Real Numbers Class 10: Concepts, Formulas and Proofs

What Real Numbers is about

Weightage: This chapter is Unit I, Number Systems, which carries 6 of the 80 theory marks.Source: CBSE Mathematics syllabus 2026-27, Unit I Number Systems

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper. See the questions

Key concepts

The Fundamental Theorem of Arithmetic

NCERT section 1.2 The Fundamental Theorem of Arithmetic

A prime number has exactly two factors, 1 and itself. A composite number has more than two. The number 1 is neither prime nor composite.

The Fundamental Theorem of Arithmetic says: every composite number can be written as a product of primes, and this factorisation is unique, apart from the order of the factors.

Why must a factorisation exist? Keep splitting a composite number into two smaller factors. The numbers keep getting smaller, so you must eventually reach factors that cannot be split. Those are primes. Uniqueness means that however you start, say a factor tree from $2\times 180$ or from $12\times 30$, you always finish with the same primes and the same powers: $360=2^3\times 3^2\times 5$.

Use this whenever you need a number's building blocks: HCF, LCM, last digits or divisibility. Always write the final answer in power form, primes in increasing order, with no composite number left inside.

Worked example

If $1350=2^x\times 3^y\times 5^z$, where x, y and z are whole numbers, then $x+y+z$ equals
(a) 5
(b) 6
(c) 7
(d) 8

Try it first, then show the solutionSolution

1. $1350=2\times 675$
2. $675=3\times 225=3\times 3\times 75$
3. $75=3\times 25=3\times 5\times 5$
4. So $1350=2^1\times 3^3\times 5^2$, giving $x=1$, $y=3$, $z=2$ (unique by the Fundamental Theorem of Arithmetic)
5. Check: $2\times 27\times 25=2\times 675=1350$
6. $x+y+z=1+3+2=6$

Answer: (b) 6

Using uniqueness: can a number end in 0?

NCERT section 1.2 The Fundamental Theorem of Arithmetic

A number ends in the digit 0 exactly when it is divisible by 10, that is, when its prime factorisation contains both 2 and 5.

Now think about a power $a^n$. Its prime factors are exactly the primes of $a$, only with larger powers. Raising a number to a power can never create a new prime.

So, to decide whether $a^n$ can end in 0, factorise $a$. If 2 or 5 is missing from $a$, it is missing from $a^n$ too. Because the factorisation is unique, there is no other way of writing $a^n$ that could bring the missing prime in. So $a^n$ never ends in 0.

The same reasoning answers questions like whether $4^n$ can end in 5 (no, 5 is not a factor of 4) or whether $6^n$ can ever be odd (no, 2 always divides it). In your final line, name the Fundamental Theorem of Arithmetic as the reason.

Worked example

Riya says that for some natural number n, $15^n$ ends with the digit 0, because 15 is a multiple of 5. Is she correct? Justify your answer using the Fundamental Theorem of Arithmetic.

Try it first, then show the solutionSolution

1. A number ending in 0 is divisible by $10=2\times 5$, so its prime factorisation must contain both 2 and 5.
2. $15=3\times 5$, so $15^n=3^n\times 5^n$.
3. The only primes in the factorisation of $15^n$ are 3 and 5; the prime 2 does not appear.
4. By the Fundamental Theorem of Arithmetic this factorisation is unique, so no other factorisation of $15^n$ can contain 2.
5. Hence $15^n$ is not divisible by 2, so it is not divisible by 10.

Answer: Riya is wrong. $15^n$ never ends in 0 for any natural number n. (It is odd and a multiple of 5, so it always ends in 5.)

Showing that an expression is composite

NCERT section 1.2 The Fundamental Theorem of Arithmetic

A common board question asks whether an expression like $3\times 5\times 17+17$ is prime or composite. Do not multiply everything out and then hunt for factors. That wastes time and invites errors.

Instead, look for a common factor in all the terms and take it out. If you can write the number as a product of two natural numbers, each greater than 1, it is composite by definition. The common factor is one factor, and the bracket left behind is the other. Just check that the bracket is bigger than 1, which it clearly is when it looks like $(\cdots +1)$.

This connects to the Fundamental Theorem of Arithmetic: a prime has no factorisation other than itself. As soon as you show two factors greater than 1, the number cannot be prime. If the bracket simplifies to a prime, you have also found the full prime factorisation.

Worked example

Show that $2\times 3\times 5\times 7\times 11+7\times 11$ is a composite number, and write its prime factorisation.

Try it first, then show the solutionSolution

1. Both terms contain the factor $7\times 11$.
2. $2\times 3\times 5\times 7\times 11+7\times 11=7\times 11\times (2\times 3\times 5+1)$
3. $=7\times 11\times (30+1)=7\times 11\times 31$
4. The number has factors other than 1 and itself (for example 7), so it is composite.
5. 7, 11 and 31 are all prime, so this is its prime factorisation; value $=77\times 31=2387$.
6. Check: $2310+77=2387$

Answer: The number is composite; $2387=7\times 11\times 31$.

HCF and LCM by prime factorisation

NCERT section 1.2 The Fundamental Theorem of Arithmetic

Factorise each number into primes first. Then:

• HCF = product of the common primes, each to its smallest power.
• LCM = product of all the primes that appear, each to its greatest power.

Why? The HCF must divide every number, so it can only use primes that all the numbers share. It also cannot use a higher power than the smallest one present, or it would fail to divide that number. The LCM must be a multiple of every number, so it needs every prime that appears anywhere, at the highest power needed by any of the numbers.

This method works for two, three or more numbers. Set out each factorisation on its own line in power form, so the examiner can see where each power came from. Then check: the HCF should divide each number and each number should divide the LCM.

Worked example

Find the HCF and LCM of 72, 120 and 180 by prime factorisation. Then show that HCF × LCM is not equal to the product of the three numbers.

Try it first, then show the solutionSolution

1. $72=2^3\times 3^2$
2. $120=2^3\times 3\times 5$
3. $180=2^2\times 3^2\times 5$
4. HCF: common primes are 2 and 3, smallest powers $2^2$ and $3^1$, so HCF $=4\times 3=12$
5. LCM: primes 2, 3, 5 with greatest powers, so LCM $=2^3\times 3^2\times 5=360$
6. Check: 12 divides 72, 120 and 180; 360 is divisible by 72, 120 and 180.
7. HCF × LCM $=12\times 360=4320$
8. Product $=72\times 120\times 180=8640\times 180=1555200$, which is not 4320.

Answer: HCF = 12, LCM = 360. HCF × LCM = 4320, which is not equal to the product 1555200, so the product rule fails for three numbers.

HCF × LCM = product of two numbers

NCERT section 1.2 The Fundamental Theorem of Arithmetic

For any two natural numbers a and b: $\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$.

Why it works: take any prime. Say its power is m in a and n in b. The HCF takes the smaller of m and n, the LCM takes the larger. Smaller plus larger equals $m+n$, which is exactly the power of that prime in $a\times b$. This happens for every prime, so the two products match.

Use it to find a missing number when you know the HCF, the LCM and the other number. It holds only for two numbers; with three numbers, a shared prime gets counted the wrong number of times.

Two quick checks follow. The HCF always divides the LCM, so HCF 18 with LCM 380 is impossible. And if two numbers are in the ratio $m:n$ with m and n co-prime, write them as $mk$ and $nk$: then HCF $=k$ and LCM $=mnk$.

Worked example

Two numbers are in the ratio $3:5$ and their LCM is 360. Find their HCF and the two numbers. Verify that HCF × LCM equals the product of the numbers.

Try it first, then show the solutionSolution

1. Let the numbers be $3k$ and $5k$, where k is a natural number.
2. Since 3 and 5 are co-prime, HCF$(3k,5k)=k$ and LCM$(3k,5k)=3\times 5\times k=15k$.
3. Given LCM = 360, so $15k=360$, giving $k=24$.
4. The numbers are $3\times 24=72$ and $5\times 24=120$, and HCF = 24.
5. Check: $72=2^3\times 3^2$, $120=2^3\times 3\times 5$; HCF $=2^3\times 3=24$, LCM $=2^3\times 3^2\times 5=360$.
6. Verify: HCF × LCM $=24\times 360=8640$ and $72\times 120=8640$.

Answer: HCF = 24; the numbers are 72 and 120. Both products equal 8640.

Real-life problems: choosing HCF or LCM

NCERT section 1.2 The Fundamental Theorem of Arithmetic

Many board questions hide HCF or LCM inside a story. Decide which one by asking what the answer must do.

• If you are splitting quantities into equal groups, pieces, rows or bags, and want the greatest size, the size must divide every quantity. Use HCF.
• If events repeat at different intervals and you want when they coincide again, or the least quantity that every group size divides, use LCM.

Remainders need one extra step. If a number must leave remainder r when it divides x and y, then it divides $x-r$ and $y-r$ exactly, so find the HCF of those. The answer must be greater than each remainder; always check this at the end. If the least number must leave remainder r when divided by a, b and c, find the LCM and add r.

Sanity check: an HCF answer is never bigger than the smallest quantity; an LCM answer is never smaller than the largest. Finish with units and a full sentence.

Worked example

A shopkeeper has 245 red marbles and 1029 blue marbles. He keeps 5 marbles of each colour aside. He packs the rest into bags so that every bag holds marbles of one colour only, every bag holds the same number of marbles, and no marble is left over. Find the greatest possible number of marbles in each bag and the number of bags he needs.

Try it first, then show the solutionSolution

1. Marbles to be packed: red $245-5=240$, blue $1029-5=1024$.
2. The bag size must divide both 240 and 1024 exactly, and we want the greatest such number, so we need HCF(240, 1024).
3. $240=2^4\times 3\times 5$
4. $1024=2^{10}$
5. The only common prime is 2, smallest power $2^4$, so HCF $=16$.
6. Red bags $=240÷16=15$; blue bags $=1024÷16=64$; total $=15+64=79$.
7. Check: $245=16\times 15+5$ and $1029=16\times 64+5$.

Answer: Each bag holds 16 marbles, and he needs 79 bags.

Proving that √2, √3 and √5 are irrational

NCERT section 1.3 Revisiting Irrational Numbers

A number is irrational if it cannot be written as $\frac{p}{q}$ with p, q integers and $q\ne 0$.

The key tool is a theorem: **if a prime p divides $a^2$, then p divides a** (a a positive integer). It follows from unique factorisation: the primes of $a^2$ are exactly the primes of a, each doubled. So if p appears in $a^2$, it must already appear in a.

The proof is by contradiction. Assume $\sqrt{p}$ is rational and write it as $\frac{a}{b}$ in lowest terms, so a and b are co-prime. Squaring gives $a^2=p{b}^2$, so p divides $a^2$, hence p divides a. Substituting $a=pc$ shows p divides b too. Then p is a common factor of a and b, contradicting co-primeness.

The same steps work for $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ and any prime. They fail for $\sqrt{4}$, because 4 is not prime and $\sqrt{4}=2$ is rational.

Worked example

Prove that $\sqrt{3}$ is irrational.

Try it first, then show the solutionSolution

1. Assume, to the contrary, that $\sqrt{3}$ is rational.
2. Then $\sqrt{3}=\frac{a}{b}$, where a and b are co-prime integers and $b\ne 0$.
3. Squaring: $3=\frac{a^2}{b^2}$, so $a^2=3b^2$.
4. So 3 divides $a^2$. Since 3 is prime, 3 divides a (if a prime divides $a^2$, it divides a).
5. Write $a=3c$ for some integer c. Then $9c^2=3b^2$, so $b^2=3c^2$.
6. So 3 divides $b^2$, and hence 3 divides b.
7. Now 3 divides both a and b, which contradicts the fact that a and b are co-prime.
8. So our assumption was wrong.

Answer: $\sqrt{3}$ is irrational.

Proving expressions like 3 + 2√5 are irrational

NCERT section 1.3 Revisiting Irrational Numbers

Once you know $\sqrt{5}$ is irrational, you can prove that expressions such as $3+2\sqrt{5}$ or $4-5\sqrt{3}$ are irrational.

Method: assume the expression equals $\frac{a}{b}$ with a, b co-prime integers and $b\ne 0$. Rearrange to get the surd alone on one side. The other side is built from integers by adding, subtracting, multiplying and dividing (never by zero), so it is rational. That makes the surd rational, which contradicts the known fact.

The general results behind this are: rational + irrational is irrational, and non-zero rational × irrational is irrational. But when a question says prove, write the full contradiction argument; do not just quote the rule.

Be careful: two irrationals can combine to give a rational. For example $\sqrt{2}\times \sqrt{2}=2$ and $(3+\sqrt{5})+(3-\sqrt{5})=6$. This is a favourite trap in assertion-reason questions.

Worked example

Given that $\sqrt{3}$ is irrational, prove that $4-5\sqrt{3}$ is irrational.

Try it first, then show the solutionSolution

1. Assume, to the contrary, that $4-5\sqrt{3}$ is rational.
2. Then $4-5\sqrt{3}=\frac{a}{b}$, where a and b are co-prime integers and $b\ne 0$.
3. Rearranging: $5\sqrt{3}=4-\frac{a}{b}=\frac{4b-a}{b}$
4. So $\sqrt{3}=\frac{4b-a}{5b}$
5. Since a and b are integers and $b\ne 0$, $4b-a$ and $5b$ are integers with $5b\ne 0$, so $\frac{4b-a}{5b}$ is rational.
6. Then $\sqrt{3}$ is rational, which contradicts the given fact that $\sqrt{3}$ is irrational.
7. So our assumption was wrong.

Answer: $4-5\sqrt{3}$ is irrational.

Maths in real life: Real Numbers

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Fundamental Theorem of Arithmetic
• HCF and LCM by Prime Factorisation
• Proving Numbers Irrational

Real Numbers formula sheet

Fundamental Theorem of Arithmetic

$n=p_1^{a_1}\times p_2^{a_2}\times \cdots \times p_k^{a_k},p_1<p_2<\cdots <p_k\text{ primes (unique)}$

Every composite number factorises into primes in exactly one way. Quote it for last-digit and divisibility arguments.

Prime divides a square

$p\text{ prime},p\mid a^2\Rightarrow p\mid a$

The key step in every proof that a square root of a prime is irrational. State it; you do not need to prove it.

Rational number

$\frac{p}{q},p,q\text{ integers},q\ne 0$

In proofs, take p and q co-prime (fraction in lowest terms). A number that cannot be written this way is irrational.

HCF by prime factorisation

$\text{HCF}=\text{product of common primes, each to its smallest power}$

Works for any number of numbers. HCF divides every given number.

LCM by prime factorisation

$\text{LCM}=\text{product of all primes present, each to its greatest power}$

Works for any number of numbers. Every given number divides the LCM.

Product relation (two numbers only)

$\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$

Use to find a missing number: other number = (HCF × LCM) ÷ known number. Never use for three numbers.

Three numbers

$\text{HCF}(a,b,c)\times \text{LCM}(a,b,c)\ne a\times b\times c\text{ in general}$

For three numbers always use prime factorisation directly.

Size and divisibility checks

$\text{HCF}\mid \text{LCM},\text{HCF}\le \text{smallest number},\text{LCM}\ge \text{largest number}$

Quick way to reject impossible HCF-LCM pairs and catch errors.

Co-prime numbers

$\text{HCF}(a,b)=1\Rightarrow \text{LCM}(a,b)=a\times b$

Two consecutive natural numbers are always co-prime.

One number divides the other

$a\mid b\Rightarrow \text{HCF}(a,b)=a,\text{LCM}(a,b)=b$

Saves time in MCQs, e.g. HCF and LCM of 15 and 45.

Numbers in a co-prime ratio

$\text{numbers }mk,nk(\text{HCF}(m,n)=1)\Rightarrow \text{HCF}=k,\text{LCM}=mnk$

Use when the ratio and the HCF or LCM are given.

Ending in zero

$n\text{ ends in }0⟺2\mid n\text{ and }5\mid n$

A power a to the n has only the primes of a, so check whether a contains both 2 and 5.

Greatest divisor with remainders

$\text{greatest number}=\text{HCF}(x-r_1,y-r_2)$

When the number divides x and y leaving remainders r1 and r2. The answer must be greater than each remainder; always check this at the end.

Least number with the same remainder

$\text{least number}=\text{LCM}(a,b,c)+r$

When the number, divided by a, b and c, leaves remainder r each time (r smaller than each divisor).

Rational and irrational combinations

$r+s\sqrt{p}\text{ is irrational for rationals }r,s(s\ne 0),p\text{ prime}$

Rational + irrational is irrational; non-zero rational × irrational is irrational. Two irrationals may give a rational.

How the board asks Real Numbers

The question patterns to expect, and the method that works for each.

Tricks to remember Real Numbers

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Low-Common, High-All

  HCF takes the LOW powers of the COMMON primes. LCM takes the HIGH powers of ALL the primes.

  Use it when: Say it to yourself before writing any HCF or LCM from factorisations, so you never swap the rules.

• Split or Sync

  Splitting into the largest equal parts means HCF. Syncing events or finding the least common amount means LCM.

  Use it when: Decide the operation in word problems before you start calculating.

• The sandwich check

  HCF ≤ smallest number ≤ largest number ≤ LCM, and the HCF must divide the LCM.

  Use it when: A 10-second check on every HCF and LCM answer, and a quick way to spot impossible pairs in MCQs.

• Proof skeleton: A-C-S-D-S-D-C

  Assume rational, Co-prime a and b, Square, Divides a, Substitute a = pc, Divides b, Contradiction.

  Use it when: Write the irrationality proof in this order so no mark-earning step is skipped.

• Zero needs a pair

  To end in 0, a number needs both a 2 and a 5 in its prime factorisation. Powers never add new primes.

  Use it when: Instant answer to any question asking whether a power can end in 0.

• Surd alone, then judge

  In proofs for expressions like $3+2\sqrt{5}$, move everything except the root to the other side, then say that side is rational.

  Use it when: Keeps the expression proofs short and clear.

Real Numbers: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Real Numbers

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using HCF × LCM = product for three numbers

  The relation is learnt as a formula without the condition that it works only for two numbers.

  Wrong working

  HCF(6, 10, 15) = 1, so LCM $=\frac{6\times 10\times 15}{1}=900$

  Right working

  $6=2\times 3$, $10=2\times 5$, $15=3\times 5$, so LCM $=2\times 3\times 5=30$

  ✓Do this: For three or more numbers, always find the LCM by prime factorisation. Use the product rule only for exactly two numbers.

  Usually costs: The whole LCM part, usually 1 to 2 marks.

• ✗Mistake: Swapping the HCF and LCM rules

  Students remember that powers matter but mix up smallest and greatest under exam pressure.

  Wrong working

  $24=2^3\times 3$, $36=2^2\times 3^2$, so HCF $=2^3\times 3^2=72$

  Right working

  HCF $=2^2\times 3=12$ and LCM $=2^3\times 3^2=72$; check $12\times 72=864=24\times 36$

  ✓Do this: Use Low-Common, High-All, then run the sandwich check: the HCF can never exceed the smaller number.

  Usually costs: ½ to 1 mark in a 2-mark question, plus any later parts that use the wrong value.

• ✗Mistake: Leaving out 'a and b are co-prime' in an irrationality proof

  Students write $\frac{a}{b}$ without saying it is in lowest terms, so the final contradiction has nothing to contradict.

  Wrong working

  Let $\sqrt{2}=\frac{a}{b}$. Then $a^2=2b^2$ ... so 2 divides a and b. This is a contradiction.

  Right working

  Let $\sqrt{2}=\frac{a}{b}$, where a and b are co-prime integers and $b\ne 0$. ... So 2 divides both a and b, which contradicts that a and b are co-prime.

  ✓Do this: Write the full assumption line every time, and name exactly what is contradicted in the final line.

  Usually costs: Usually ½ to 1 mark.

• ✗Mistake: Not using the prime-divides-square theorem correctly

  Students write the square as a multiple of 3 instead of the number itself, and the proof goes in circles.

  Wrong working

  Since 3 divides $a^2$, write $a^2=3c$. Then $3c=3b^2$, so $b^2=c$.

  Right working

  Since 3 divides $a^2$ and 3 is prime, 3 divides a. Write $a=3c$. Then $9c^2=3b^2$, so $b^2=3c^2$, and 3 divides b.

  ✓Do this: Always go from 'p divides $a^2$' to 'p divides a', quoting the theorem, then substitute $a=pc$.

  Usually costs: 1 to 1½ marks of a 3-mark proof.

• ✗Mistake: Quoting a rule instead of proving

  Students know rational + irrational is irrational and think stating it is enough.

  Wrong working

  3 is rational and $2\sqrt{5}$ is irrational, so $3+2\sqrt{5}$ is irrational.

  Right working

  Assume $3+2\sqrt{5}=\frac{a}{b}$ (a, b co-prime, $b\ne 0$). Then $\sqrt{5}=\frac{a-3b}{2b}$, which is rational, contradicting that $\sqrt{5}$ is irrational.

  ✓Do this: When the question says prove or show, use the contradiction method with every step written.

  Usually costs: Most of the marks for that part; often only ½ to 1 mark is given.

• ✗Mistake: Ignoring remainders in HCF word problems

  Students take the HCF of the given numbers directly without asking what must divide exactly.

  Wrong working

  Greatest number dividing 245 and 1029 leaving remainder 5: $245=5\times 7^2$, $1029=3\times 7^3$, so HCF $=49$

  Right working

  Subtract the remainder first: $240=2^4\times 3\times 5$, $1024=2^{10}$, so HCF $=2^4=16$. Check: $245=16\times 15+5$

  ✓Do this: Remove the remainders, then find the HCF. Always check by dividing back.

  Usually costs: Usually 2 or more marks, since everything after the first step is wrong.

• ✗Mistake: Leaving composite numbers in a factorisation

  The factor tree is stopped early, often at 9, 15 or 25.

  Wrong working

  $360=2^3\times 9\times 5$

  Right working

  $360=2^3\times 3^2\times 5$

  ✓Do this: Before moving on, check that every base in your factorisation is prime.

  Usually costs: ½ mark for the factorisation, and the HCF or LCM built on it may also be wrong.

NCERT exercise guide: Real Numbers

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Real Numbers: previous year questions

• Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; 83 questions in all, counting OR alternatives. In 4 of 4 sample papers.
• Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; 82 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Real Numbers

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Real Numbers question, by year

Practice questions on Real Numbers

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   If $p$ and $q$ are two different prime numbers, then LCM$(p^{2}q,p{q}^3)$ is
   (a) $pq$
   (b) $p^2{q}^3$
   (c) $p^3{q}^4$
   (d) $p{q}^3$

   Hint

   For LCM, take each prime with its greatest power.

   Answer

   (b) $p^2{q}^3$. The greatest power of $p$ is $p^2$ and the greatest power of $q$ is $q^3$.

2. Q2 · 2 marks · Very short answerBoard level

   The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other number.

   Hint

   For two numbers, HCF × LCM = product of the numbers.

   Answer

   Other number $=\frac{23\times 1449}{161}=23\times 9=207$, since $1449=161\times 9$. Check: $161=7\times 23$ and $207=3^2\times 23$, so HCF = 23 and LCM $=3^2\times 7\times 23=1449$.

3. Q3 · 2 marks · Very short answerBoard level

   Explain why $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ is a composite number.

   Hint

   Both terms have a common factor. Take it out.

   Answer

   $7\times 6\times 5\times 4\times 3\times 2\times 1+5=5(7\times 6\times 4\times 3\times 2\times 1+1)=5\times 1009$. It has factors other than 1 and itself, so it is composite.

4. Q4 · 3 marks · Short answerHarder board level

   Given that $\sqrt{2}$ is irrational, prove that $\frac{2+3\sqrt{2}}{5}$ is irrational.

   Hint

   Assume it equals a rational number $r$, then make $\sqrt{2}$ the subject.

   Answer

   Suppose $\frac{2+3\sqrt{2}}{5}=r$, with $r$ rational. Then $3\sqrt{2}=5r-2$, so $\sqrt{2}=\frac{5r-2}{3}$, which is rational. This contradicts the irrationality of $\sqrt{2}$, so the number is irrational.

5. Q5 · 5 marks · Long answerHarder board level

   Prove that $\sqrt{3}$ is irrational. Hence show that $\frac{1}{2-\sqrt{3}}$ is irrational.

   Hint

   For the second part, rationalise the denominator first.

   Answer

   Assume $\sqrt{3}=\frac{a}{b}$ with $a$, $b$ coprime and $b\ne 0$. Then $a^2=3b^2$, so 3 divides $a$. Write $a=3c$; then $b^2=3c^2$, so 3 divides $b$. This contradicts coprimality, so $\sqrt{3}$ is irrational (3 marks).

   $\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$. If $2+\sqrt{3}=r$ were rational, then $\sqrt{3}=r-2$ would be rational, a contradiction. So the number is irrational (2 marks).

6. Q6 · 3 marks · Short answerHarder board level

   Find the smallest four-digit number that leaves a remainder of 5 when divided by each of 28, 42 and 63.

   Hint

   The number minus 5 must be a common multiple of 28, 42 and 63.

   Answer

   $28=2^2\times 7$, $42=2\times 3\times 7$, $63=3^2\times 7$, so LCM $=2^2\times 3^2\times 7=252$. The smallest four-digit multiple of 252 is $252\times 4=1008$ (since $252\times 3=756$). Required number $=1008+5=1013$.

7. Q7 · 3 marks · Short answerHarder board level

   Let $a=2^3\times 3^m\times 5$ and $b=2^n\times 3^2\times 7$, where $m$ and $n$ are natural numbers. If HCF$(a,b)=36$ and LCM$(a,b)=7560$, find $m$ and $n$, and hence find $a$ and $b$.

   Hint

   Write 36 and 7560 in prime factorised form and compare the powers of 2 and of 3.

   Answer

   $36=2^2\times 3^2$ and $7560=2^3\times 3^3\times 5\times 7$. From HCF, the smaller power of 2 is 2, so $n=2$. From LCM, the greater power of 3 is 3, so $m=3$. Then $a=8\times 27\times 5=1080$ and $b=4\times 9\times 7=252$.

8. Q8 · 5 marks · Long answerHarder board level

   A tailor has three pieces of cloth of lengths 360 cm, 480 cm and 672 cm.
   (i) He wants to cut all of them into pieces of equal length, as long as possible, with no cloth wasted. Find the length of each piece and the total number of pieces.
   (ii) Find the shortest length of a single roll of cloth from which pieces of exactly 360 cm, or exactly 480 cm, or exactly 672 cm can each be cut with nothing left over. Give your answer in metres.

   Hint

   Part (i) needs the HCF and part (ii) needs the LCM. Factorise all three lengths first.

   Answer

   $360=2^3\times 3^2\times 5$, $480=2^5\times 3\times 5$, $672=2^5\times 3\times 7$. (i) HCF $=2^3\times 3=24$ cm; pieces $=15+20+28=63$. (ii) LCM $=2^5\times 3^2\times 5\times 7=10080$ cm $=100.8$ m.

9. Q9 · 1 mark · Assertion-reasonHOTS / competency

   Assertion (A): $\sqrt{2}+\sqrt{8}$ is an irrational number.
   Reason (R): The sum of two irrational numbers is always irrational.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Simplify $\sqrt{8}$ first. Then try $\sqrt{2}$ added to $-\sqrt{2}$.

   Answer

   (c) A is true but R is false. $\sqrt{8}=2\sqrt{2}$, so the sum is $3\sqrt{2}$, which is irrational. But $\sqrt{2}+(-\sqrt{2})=0$ is rational, so R is false.

10. Q10 · 3 marks · Short answerHOTS / competency

    Aman says: 'If 15 divides $n^2$ for a natural number $n$, then 225 must divide $n^2$.' Is he right? Justify using prime factorisation.

    Hint

    15 is not prime, but 3 and 5 are. Apply 'prime divides $n^2$ implies prime divides $n$' to each.

    Answer

    Yes. 15 divides $n^2$, so 3 divides $n^2$ and 5 divides $n^2$. Since 3 and 5 are primes, 3 divides $n$ and 5 divides $n$. Both distinct primes appear in the factorisation of $n$, so 15 divides $n$. Write $n=15k$; then $n^2=225k^2$, which is divisible by 225.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the Fundamental Theorem of Arithmetic.Show answer

   Every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order of the factors.

2. 2.Is 1 prime, composite or neither?Show answer

   Neither. It has only one factor.

3. 3.Which powers does the HCF use, and which does the LCM use?Show answer

   HCF: smallest powers of the common primes. LCM: greatest powers of all the primes.

4. 4.For two numbers a and b, what is HCF × LCM?Show answer

   $a\times b$. It holds only for two numbers.

5. 5.If a prime p divides $a^2$, what can you conclude?Show answer

   p divides a.

6. 6.Can two numbers have HCF 18 and LCM 380?Show answer

   No. The HCF must divide the LCM, and 18 does not divide 380.

7. 7.What is the LCM of two co-prime numbers?Show answer

   Their product, since their HCF is 1.

8. 8.Is the sum of two irrational numbers always irrational?Show answer

   No. For example $(2+\sqrt{3})+(2-\sqrt{3})=4$, which is rational.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state the Fundamental Theorem of Arithmetic in one sentence?
• ☐Can you write 4620 as a product of prime powers?
• ☐Can you find the HCF and LCM of three numbers by prime factorisation and check them?
• ☐Can you find a missing number using HCF × LCM = product, and say why it fails for three numbers?
• ☐Can you explain, citing the theorem, why $15^n$ never ends in 0?
• ☐Can you show that an expression like $2\times 3\times 5\times 7\times 11+7\times 11$ is composite?
• ☐Can you decide HCF or LCM from the words of a problem and adjust for remainders?
• ☐Can you write the full proof that $\sqrt{5}$ is irrational, naming the contradiction?
• ☐Can you prove that $3+2\sqrt{5}$ is irrational using the known irrationality of $\sqrt{5}$?
• ☐Can you give an example of two irrational numbers whose sum or product is rational?

Real Numbers: frequently asked questions