division when divisors are small and digits are handled with carrying of remainders. Practice both so you can check answers in exams quickly using the division statement.
Applications
Division algorithm underlies tests of divisibility, HCF algorithms like Euclid's method, and many word problems about sharing or grouping. It is also used in converting improper fractions to mixed numbers and in modular arithmetic ideas introduced later.
Practice tips
- Always write the check a = bq + r to verify your division.
- Use divisibility tests to find if remainder will be zero before performing full division.
- Remember remainder must be less than divisor; if not, adjust quotient.
- Divide 23 by 4: quotient 5, remainder 3 because 4×5+3=23.
- Divide 100 by 10: quotient 10, remainder 0 because 10×10+0=100.
- For integers a (dividend) and b (positive divisor): a = bq + r where 0 ≤ r < b
- Remainder r is less than divisor b
Using HCF in division and grouping problems
Problem type and goal
Many problems ask you to divide objects into the largest possible equal groups without any leftover. Other problems ask for the largest size of pieces when cutting lengths into equal parts. These are solved by finding the HCF of the given quantities. HCF gives the greatest number that divides all the given amounts exactly, so it tells the maximum size of each group or piece.
How to approach such problems
Step 1: Identify the numbers that must be divided equally (counts, lengths, or weights). Step 2: Compute their HCF using prime factorisation or listing factors. Step 3: Divide each original number by the HCF to find how many groups or pieces result. Always give the size of each group (the HCF) and the number of groups for each quantity if the question asks.
Worked examples and reasoning
Example: three ropes of lengths 24 m, 36 m and 60 m are to be cut into equal longest pieces. Compute prime factors: 24 = 2^3 × 3, 36 = 2^2 × 3^2, 60 = 2^2 × 3 × 5. Common prime powers are 2^2 and 3^1 so HCF = 4×3 = 12. Each piece is 12 m long. Another example: dividing 18, 30 and 42 sweets into maximum equal packets—HCF is 6 so packets of 6 sweets.
Checking and common mistakes
After finding HCF, check that it divides each number exactly. Do not confuse HCF with LCM—HCF gives the largest equal share size, while LCM gives the smallest common multiple. For more than two numbers, use prime factorisation across all numbers to pick the smallest power for each common prime.
Practical uses
- Sharing sweets equally into largest identical packets, cutting ribbons into longest equal pieces, or making identical groups for decoration.
- Always state both the size of each group (HCF) and how many groups each original quantity yields.
- Three children have 18, 30 and 42 apples and want identical baskets with no apple leftover; HCF(18,30,42)=6 so each basket has 6 apples.
- Lengths 36 cm and 54 cm are cut into equal longest pieces: HCF(36,54)=18 cm per piece.
- Use HCF to find largest equal group size: group size = HCF(numbers)
- Number of groups for each = original number ÷ HCF
Using LCM in scheduling and periodic events
Typical problems
Problems that ask when repeating events will happen together involve LCM. For example, if two bells ring every 6 and 8 minutes we ask when they will next ring at the same time. LCM gives the smallest positive time interval after which all given cycles coincide.
Steps to solve timing problems
Step 1: Make sure the units (seconds, minutes, hours, days) are the same for all cycles. Step 2: Compute the LCM of the cycle lengths using prime factorisation or listing multiples. Step 3: The first common time after the start is the LCM; subsequent common times occur at multiples of the LCM. If events start at non-zero offsets, find times from each start and compare or shift cycles accordingly.
Worked examples and explanation
If lights flash every 9 seconds and 12 seconds, LCM(9,12)=36 seconds so they flash together at 36s, 72s, 108s, ... If two buses come every 12 and 15 minutes and both arrive at 9:00, LCM(12,15)=60 minutes, so they next arrive together at 10:00. When start times differ, convert to times from a common zero and check the first matching value that is ≥ both start times.
Complex cases
For three or more cycles use prime factor method: write prime powers for each interval and take the highest power of each prime. This avoids long lists. Also check answers by dividing the LCM by each interval to ensure integer multiples. Where problems ask how many times events coincide in a time period, divide the period length by the LCM to get the count.
Practical uses
- Planning schedules so tasks align, traffic signal cycles, festival timings, and repeating maintenance schedules.
- Remember to express final answer clearly with units and the actual time (e.g., 9:36 AM).
- Two lights flash every 6 s and 10 s. LCM(6,10)=30 s → they flash together every 30 seconds.
- Buses arrive every 12 and 15 minutes; LCM=60 minutes means they arrive together after 1 hour.
- To find when events coincide: time = LCM(intervals)
- For two numbers a,b: LCM(a,b) = (a × b)/HCF(a,b)
Number properties and puzzles (even/odd, commutative, associative, distributive)
Even and odd rules
Even numbers are divisible by 2; odd numbers are not. Key rules: even + even = even, odd + odd = even, even + odd = odd. For multiplication, if any factor is even, the product is even. These simple parity rules help solve many short puzzles about digits and sums without full calculation.
Arithmetic laws
Commutative law: the order of addition or multiplication does not change the result (a + b = b + a, a × b = b × a). Associative law: grouping of addends or factors does not change the result ((a + b) + c = a + (b + c)). Distributive law: multiplication distributes over addition (a × (b + c) = a × b + a × c). These laws allow us to rearrange calculations for ease and to simplify expressions mentally.
Puzzles and problem solving
Puzzles often ask for the smallest or largest number meeting certain digit conditions, or to find digits satisfying divisibility. Use laws to rearrange sums to make tens and use parity to reduce cases. For example, to check if the sum of three numbers is odd or even, use parity rules rather than adding exactly. For digit puzzles use place-value understanding together with divisibility tests.
Practice tips
- Rearrange addends to form tens quickly when adding many numbers (associative law).
- Use distributive law to simplify multiplication like 12×15 = (10+2)×15 = 150+30 = 180.
- Apply parity to rule out impossible answers in multiple choice questions.
Link to future topics
These properties are the foundation for algebra and higher arithmetic. Practising them now builds fluency and speed for exam problems and mental calculations.
- Sum of 7 + 9 + 4 = (7 + 9) + 4 or (7 + 4) + 9 → both give 20 by associative law.
- 2 × (3 + 5) = 2×3 + 2×5 = 6 + 10 = 16 by distributive law.
- Commutative: a + b = b + a, a × b = b × a
- Associative: (a + b) + c = a + (b + c), (a × b) × c = a × (b × c)
- Distributive: a × (b + c) = a × b + a × c
Mixed practice: combining methods in problems
Why mixed practice matters
Real exam problems rarely require only one small idea. They often combine divisibility tests, prime factorisation, HCF and LCM methods, and arithmetic properties. This topic trains you to choose the best tool and use methods together efficiently to reach the answer.
General approach
1. Read the problem carefully and identify the numbers involved and what is asked (HCF, LCM, grouping, timing, simplification). 2. Decide the quickest method: for small numbers listing factors or multiples may be enough; for larger numbers, use prime factorisation. 3. Use divisibility tests early to reduce work: check for small prime divisors before factoring. 4. Solve step by step and always check via multiplication or the division statement.
Common combined problem types
Problems include splitting objects into largest equal groups (HCF), finding the next common event time (LCM), simplifying fractions (divide numerator and denominator by HCF), and finding the smallest number divisible by many numbers (LCM). Some questions may ask to show work: give prime factors, highlight common primes for HCF and highest primes for LCM, and compute the result.
Worked strategy examples
Example 1: Simplify 90/126. Prime factors: 90=2×3^2×5, 126=2×3^2×7. Common primes 2 and 3^2 give HCF 18; divide numerator and denominator by 18 to get 5/7. Example 2: Smallest number divisible by 6, 10 and 15: prime factors 6=2×3, 10=2×5, 15=3×5; take highest powers 2×3×5=30.
Exam tips
- Write clear steps: factorisation, selection of common primes, HCF/LCM computation and check.
- If unsure, test small numbers or use the HCF×LCM check for pairs.
- Practice many mixed problems to recognise which method is faster in each case.
- Find smallest number divisible by 6 and 15 and 10: compute LCM(6,15,10)=30.
- Simplify 90/126: prime factors 90=2×3^2×5, 126=2×3^2×7 ⇒ HCF=2×3^2=18 ⇒ simplified 90/126=5/7.
- Use prime factors for HCF and LCM in combined problems
- Simplify fraction a/b by dividing numerator and denominator by HCF(a,b)
Key Concepts
- Natural numbers
- Counting numbers 1,2,3,... used for counting objects.
- Whole numbers
- Numbers including 0 and natural numbers: 0,1,2,3,...
- Factor
- A whole number that divides another exactly with zero remainder.
- Multiple
- A number obtained by multiplying a given number by an integer.
- Prime number
- A number greater than 1 having exactly two factors: 1 and itself.
- Composite number
- A number greater than 1 that has more than two factors.
- Prime factorisation
- Writing a number as a product of prime numbers.
- HCF (Highest Common Factor)
- The largest number that divides two or more numbers exactly.
- LCM (Least Common Multiple)
- The smallest positive integer that is a multiple of given numbers.
- Divisibility test
- A quick rule to check whether one number divides another without full division.
- Quotient and remainder
- Result and leftover when one number is divided by another.
- Even and odd
- Even numbers are divisible by 2; odd numbers are not.
- Commutative law
- The order of addition or multiplication does not change the result.
- Associative law
- Grouping of addition or multiplication does not change the result.
- Distributive law
- Multiplication distributes over addition: a(b+c)=ab+ac.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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List all factors of 28. / 28 के सभी गुणक (फैक्टर्स) बताइए।
Show answer
Factors of 28 are 1, 2, 4, 7, 14, 28. / 28 के गुणक हैं 1, 2, 4, 7, 14, 28।
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Give three multiples of 7. / 7 के तीन गुणनफल दीजिए।
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Three multiples of 7 are 7, 14 and 21. / 7 के तीन गुणनफल हैं 7, 14 और 21।
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State whether 29 is prime or composite and justify. / बताइए कि 29 अभाज्य है या समघटी और कारण बताइए।
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29 is prime because no integer from 2 to 5 divides it evenly. / 29 अभाज्य है क्योंकि 2 से 5 तक कोई भी पूर्णांक इसे पूर्ण रूप से नहीं विभाजित कर सकता।
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Find the prime factorisation of 90. / 90 का अभाज्य गुणनखंडन (प्राइम फैक्टराइज़ेशन) कीजिए।
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Prime factorisation of 90 is 2 × 3^2 × 5. / 90 का अभाज्य गुणनखंडन है 2 × 3^2 × 5।
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Find HCF of 36 and 60 by prime factorisation. / अभाज्य गुणनखंडन से 36 और 60 का HCF बताइए।
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36 = 2^2 × 3^2, 60 = 2^2 × 3 × 5. Common primes with least powers: 2^2 and 3^1. HCF = 4 × 3 = 12. / 36 = 2^2 × 3^2, 60 = 2^2 × 3 × 5. सामान्य अभाज्य और न्यूनतम घात 2^2 और 3^1 → HCF = 4 × 3 = 12।
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Find LCM of 8 and 12. / 8 और 12 का LCM निकालिए।
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8 = 2^3, 12 = 2^2 × 3. Take highest powers 2^3 and 3 so LCM = 2^3 × 3 = 8 × 3 = 24. / 8 = 2^3, 12 = 2^2 × 3. उच्चतम घातों के अनुसार LCM = 2^3 × 3 = 24।
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Check the relation a × b = HCF × LCM for a = 14 and b = 21. / a = 14 तथा b = 21 के लिए a × b = HCF × LCM समीकरण सत्यापित कीजिए।
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HCF(14,21)=7, LCM(14,21)=42. a × b = 14 × 21 = 294. HCF × LCM = 7 × 42 = 294. Relation holds. / HCF(14,21)=7, LCM(14,21)=42. a × b = 14 × 21 = 294. HCF × LCM = 7 × 42 = 294. सम्बन्ध सत्य है।
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Is 4,605 divisible by 9? Show working. / क्या 4,605 को 9 से विभाजित किया जा सकता है? काम दिखाइए।
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Sum of digits = 4 + 6 + 0 + 5 = 15. 15 is not divisible by 9, so 4,605 is not divisible by 9. / अंकों का योग = 4+6+0+5 = 15. 15 को 9 से विभाजित नहीं किया जा सकता, अतः 4,605 9 से विभाज्य नहीं है।
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Divide 97 by 8. State quotient and remainder. / 97 को 8 से भाग दीजिए। भागफल और शेष बताइए।
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97 ÷ 8 gives quotient 12 and remainder 1 because 8 × 12 + 1 = 97. / 97 ÷ 8 का भागफल 12 और शेष 1 है क्योंकि 8 × 12 + 1 = 97।
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Three ropes measure 24 m, 36 m and 60 m. Cut them into equal longest pieces. What is the length of each piece? / तीन रस्सियाँ 24 मी., 36 मी., और 60 मी. लम्बी हैं। उन्हें बराबर और संभवत: सबसे लंबी टुकड़ों में काटना है। प्रत्येक टुकड़े की लम्बाई क्या होगी?
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Required length = HCF(24,36,60). Prime factors: 24=2^3×3, 36=2^2×3^2, 60=2^2×3×5. Common minimum powers: 2^2×3 = 4×3 = 12. So each piece is 12 m. / आवश्यक लम्बाई = HCF(24,36,60). 24=2^3×3, 36=2^2×3^2, 60=2^2×3×5. सामान्य न्यूनतम घात = 2^2×3 = 12. अतः प्रत्येक टुकड़ा 12 मी. होगा।
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Two bells ring every 9 minutes and 12 minutes. When will they ring together next after 9:00 AM? / दो घंटियाँ हर 9 मिनट और 12 मिनट पर बजती हैं। वे सुबह 9:00 बजे के बाद अगली बार कब साथ बजेंगी?
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LCM(9,12)=36 minutes. So after 9:00 AM they ring together at 9:36 AM. / LCM(9,12)=36 मिनट। अतः 9:00 के बाद वे 9:36 बजे साथ बजेंगे।
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