Every whole number bigger than $1$ is either a prime or a product of primes, and it can be split into primes in only one way. That single fact runs through this whole chapter.
Think of the numbers $2, 3, 5, 7, 11, \dots$ as building blocks, a bit like LEGO bricks. Multiply some of them together and you can build any whole number you like: $12=2\times2\times3$, $90=2\times3\times3\times5$, $1001=7\times11\times13$. These special numbers are the primes, and splitting a number into primes is called prime factorisation.
Here is the surprising part. However you start splitting a number, you always finish with exactly the same primes. You can break $360$ as $36\times10$ or as $8\times45$, and both routes end at $2\times2\times2\times3\times3\times5$. This is the Fundamental Theorem of Arithmetic. It lets us treat a prime factorisation as a number's unique "fingerprint". In the next section we use it to find HCF and LCM. In Section 1.3 it helps us prove that $\sqrt2$ is not a fraction. Outside school, the same idea keeps your UPI payments and online banking safe.
Primes and composite numbers
You met factors and multiples in earlier classes. Let us fix the words we will use all through this chapter.
📘 Definition 1.1 — Prime and composite numbers
A natural number $p\gt1$ is called a prime number (or simply a prime) if its only factors are $1$ and $p$ itself.
A natural number $n\gt1$ that is not prime is called a composite number. A composite number has at least one factor other than $1$ and itself, so it can be written as $n=a\times b$ with $1\lt a\lt n$ and $1\lt b\lt n$.
The number $1$ is neither prime nor composite.
The primes below $50$ are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$. Notice that $2$ is the only even prime: every other even number has $2$ as a factor, so it is composite.
🧭 Problem-solving strategy — is $n$ prime?
Find the largest whole number $k$ with $k^2\le n$. (For $n=97$: $9^2=81\le97$ but $10^2=100\gt97$, so $k=9$.)
Try dividing $n$ by each prime up to $k$: $2, 3, 5, 7, \dots$ Use the divisibility rules (even, digit sum, last digit $0$ or $5$) to go faster.
If none of them divides $n$, then $n$ is prime. If one does, you have found a factor and $n$ is composite.
Why stop at $k$? If $n=a\times b$ with both $a$ and $b$ bigger than $k$, then both are at least $k+1$, so $a\times b\ge(k+1)^2\gt n$, which is impossible. So a composite $n$ always has a factor no bigger than $k$, and that factor has a prime factor no bigger than $k$.
✏️ Example 1 — Prime or composite?
Decide whether each number is prime or composite: (a) $51$ (b) $97$ (c) $91$ (d) $221$.
💡 Show solution▶
(a) 51 The digit sum is $5+1=6$, which is divisible by $3$, so $3$ divides $51$: $51=3\times17$. Composite.
(b) 97 Here $k=9$ (since $9^2=81\le97\lt100$). Test the primes $2, 3, 5, 7$: $97$ is odd; digit sum $16$ is not a multiple of $3$; it does not end in $0$ or $5$; $97=7\times13+6$, so $7$ does not divide it. Prime.
(c) 91 It looks prime, but test $7$: $91=7\times13$. Composite. (A common slip: "it is odd and does not end in 5, so it must be prime". Always test every prime up to $k$.)
(d) 221 $14^2=196\le221\lt225=15^2$, so test the primes up to $13$. $2, 3, 5, 7, 11$ fail, but $221=13\times17$. Composite.
Prime factorisation: factor trees and division ladders
To factorise a composite number into primes, keep splitting until every piece is prime. There are two neat ways to record the work.
Factor tree. Write the number at the top. Split it into any two factors, then split each composite factor again. The leaves (the ends of the branches) are the primes.
Division ladder. Divide by the smallest prime that goes in exactly, write the quotient underneath, and repeat until you reach $1$. The primes down the left side are the prime factors.
📘 Definition 1.2 — Prime factorisation
Writing a natural number $n\gt1$ as a product of primes is called the prime factorisation of $n$. We usually list the primes in increasing order and collect repeated primes as powers, for example
$$360=2\times2\times2\times3\times3\times5=2^3\times3^2\times5 .$$
The power to which a prime appears is called its exponent; in $360$ the exponent of $2$ is $3$, of $3$ is $2$ and of $5$ is $1$.
✏️ Example 2 — A factor tree and a division ladder
Find the prime factorisation of $4620$ in two ways: with a factor tree, and with a division ladder.
💡 Show solution▶
Factor tree $4620$ ends in $0$, so start with $4620=10\times462$. Then $10=2\times5$ and $462=2\times231$; next $231=3\times77$ (digit sum $6$) and $77=7\times11$. The leaves are $2, 5, 2, 3, 7, 11$.
Division ladder $$\begin{array}{r|r}2&4620\\2&2310\\3&1155\\5&385\\7&77\\11&11\\&1\end{array}$$ The primes down the left are $2, 2, 3, 5, 7, 11$.
Answer Both methods give the same primes: $$4620=2^2\times3\times5\times7\times11 .$$
Choose a number and a first split. Press Play (or Next split) to split one composite number at a time; green circles are primes and can't be split. When the tree is finished, it starts again with a different first split. Watch the leaves: whichever way you start, you end with the same primes.
splits done—
product of the pieces so far—
prime factorisation—
✏️ Example 3 — A bigger number
Find the prime factorisation of $9828$.
💡 Show solution▶
Divide by 2 $9828$ is even: $9828\div2=4914$, and $4914\div2=2457$. Now $2457$ is odd, so we are done with $2$.
Divide by 3 Digit sum of $2457$ is $18$, so divide by $3$: $2457\div3=819$; digit sum $18$ again: $819\div3=273$; digit sum $12$: $273\div3=91$. Digit sum of $91$ is $10$, so $3$ no longer divides.
Finish $91$ does not end in $0$ or $5$, but $91=7\times13$, and both are prime.
1. Stopping too early. $4620=2^2\times1155$ is not a prime factorisation, because $1155$ is not prime. Every number in the final answer must be prime.
2. Writing $1$ as a factor. $1$ is not a prime, so it never appears in a prime factorisation.
3. Losing a factor in a long ladder. Always multiply your answer back out (the "Check" step). It takes thirty seconds and catches most slips.
The Fundamental Theorem of Arithmetic
Two questions are hiding in what we have done. Can every composite number be broken into primes? And could two different people get two different lists of primes for the same number? The theorem answers both.
📐 Theorem 1.1 (Fundamental Theorem of Arithmetic)
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
In other words, every composite number $n$ can be written in exactly one way as $n=p_1^{a_1}\times p_2^{a_2}\times\cdots\times p_k^{a_k}$ with primes $p_1\lt p_2\lt\cdots\lt p_k$ and natural-number exponents $a_1, a_2, \dots, a_k$.
Why every composite number splits into primes. If $n$ is composite, write $n=a\times b$ with both factors smaller than $n$. Any factor that is still composite splits again into smaller numbers. The numbers keep getting smaller, and they can't go below $2$, so the splitting must stop. It can only stop when every piece is prime.
Why the answer is always the same. The full proof of uniqueness is beyond Class 10, and you are not asked to prove it. It rests on one more fact about primes: if a prime divides a product $a\times b$, then it divides $a$ or it divides $b$. Interactive 1 lets you see uniqueness at work: four different first splits of $360$, one set of leaves.
✏️ Example 4 — Missing numbers in a factor tree
In a factor tree, the top number splits as $2\times A$, then $A$ splits as $3\times B$, and $B$ splits as $7\times13$. Find $B$, $A$ and the top number, and write its prime factorisation.
💡 Show solution▶
Work upwards Start at the bottom, where both numbers are known: $B=7\times13=91$.
Next level $A=3\times B=3\times91=273$.
Top Top $=2\times A=2\times273=546$.
Answer The leaves are $2, 3, 7, 13$, so $546=2\times3\times7\times13$.
✏️ Example 5 — Can it end with the digit 0?
(a) Check whether $15^n$ can end with the digit $0$ for any natural number $n$. (b) Do the same for $6^n$.
💡 Show solution▶
Key idea A number ends with the digit $0$ exactly when it is divisible by $10=2\times5$. So its prime factorisation must contain both the prime $2$ and the prime $5$.
(a) $15=3\times5$, so $15^n=3^n\times5^n$. By the Fundamental Theorem of Arithmetic this is the only prime factorisation of $15^n$, and it contains no $2$. So $15^n$ is never divisible by $2$, and $15^n$ can never end with $0$.
(b) $6=2\times3$, so $6^n=2^n\times3^n$. It contains $2$ but no $5$, so again $6^n$ never ends with $0$.
Check The powers of $15$ are $15, 225, 3375, 50625,\dots$ (all end in $5$), and the powers of $6$ are $6, 36, 216, 1296,\dots$ (all end in $6$). ✓ Uniqueness matters here: without it, someone might hope for a different factorisation of $6^n$ that does contain a $5$.
✏️ Example 6 — Spotting a hidden common factor
Explain why (a) $2\times3\times5\times7+5$ and (b) $3\times5\times7\times11+11$ are composite numbers.
💡 Show solution▶
(a) Both terms contain the factor $5$. Take it out: $2\times3\times5\times7+5=5\times(2\times3\times7+1)=5\times43=215$. It has the factor $5$ (as well as $1$ and $215$), so it is composite.
(b) Take out $11$: $3\times5\times7\times11+11=11\times(105+1)=11\times106=1166$. Composite.
Lesson Do not multiply out first. Look for a factor common to both terms; the bracket shows the number is a product of two numbers bigger than $1$.
Interactive 2 — Prime Fingerprints: Multiplying Adds Exponents
Each bar shows the exponent of one prime. The blue blocks belong to $a$ and the red blocks to $b$. Stacked together they give the fingerprint of $a\times b$: when you multiply numbers, the exponents add. The shaded columns $2$ and $5$ decide whether the product ends in $0$. Press Play to step $b$ upwards.
1215
$a$—
$b$—
$a\times b$—
ends in 0?—
Exercises
I
Write each number as a product of its prime factors, using powers for repeated primes.
1 $198$
2 $252$
3 $945$
4 $1001$
5 $2310$
6 $4096$
7 $6435$
8 $3332$
9 $8987$
10 $10010$
Check: type any number from 2 to 1 000 000 and see its division ladder
9 $8987$ is odd, digit sum $32$, does not end in $0$/$5$; $7$ fails; $11$ works: $8987=11\times817$, and $817=19\times43$. So $8987=11\times19\times43$.
1 In a factor tree the top number splits as $2\times A$, $A=2\times B$, $B=3\times C$ and $C=5\times7$. Find $C$, $B$, $A$ and the top number.
2 Which number has the prime factorisation $2^3\times3^2\times5^2$?
3 If $72=2^a\times3^b$, find $a$ and $b$. Why can there be only one answer?
4 Find the smallest natural number by which $1176$ must be multiplied to give a perfect square, and find the square root of the result.
5 Find the smallest natural number by which $1323$ must be multiplied to give a perfect cube, and find the cube root of the result.
6Kite festival. For a Makar Sankranti display, $1001$ kites are to be arranged in a rectangular block with the same number in every row, and more than one row and more than one kite per row. Which prime numbers of rows are possible?
7Locker code. A school locker code is a 4-digit number between $2000$ and $3000$ equal to $7\times11\times13\times p$, where $p$ is a prime. Find the code.
8Challenge. Using $72=2^3\times3^2$, explain why every factor of $72$ has the form $2^i\times3^j$ with $0\le i\le3$ and $0\le j\le2$. How many factors does $72$ have?
9Challenge. Explain why, in the prime factorisation of a perfect square, every exponent is even. Use this to show that $2\times m^2$ is never a perfect square for any natural number $m$.
💡 Show solutions▶
1 Work upwards: $C=35$, $B=3\times35=105$, $A=2\times105=210$, top $=2\times210=420=2^2\times3\times5\times7$.
2 $8\times9\times25=72\times25=1800$.
3 $72=8\times9=2^3\times3^2$, so $a=3$, $b=2$. By the Fundamental Theorem of Arithmetic the prime factorisation of $72$ is unique, so no other exponents are possible.
4 $1176=2^3\times3\times7^2$. In a perfect square every exponent is even, so we need one more $2$ and one more $3$: multiply by $2\times3=6$. Then $1176\times6=7056=2^4\times3^2\times7^2=(2^2\times3\times7)^2=84^2$. Smallest multiplier $6$, square root $84$.
5 $1323=3^3\times7^2$. For a cube every exponent must be a multiple of $3$, so multiply by one more $7$: $1323\times7=9261=3^3\times7^3=21^3$. Multiplier $7$, cube root $21$.
6 The number of rows must be a factor of $1001=7\times11\times13$. The prime factors are $7$, $11$ and $13$, so $7$ rows (of $143$), $11$ rows (of $91$) or $13$ rows (of $77$).
7 $7\times11\times13=1001$, so the code is $1001p$. We need $2000\lt1001p\lt3000$, i.e. $p$ between about $2.0$ and $3.0$; $p=2$ gives $2002$ (in range) and $p=3$ gives $3003$ (too big). Code: $2002$.
8 If $d$ divides $72$, every prime in $d$ must also be in $72$ (otherwise $72$ would have a second factorisation containing that prime), and its exponent cannot be larger than in $72$. So $d=2^i3^j$ with $i\in\{0,1,2,3\}$ and $j\in\{0,1,2\}$: $4\times3=12$ factors ($1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$).
9 If $n=p_1^{a_1}\cdots p_k^{a_k}$ then $n^2=p_1^{2a_1}\cdots p_k^{2a_k}$, and by uniqueness these are the exponents of $n^2$ — all even. In $2m^2$ the exponent of $2$ is $1+(\text{an even number})$, which is odd, so $2m^2$ is not a perfect square. (Section 1.3 turns this idea into the proof that $\sqrt2$ is irrational.)
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study
How this is examined: board papers ask you to write a prime factorisation, to read exponents from it, and to use uniqueness to decide whether a number like $a^n$ can end in a given digit. Show the factorisation clearly — that line carries the marks.
E1.Board · 1 mark The exponent of $2$ in the prime factorisation of $1440$ is (A) $3$ (B) $4$ (C) $5$ (D) $6$
💡 Show solution▶
Idea $1440=144\times10=2^4\times3^2\times2\times5=2^5\times3^2\times5$.
Trap Option (B) comes from factorising $144=2^4\times3^2$ and forgetting the $2$ hidden inside the $10$.
Answer (C) $5$.
E2.Board · 1 markAssertion (A): $12^n$ cannot end with the digit $0$ for any natural number $n$. Reason (R): The prime factorisation of $12^n$ contains only the primes $2$ and $3$. (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.
💡 Show solution▶
Check R $12^n=(2^2\times3)^n=2^{2n}\times3^n$. True.
Check A To end in $0$ a number must have both $2$ and $5$ as prime factors. $12^n$ has no $5$, and by uniqueness it cannot gain one. True — and R is exactly the reason.
Answer (A). Option (B) is tempting when a student checks each statement separately but does not ask whether R explains A.
E3.Board · 2 marks Express $3150$ as a product of its prime factors. Hence find the sum of the exponents in its prime factorisation.
E4.Board · 2 marks Check whether $14^n$ can end with the digit $0$ for any natural number $n$.
💡 Show solution▶
Condition If $14^n$ ended in $0$, it would be divisible by $10=2\times5$, so $5$ would appear in its prime factorisation.
Factorise $14^n=(2\times7)^n=2^n\times7^n$. By the Fundamental Theorem of Arithmetic this factorisation is unique, and it has no $5$.
Answer So $14^n$ can never end with the digit $0$.
E5.Board · 3 marks If $2^x\times3^y\times5^z=2250$, find $x$, $y$ and $z$, and hence the value of $2^x+3^y+5^z$. Explain why your values are the only possible ones.
Compare So $x=1$, $y=2$, $z=3$. They are the only values because the prime factorisation of $2250$ is unique (Fundamental Theorem of Arithmetic).
Answer $2^1+3^2+5^3=2+9+125=136$.
E6.Board · Case studyColour-coded crates. A warehouse in Chennai labels each crate with a number made by multiplying "colour tags": every red tag counts as $2$, blue as $3$, green as $5$ and yellow as $7$. A crate with $2$ red tags and $1$ green tag is labelled $2\times2\times5=20$. (i) A crate is labelled $2940$. How many tags of each colour does it carry? (2 marks) (ii) Can two crates with different sets of tags ever have the same label? Give a reason. (1 mark) (iii) One more red tag and one more green tag are added to the crate in (i). What is its new label, and why must it end in $0$? (1 mark)
💡 Show solution▶
(i) $2940=2^2\times3\times5\times7^2$ (check: $4\times3\times5\times49=2940$). So $2$ red, $1$ blue, $1$ green and $2$ yellow tags.
(ii) No. The label is the product of the tags, and by the Fundamental Theorem of Arithmetic a number has only one prime factorisation, so the label tells us the tags exactly.
(iii) New label $=2940\times2\times5=29400$. It now contains the factor $2\times5=10$, so it ends in $0$.
🔬 Research–Teaching NexusMini project · optional, about 45 minutes
Research question. How common are primes among the 4-digit numbers you see every day — on bus routes, vehicle number plates, house numbers and train numbers?
Why it matters. Multiplying two big primes is easy, but splitting the product back into its primes is extremely slow, even for computers. The RSA system that protects online banking, UPI apps and secure websites all over the world relies on this. Mathematicians are still hunting for patterns in the primes: the volunteer project GIMPS keeps finding record-breaking primes with millions of digits.
What to do.
Collect $30$ different 4-digit numbers from the world around you (number plates, bus or train numbers, PIN-code parts, page numbers…). Write down where each came from.
Test each number for primality using the strategy box in this section (try the primes up to $\sqrt n$; for a 4-digit number that is never more than the primes up to $97$). A spreadsheet can do the division for you: in column B type =MOD(A1,7) and look for a $0$.
For each composite number, write its prime factorisation.
Work out the fraction of your numbers that are prime, and compare it with the true fraction among all 4-digit numbers: there are $1061$ primes between $1000$ and $9999$, out of $9000$ numbers.
Find the composite number in your list with the largest smallest prime factor. Why do such numbers take the longest to factorise?
Tools: Notebook and calculator, or a spreadsheet (Google Sheets / Excel)
Hand in: A table of your $30$ numbers with their source and prime factorisation, your percentage of primes compared with the true percentage, and two sentences on why big numbers are hard to factorise.
Going further. Repeat with 2-digit numbers (there are $21$ primes between $10$ and $99$). Is the fraction of primes going up or down as the numbers get bigger?
Teacher/parent note: the true fraction of primes among 4-digit numbers is $1061/9000\approx11.8\%$, and among 2-digit numbers $21/90\approx23.3\%$ — primes thin out as numbers grow. A sample of $30$ will typically contain $2$–$6$ primes; the point is the method, not an exact match.
🧪 Self-check quiz
0/5 correct
Q1. The prime factorisation of $120$ is
Q2. The Fundamental Theorem of Arithmetic says that every composite number
Q3. For a natural number $n$, which of these can never end with the digit $0$?
Q4. Counting repeated primes, how many prime factors does $72$ have?