MathsCBSE Class 10 · Real Numbers

Fundamental Theorem of Arithmetic for Class 10

Q: Is Euclid's division lemma in the Class 10 syllabus for 2026-27?

No. Euclid's division lemma and algorithm are not in the 2026-27 syllabus, so you do not need to study them for this chapter.

Q: Is 1 a prime number, and does it have a prime factorisation?

No. 1 is neither prime nor composite, so it has no prime factorisation. The theorem is about natural numbers greater than 1.

Q: Does the order of the primes matter in a prime factorisation?

No. The theorem says the factorisation is unique apart from order, so 2 × 3 × 5 and 5 × 2 × 3 are the same factorisation. In the exam, writing the primes in ascending order with powers is neat and clear.

Q: How do I prove that a number like 6 to the power n never ends in 0?

Write the base as a product of primes, so 6 to the power n becomes 2 to the power n times 3 to the power n. By uniqueness of prime factorisation, 5 is not a factor, and a number ending in 0 must be divisible by both 2 and 5. So it can never end in 0.

Q: Why do I need this theorem for the irrationality proofs?

Those proofs use the fact that if a prime divides the square of a number, it divides the number itself. That fact follows from the Fundamental Theorem of Arithmetic, because the primes in a square are exactly the primes of the original number.

Updated 13 October 2026 · Fact-checked

The Fundamental Theorem of Arithmetic says every composite number can be written as a product of primes in exactly one way, apart from the order of the factors. Board questions ask you to prime-factorise numbers, show a number is composite, decide whether a power like 6^n can end in 0, and use factorisation for HCF, LCM and real-life problems.

Understand Fundamental Theorem of Arithmetic

Every natural number greater than 1 is either prime or composite. A prime has exactly two factors, 1 and itself. A composite number has more. Take a composite number and keep splitting it into factors: $360 = 2 \times 180 = 2 \times 2 \times 90 = \dots = 2^{3} \times 3^{2} \times 5$ . The splitting stops when every piece is a prime.

The Fundamental Theorem of Arithmetic says two things. First, every composite number can be written as a product of primes. Second, this factorisation is unique, apart from the order in which the primes are written. So whatever factor tree you draw for 360, you always end with three 2s, two 3s and one 5. The usual way to write it is in ascending order with powers: $2^{3} \times 3^{2} \times 5$ .

Why does it work? Each split gives you smaller numbers, so the splitting cannot go on forever. It must end in primes. Uniqueness means primes behave like the atoms of numbers: each number has exactly one 'formula' in terms of primes. You are not asked to prove the theorem in Class 10, but you must use it. Uniqueness is what lets you compare powers: if $2^{a} \times 3^{b} = 2^{4} \times 3^{2}$ , then $a = 4$ and $b = 2$ .

The theorem powers the rest of the chapter. The HCF of two numbers is the product of the smallest powers of their common primes, and the LCM is the product of the greatest powers of all the primes involved. A number ends in 0 only when it is divisible by $10 = 2 \times 5$ , so its factorisation must contain both 2 and 5. That is why $6^{n} = 2^{n} \times 3^{n}$ can never end in 0. The irrationality proofs use one more consequence: if a prime $p$ divides $a^{2}$ , then $p$ divides $a$ . This holds because the primes in $a^{2}$ are exactly the primes in $a$ .

Know the conditions. The theorem is about natural numbers greater than 1. The number 1 is neither prime nor composite and has no prime factorisation. Uniqueness is only 'up to order'. Every factor must be prime, so $42 = 6 \times 7$ is a factorisation but not a prime factorisation. Finally, the result $HCF \times LCM = product$ is true for two numbers only, not for three.

Fundamental Theorem of Arithmetic: formulas to remember

Fundamental Theorem of Arithmetic: $n = p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \dots \times p_{k}^{a_{k}}$; Every composite n can be written like this, with distinct primes p_1 < p_2 < ... < p_k and natural number powers. The form is unique apart from order.
Comparing powers (uniqueness): $2^{a} \times 3^{b} \times 5^{c} = 2^{x} \times 3^{y} \times 5^{z} \Rightarrow a = x, b = y, c = z$; Because the factorisation is unique, you can equate the powers of each prime. Use this when the unknowns are in the exponents.
HCF by prime factorisation: $HCF = product of common primes, each with its smallest power$; Only primes that appear in every number are used.
LCM by prime factorisation: $LCM = product of all primes involved, each with its greatest power$; Every prime that appears in any of the numbers is used.
HCF-LCM product rule (two numbers only): $HCF (a, b) \times LCM (a, b) = a \times b$; True for any two positive integers. It fails in general for three numbers.
Test for ending in 0: $n ends in 0 ⟺ 2 and 5 both appear in the prime factorisation of n$; This is because 10 = 2 × 5. If either prime is missing, the number cannot end in 0.
Prime dividing a square: $p prime, p ∣ a^{2} \Rightarrow p ∣ a$; A consequence of the theorem, used in the irrationality proofs of square roots such as √2, √3 and √5.

How to solve Fundamental Theorem of Arithmetic questions

Almost every question on this topic comes down to writing a number as a product of primes and then reading information off the powers. Use this method every time.

1Identify what is asked: a prime factorisation, a composite proof, an 'ends in 0' test, unknown powers, or an HCF/LCM in context.
2Prime-factorise each number by repeated division by the smallest prime (2, then 3, then 5, then 7, ...). Write the result in ascending order with powers.
3Check your factorisation by multiplying back. Make sure every factor is prime.
4For a 'composite' question, take out a common factor first, for example $a \times b + b = b (a + 1)$ . Then state that the number has a factor other than 1 and itself.
5For powers like $k^{n}$ , write $k$ as a product of primes and raise each prime to the power $n$ . By uniqueness, no other primes can appear. Use this to decide whether 2 and 5 are both present.
6For unknown exponents, factorise the given number and equate the powers of each prime, quoting uniqueness of prime factorisation.
7For HCF or LCM in a real-life problem, decide which one the context needs. Use HCF for the 'largest size that divides exactly' and LCM for 'the next time together' or 'the smallest common multiple'. Then answer in the units of the question.

Fundamental Theorem of Arithmetic in real life

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.

Fundamental Theorem of Arithmetic

Suppose your family runs a small sweet shop and, before Diwali, has 336 laddoos, 240 barfis and 192 pieces of kaju katli to pack. Every box must hold the same number of sweets and only one kind of sweet, so that the boxes look identical on the shelf. You want to use as few boxes as possible.

Using prime factorisation, find the greatest number of sweets that can go in each box and the least total number of boxes needed. If each empty box costs Rs 15, find the total cost of the boxes.

Try it first, then show the solutionSolution

Each box must hold a number that divides 336, 240 and 192 exactly. The fewest boxes means the largest such number, so you need HCF(336, 240, 192).
$336 = 2 \times 168 = 2^{2} \times 84 = 2^{3} \times 42 = 2^{4} \times 21 = 2^{4} \times 3 \times 7$
$240 = 2 \times 120 = 2^{2} \times 60 = 2^{3} \times 30 = 2^{4} \times 15 = 2^{4} \times 3 \times 5$
$192 = 2 \times 96 = 2^{2} \times 48 = 2^{3} \times 24 = 2^{4} \times 12 = 2^{6} \times 3$
The common primes are 2 and 3. The smallest power of 2 is $2^{4}$ and the smallest power of 3 is $3^{1}$ .
$HCF = 2^{4} \times 3 = 16 \times 3 = 48$ , so each box holds 48 sweets.
Laddoo boxes $= \frac{336}{48} = 7$ , barfi boxes $= \frac{240}{48} = 5$ , kaju katli boxes $= \frac{192}{48} = 4$ .
Total boxes $= 7 + 5 + 4 = 16$ .
Cost of boxes $= 16 \times 15 = 240$ , that is Rs 240.
Check: $48 \times 7 = 336$ , $48 \times 5 = 240$ , $48 \times 4 = 192$ . All sweets are packed with nothing left over.

Answer: Each box holds 48 sweets; the least number of boxes is 16 (7 laddoo, 5 barfi, 4 kaju katli), costing Rs 240 in total.

Memory hook: Equal boxes, fewest boxes: take the smallest powers of the common primes, the HCF.

Fundamental Theorem of Arithmetic

Imagine three friends, Asha, Bhavesh and Chitra, practising on the school's running track one morning. Asha takes 72 seconds for one lap, Bhavesh takes 108 seconds and Chitra takes 90 seconds. They start together from the same starting line at 7:00 am and keep running at the same steady pace.

Using prime factorisation, find the earliest time at which all three will be together again at the starting line. How many laps will each friend have completed by then?

Try it first, then show the solutionSolution

They meet at the start after a time that is a multiple of 72, 108 and 90 seconds. The earliest such time is LCM(72, 108, 90).
$72 = 2 \times 36 = 2^{2} \times 18 = 2^{3} \times 9 = 2^{3} \times 3^{2}$
$108 = 2 \times 54 = 2^{2} \times 27 = 2^{2} \times 3^{3}$
$90 = 2 \times 45 = 2 \times 3^{2} \times 5$
The primes involved are 2, 3 and 5. The greatest powers are $2^{3}$ , $3^{3}$ and $5^{1}$ .
$LCM = 2^{3} \times 3^{3} \times 5 = 8 \times 27 \times 5 = 1080$ seconds.
$1080$ seconds $= \frac{1080}{60} = 18$ minutes, so they meet at 7:18 am.
Asha's laps $= \frac{1080}{72} = 15$ , Bhavesh's laps $= \frac{1080}{108} = 10$ , Chitra's laps $= \frac{1080}{90} = 12$ .
Check: each lap count is a whole number, and no smaller common multiple exists because every prime power used is the least needed to cover all three numbers.

Answer: They are next together at the starting line at 7:18 am (after 1080 seconds); Asha completes 15 laps, Bhavesh 10 laps and Chitra 12 laps.

Memory hook: Runners meeting again at the start: take the greatest powers of all the primes, the LCM.

Tricks to remember Fundamental Theorem of Arithmetic

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

Primes are atoms
Think of primes as atoms and numbers as molecules. A molecule has one fixed formula, and so does a number: 360 is always $2^{3} \times 3^{2} \times 5$ .
Use it when: Reminds you that powers can be compared, which is how you find unknown exponents.
HCF hugs the small, LCM loves the large
HCF takes the common primes with the smallest powers. LCM takes all the primes with the largest powers.
Use it when: Stops you mixing up min and max powers when finding HCF and LCM from factorisations.
Zero needs a 2-5 pair
A trailing 0 comes from a factor of 10, and $10 = 2 \times 5$ . No pair, no zero.
Use it when: Instantly answers 'can $k^{n}$ end in 0?' questions: just check whether $k$ contains both 2 and 5.
Product rule is a pair rule
HCF × LCM = product works for a pair: two numbers, never three.
Use it when: Prevents a common error in assertion-reason and three-number questions.

Fundamental Theorem of Arithmetic: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Showing a number is composite · 2 marks · Very short answerBoard level

Explain why $7 \times 11 \times 13 + 13$ is a composite number. Also write its prime factorisation.

Try it first, then show the solutionSolution

$7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1)$ , taking 13 as a common factor. (½ mark)
$= 13 \times 78$ (½ mark)
Since $78 = 2 \times 3 \times 13$ , the number $= 2 \times 3 \times 1 3^{2} = 1014$ . (½ mark)
The number has factors such as 2, 3 and 13 other than 1 and itself, so it is composite. (½ mark)

Answer: It is composite because it equals $13 \times 78$ . Its prime factorisation is $2 \times 3 \times 1 3^{2}$ (= 1014).

Example 2: Finding unknown powers using uniqueness · 3 marks · Short answerBoard level

If $2^{x} \times 3^{y} \times 5^{z} = 2160$ , where $x$ , $y$ , $z$ are whole numbers, find $x$ , $y$ and $z$ . Hence find the value of $2^{- x} \times 3^{y} \times 5^{z}$ .

Try it first, then show the solutionSolution

Divide repeatedly by 2: $2160 \to 1080 \to 540 \to 270 \to 135$ , so $2160 = 2^{4} \times 135$ .
Divide 135 by 3: $135 \to 45 \to 15 \to 5$ , so $135 = 3^{3} \times 5$ and $2160 = 2^{4} \times 3^{3} \times 5$ . (1 mark)
So $2^{x} \times 3^{y} \times 5^{z} = 2^{4} \times 3^{3} \times 5^{1}$ . By the uniqueness of prime factorisation (FTA), $x = 4$ , $y = 3$ , $z = 1$ . (1 mark)
$2^{- x} \times 3^{y} \times 5^{z} = \frac{3 ^{3} \times 5}{2 ^{4}} = \frac{27 \times 5}{16} = \frac{135}{16}$ (1 mark)
Check: $16 \times 135 = 2160$ , which matches.

Answer: $x = 4$ , $y = 3$ , $z = 1$ , and $2^{- x} \times 3^{y} \times 5^{z} = \frac{135}{16}$ .

Example 3: HCF and LCM from prime powers, then working backwards · 3 marks · Short answerHarder board level

Two positive integers are $a = x^{3} y^{2}$ and $b = x y^{3}$ , where $x$ and $y$ are distinct primes.
(i) Find $HCF (a, b)$ and $LCM (a, b)$ in terms of $x$ and $y$ .
(ii) For one such pair, HCF $= 18$ and LCM $= 216$ . Find $x$ , $y$ , $a$ and $b$ .

Try it first, then show the solutionSolution

(i) The common primes are $x$ and $y$ . Their smallest powers are $x^{1}$ and $y^{2}$ , so HCF $= x y^{2}$ . (1 mark)
The greatest powers are $x^{3}$ and $y^{3}$ , so LCM $= x^{3} y^{3}$ . (½ mark)
(ii) $x y^{2} = 18 = 2 \times 3^{2}$ . By the uniqueness of prime factorisation, $x = 2$ and $y = 3$ . (1 mark)
Check with the LCM: $x^{3} y^{3} = 2^{3} \times 3^{3} = 8 \times 27 = 216$ , which matches. (½ mark)
So $a = 2^{3} \times 3^{2} = 72$ and $b = 2 \times 3^{3} = 54$ . Check: $72 \times 54 = 3888 = 18 \times 216$ .

Answer: (i) HCF $= x y^{2}$ , LCM $= x^{3} y^{3}$ . (ii) $x = 2$ , $y = 3$ , so $a = 72$ and $b = 54$ .

Example 4: Can a power end in 0 or 5? · 3 marks · Short answerHarder board level

Show that $1 2^{n}$ cannot end with the digit 0 or the digit 5 for any natural number $n$ . Also find the smallest natural number $m$ such that $1 2^{n} \times m$ ends with the digit 0 for every natural number $n$ .

Try it first, then show the solutionSolution

$12 = 2^{2} \times 3$ , so $1 2^{n} = 2^{2 n} \times 3^{n}$ . (½ mark)
A number ending in 0 is divisible by $10 = 2 \times 5$ , so 5 must appear in its prime factorisation. (½ mark)
By the uniqueness of prime factorisation, the only primes in $1 2^{n}$ are 2 and 3. There is no 5, so $1 2^{n}$ cannot end in 0. (½ mark)
$1 2^{n}$ has 2 as a factor, so it is even. Every number ending in 5 is odd, so $1 2^{n}$ cannot end in 5. (½ mark)
For $1 2^{n} \times m$ to end in 0, its factorisation needs both 2 and 5. The factor 2 is already present, so $m$ must contain 5. The numbers 1, 2, 3 and 4 do not, so the smallest is $m = 5$ . (½ mark)
Check: $12 \times 5 = 60$ and $144 \times 5 = 720$ , both ending in 0. (½ mark)

Answer: $1 2^{n} = 2^{2 n} \times 3^{n}$ has no factor 5, so it cannot end in 0. It is even, so it cannot end in 5. The smallest $m$ is 5.

Example 5: Two tankers: HCF and LCM in context · 5 marks · Long answerHarder board level

A petrol depot has two tankers holding 850 litres and 680 litres.
(a) Find the largest container capacity that can measure the petrol of each tanker exactly. How many full containers are needed in all to empty both tankers?
(b) The depot wants a storage tank whose capacity is the smallest volume that can be filled exactly using whole loads of the first tanker alone or of the second tanker alone. Find this capacity.
(c) Verify that HCF × LCM equals the product of 850 and 680.

Try it first, then show the solutionSolution

$850 = 2 \times 5^{2} \times 17$ and $680 = 2^{3} \times 5 \times 17$ . (1 mark)
(a) The largest exact measure is the HCF, using the smallest powers of the common primes: $2 \times 5 \times 17 = 170$ litres. (1 mark)
Number of containers $= \frac{850}{170} + \frac{680}{170} = 5 + 4 = 9$ . (1 mark)
(b) The smallest common multiple is the LCM, using the greatest powers of all the primes: $2^{3} \times 5^{2} \times 17 = 8 \times 25 \times 17 = 3400$ litres. (1 mark)
(c) $HCF \times LCM = 170 \times 3400 = 578000$ and $850 \times 680 = 578000$ . They are equal, so the result is verified. (1 mark)

Answer: (a) 170 litres; 9 containers. (b) 3400 litres. (c) Both products equal 578000.

Example 6: Case study: the factor tree · 4 marks · Case-basedHOTS / competency

In a class activity, Meena draws a factor tree for a number $N$ . At the top, $N$ splits into 2 and $A$ . Then $A$ splits into 3 and $B$ , $B$ splits into 3 and $C$ , and finally $C$ splits into 5 and 7. All the end branches 2, 3, 3, 5, 7 are primes.
(i) Find the value of $N$ and write its prime factorisation. (1 mark)
(ii) Find the HCF of $N$ and 84. (1 mark)
(iii) Find the smallest natural number by which $N$ must be multiplied so that the product is a perfect square. Also find the square root of that product. (2 marks)

Try it first, then show the solutionSolution

Work upwards: $C = 5 \times 7 = 35$ , $B = 3 \times 35 = 105$ , $A = 3 \times 105 = 315$ .
(i) $N = 2 \times 315 = 630 = 2 \times 3^{2} \times 5 \times 7$ . (1 mark)
(ii) $84 = 2^{2} \times 3 \times 7$ . The common primes with their smallest powers give HCF $= 2 \times 3 \times 7 = 42$ . (1 mark)
(iii) In a perfect square, every prime has an even power. In $630 = 2^{1} \times 3^{2} \times 5^{1} \times 7^{1}$ , the primes 2, 5 and 7 have odd powers. (½ mark)
So multiply by $2 \times 5 \times 7 = 70$ . Any smaller multiplier leaves at least one odd power. (½ mark)
$630 \times 70 = 44100 = 2^{2} \times 3^{2} \times 5^{2} \times 7^{2}$ (½ mark)
Square root $= 2 \times 3 \times 5 \times 7 = 210$ . Check: $21 0^{2} = 44100$ . (½ mark)

Answer: (i) $N = 630 = 2 \times 3^{2} \times 5 \times 7$ . (ii) HCF = 42. (iii) Multiply by 70 to get 44100, whose square root is 210.

Where students lose marks in Fundamental Theorem of Arithmetic

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

Mistake: Leaving a composite number inside the 'prime factorisation'
Students stop splitting too early, so the answer is a factorisation but not a prime factorisation.
Wrong working
$1014 = 6 \times 1 3^{2}$
Right working
$1014 = 2 \times 3 \times 1 3^{2}$ (every factor is prime)
Do this: After factorising, check each factor: is it prime? Multiply back to confirm the number.
Usually costs: The ½ to 1 mark for the factorisation step, and every later step that uses it.
Mistake: Weak reasoning in 'can $k^{n}$ end in 0?' questions
Students argue from the last digit, or say 'it is even, so it ends in 0', instead of using prime factorisation.
Wrong working
$4^{n}$ ends in 0 for large $n$ because $4^{n}$ is even.
Right working
$4^{n} = 2^{2 n}$ . Its only prime factor is 2, by uniqueness of prime factorisation. To end in 0 it needs 5 as a factor, so $4^{n}$ never ends in 0.
Do this: Always write $k^{n}$ in prime form, quote uniqueness, and say that 5 (or 2) is missing.
Usually costs: Most of the marks in a 2- or 3-mark question, because the reasoning is what gets marked.
Mistake: Using HCF × LCM = product for three numbers
The rule is remembered without its condition.
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 numbers, always find the LCM from greatest powers. Use the product rule only for a pair.
Usually costs: The whole answer, plus any later part that depends on the LCM.
Mistake: Taking greatest powers for HCF (or smallest for LCM)
The two rules get swapped under exam pressure.
Wrong working
$850 = 2 \times 5^{2} \times 17$ , $680 = 2^{3} \times 5 \times 17$ , HCF $= 2^{3} \times 5^{2} \times 17 = 3400$
Right working
HCF $= 2 \times 5 \times 17 = 170$ (smallest powers). LCM $= 2^{3} \times 5^{2} \times 17 = 3400$ (greatest powers).
Do this: Sense check: the HCF can never be bigger than the smaller number, and the LCM can never be smaller than the larger number.
Usually costs: 1 to 2 marks, and the context answer (containers, tiles, time) becomes wrong too.
Mistake: Calling a number composite without showing a factor
Students compute the value and simply assert 'it is composite', or say 'it is large'.
Wrong working
$7 \times 11 \times 13 + 13 = 1014$ , which is composite.
Right working
$7 \times 11 \times 13 + 13 = 13 (77 + 1) = 13 \times 78$ , so it has factors other than 1 and itself. Hence it is composite.
Do this: Take out the common factor and state the definition of a composite number in your last line.
Usually costs: 1 mark out of 2.

Exam tips

Write prime factorisations in ascending order with powers, for example $2^{3} \times 3^{2} \times 5$ . It makes HCF and LCM easy to read and shows the examiner a clean method.
When a question says 'using prime factorisation' or 'using the Fundamental Theorem of Arithmetic', show the factorisation of every number. A correct answer without it can lose method marks.
In reasoning questions (ends in 0, unknown powers), quote the reason in words: 'by the uniqueness of prime factorisation'. This is the step the examiner is looking for.
In word problems, decide first whether the context needs HCF (largest equal size, maximum number of groups) or LCM (next time together, smallest common quantity). Then give units in the final answer.
For MCQs and assertion-reason items, quickly test any claimed rule with small numbers such as 6, 10 and 15. A counterexample settles it fast.

Practice questions on Fundamental Theorem of Arithmetic

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

Q1 · 2 marks · Very short answerBoard level
The HCF of two numbers is 18 and their product is 12960. Find their LCM. Can two numbers have HCF 18 and LCM 380? Give a reason using prime factorisation.
Hint
Use HCF × LCM = product for two numbers. Remember that the HCF always divides the LCM.
Answer
LCM $= \frac{12960}{18} = 720$ . No: the HCF must divide the LCM, but $18 = 2 \times 3^{2}$ while $380 = 2^{2} \times 5 \times 19$ has no factor 3, so 18 does not divide 380.
Q2 · 3 marks · Short answerBoard level
If $2^{a} \times 3^{b} \times 7^{c} = 3528$ , where $a$ , $b$ , $c$ are whole numbers, find $a$ , $b$ , $c$ . Hence find the value of $2^{a} + 3^{b} + 7^{c}$ .
Hint
Divide by 2 until the result is odd, then by 3, then by 7. Equate powers using uniqueness.
Answer
$3528 = 2^{3} \times 441 = 2^{3} \times 3^{2} \times 7^{2}$ . By uniqueness, $a = 3$ , $b = 2$ , $c = 2$ . So $2^{a} + 3^{b} + 7^{c} = 8 + 9 + 49 = 66$ .
Q3 · 3 marks · Short answerHarder board level
Find the smallest perfect square that is divisible by each of 12, 18 and 24. Hence state its square root.
Hint
Any number divisible by all three is a multiple of their LCM. Then make every prime power even.
Answer
$12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$ , $24 = 2^{3} \times 3$ . LCM $= 2^{3} \times 3^{2} = 72$ . The required number is a multiple of 72 with every prime power even, so raise $2^{3}$ to $2^{4}$ : $2^{4} \times 3^{2} = 144$ . Square root $= 2^{2} \times 3 = 12$ . Check: $144 = 12 \times 12 = 18 \times 8 = 24 \times 6$ .
Q4 · 5 marks · Long answerHarder board level
Three runners take 36 minutes, 48 minutes and 60 minutes to complete one round of a circular track. They start together from the same point at 7:00 am and run in the same direction.
(a) Using prime factorisation, find when they will next be together at the starting point.
(b) How many rounds will each runner have completed by then?
(c) Find the HCF of 36, 48 and 60 and show that HCF × LCM is not equal to the product of the three numbers.
Hint
The meeting time is the LCM. For (c), compare $12 \times 720$ with $36 \times 48 \times 60$ .
Answer
$36 = 2^{2} \times 3^{2}$ , $48 = 2^{4} \times 3$ , $60 = 2^{2} \times 3 \times 5$ . (a) LCM $= 2^{4} \times 3^{2} \times 5 = 720$ min $= 12$ h, so they meet at 7:00 pm. (b) $\frac{720}{36} = 20$ , $\frac{720}{48} = 15$ , $\frac{720}{60} = 12$ rounds. (c) HCF $= 2^{2} \times 3 = 12$ . Then $12 \times 720 = 8640$ , but $36 \times 48 \times 60 = 103680$ , so they are not equal.
Q5 · 1 mark · Assertion-reasonBoard level
Assertion (A): For the numbers 6, 10 and 15, HCF × LCM = 6 × 10 × 15.
Reason (R): For any two positive integers $a$ and $b$ , $HCF (a, b) \times LCM (a, b) = a \times b$ .
(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
Factorise 6, 10 and 15. Find their HCF and LCM, then test A with actual numbers.
Answer
(d). $6 = 2 \times 3$ , $10 = 2 \times 5$ , $15 = 3 \times 5$ , so HCF = 1 and LCM = 30. Then HCF × LCM = 30, but the product is 900, so A is false. R is a true result, but it holds for two numbers only.
Q6 · 4 marks · Case-basedHOTS / competency
Suppose for a Diwali display your school hangs three strings of lights. All three blink together at the start. After that, the first string blinks every 8 seconds, the second every 12 seconds and the third every 20 seconds.
(i) Write the prime factorisations of 8, 12 and 20. (1 mark)
(ii) After how many seconds will all three blink together again for the first time? How many times will they blink together in the first 10 minutes, not counting the start? (1 mark)
(iii) A fourth string is added. Its blink interval is an odd number of seconds greater than 1, and now all four strings blink together for the first time after 360 seconds. Find the smallest possible interval of the fourth string. (2 marks)
Hint
The first common blink is the LCM. For (iii), compare the prime factorisations of 360 and of the LCM in (ii) to see which prime power the fourth interval must supply.
Answer
(i) $8 = 2^{3}$ , $12 = 2^{2} \times 3$ , $20 = 2^{2} \times 5$ . (ii) LCM $= 2^{3} \times 3 \times 5 = 120$ seconds; in 600 seconds they blink together $\frac{600}{120} = 5$ times. (iii) $360 = 2^{3} \times 3^{2} \times 5$ and $120 = 2^{3} \times 3 \times 5$ , so the fourth interval $k$ must divide 360 and supply $3^{2}$ . Since $k$ is odd, it has no factor 2, so $k$ is 9 or 45. The smallest is 9 seconds. Check: LCM $(120, 9) = 2^{3} \times 3^{2} \times 5 = 360$ .
Q7 · 3 marks · Short answerHOTS / competency
The HCF of two numbers is 12 and their LCM is 336.
(a) If one number is 48, find the other.
(b) Find all pairs of numbers with HCF 12 and LCM 336, and explain why there are no others.
Hint
Write the numbers as $12 m$ and $12 n$ , where $m$ and $n$ have no common prime factor. Then $12 m \times 12 n = 12 \times 336$ .
Answer
(a) Other number $= \frac{12 \times 336}{48} = 84$ . (b) Let the numbers be $12 m$ and $12 n$ , with $m$ and $n$ sharing no prime. Then $144 mn = 12 \times 336$ , so $mn = 28 = 2^{2} \times 7$ . Since $m$ and $n$ share no prime, the whole $2^{2}$ must go to one of them: $(m, n) = (1, 28)$ or $(4, 7)$ . So the pairs are 12 and 336, and 48 and 84. Check: $48 = 2^{4} \times 3$ and $84 = 2^{2} \times 3 \times 7$ give HCF $= 12$ and LCM $= 2^{4} \times 3 \times 7 = 336$ .

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.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.Write the prime factorisation of 360.Show answer
$2^{3} \times 3^{2} \times 5$
3.Which primes must appear in a number for it to end in 0?Show answer
Both 2 and 5, since $10 = 2 \times 5$ .
4.HCF uses which powers of the primes, and LCM uses which?Show answer
HCF uses the smallest powers of the common primes. LCM uses the greatest powers of all the primes.
5.Is HCF × LCM = product true for three numbers?Show answer
Not in general. It is guaranteed only for two numbers.
6.If a prime $p$ divides $a^{2}$ , what can you say?Show answer
$p$ divides $a$ . This is used in proofs of irrationality.

Fundamental Theorem of Arithmetic: frequently asked questions

Is Euclid's division lemma in the Class 10 syllabus for 2026-27?

No. Euclid's division lemma and algorithm are not in the 2026-27 syllabus, so you do not need to study them for this chapter.

Is 1 a prime number, and does it have a prime factorisation?

No. 1 is neither prime nor composite, so it has no prime factorisation. The theorem is about natural numbers greater than 1.

Does the order of the primes matter in a prime factorisation?

No. The theorem says the factorisation is unique apart from order, so 2 × 3 × 5 and 5 × 2 × 3 are the same factorisation. In the exam, writing the primes in ascending order with powers is neat and clear.

How do I prove that a number like 6 to the power n never ends in 0?

Write the base as a product of primes, so 6 to the power n becomes 2 to the power n times 3 to the power n. By uniqueness of prime factorisation, 5 is not a factor, and a number ending in 0 must be divisible by both 2 and 5. So it can never end in 0.

Why do I need this theorem for the irrationality proofs?

Those proofs use the fact that if a prime divides the square of a number, it divides the number itself. That fact follows from the Fundamental Theorem of Arithmetic, because the primes in a square are exactly the primes of the original number.