Prime factorisation

Every whole number bigger than $1$ is built from a small, unchanging set of "atoms": the prime numbers. Split $360$ apart piece by piece — halve it, halve it again, keep dividing by the smallest prime that fits — and you always land on the same six primes, however you split it: three $2$ s, two $3$ s, one $5$ . That fact, stated properly, is called the Fundamental Theorem of Arithmetic, and the everyday tool for using it is prime factorisation. In Class 8's "Power Play" chapter, prime factorisation is the bridge into exponential notation. Once a number is broken into its primes, the repeats can be grouped and written as powers — $360 = 2^3\times3^2\times5$ — turning long strings of multiplication into a compact, comparable form. This article rebuilds the factor tree and division methods from scratch, derives exactly why the exponent equals the number of divisions by a prime, and works through a full example end to end.

The core insight

Every composite number breaks down into exactly one set of prime factors, and grouping the repeats of that unique set gives the number's exponential (power) form.

The core idea

A prime number is a whole number greater than $1$ whose only factors are $1$ and itself — $2, 3, 5, 7, 11, 13, \ldots$ . A composite number has at least one other factor besides $1$ and itself, which means it can be split further. Prime factorisation is the process of splitting a composite number all the way down until every piece left is prime.

Why it always ends the same way. However you choose to split a number first, the final list of prime factors you land on is always identical (just possibly in a different order). This is the Fundamental Theorem of Arithmetic: every composite number greater than $1$ has exactly one prime factorisation. At this stage it is used as a working fact rather than proved, but it is why prime factorisation is a reliable tool — there is never a "wrong" tree that gives a different valid answer.

Two equivalent methods.

Factor tree: split the number into any two factors, then split each of those further, branch by branch, stopping only when a branch ends in a prime.
Division method: divide the number repeatedly by the smallest prime that fits — try $2$ first, then $3$ , then $5$ , then $7$ , and so on — until the quotient reaches $1$ .

Once every prime factor is listed, repeated primes are grouped using exponents, since an exponent is simply shorthand for repeated multiplication of the same base:

N = p_1^{a_1}\times p_2^{a_2}\times\cdots\times p_k^{a_k}

where: $N$ is the original composite number, $p_1, p_2,\ldots,p_k$ are the distinct primes that divide $N$ , and each exponent $a_i$ counts how many times $p_i$ appears in the full factor list — for $360$ , this reads $p_1=2,\ a_1=3$ ; $p_2=3,\ a_2=2$ ; $p_3=5,\ a_3=1$ . Quick divisibility checks make the division method fast: a number is divisible by $2$ if it is even, by $3$ if its digit sum is a multiple of $3$ , and by $5$ if it ends in $0$ or $5$ . Testing primes strictly in increasing order — $2, 3, 5, 7, 11,\ldots$ — guarantees no factor is missed and none is double-counted.

Figure

Factor tree splitting 360 down to its prime leaves two three three five

Derivation AWhy the exponent counts the number of divisions

The goal is to show, using nothing beyond repeated division, exactly why the exponent attached to a prime in the factorised form of $360$ equals the number of times that prime divides evenly before the quotient stops being a multiple of it — so that grouping factors into exponential notation is never a guess but a direct count.

Step 1. Start dividing $360$ by the smallest prime, $2$ , and keep going as long as the result stays whole: $360\div2=180$ , $180\div2=90$ , $90\div2=45$ . At this point $45$ is odd, so $2$ can no longer divide it evenly — the chain of divisions by $2$ has stopped after exactly three steps.

Step 2. Each of those three divisions removed one factor of $2$ from the number, which means $360$ contains exactly three factors of $2$ multiplied together: $360 = 2\times2\times2\times45 = 2^3\times45$ . The exponent $3$ is not chosen; it is simply the tally of how many successful divisions by $2$ happened.

Step 3. Repeat the same counting on the leftover $45$ with the next prime, $3$ : $45\div3=15$ , $15\div3=5$ . Two successful divisions, and $5$ is not divisible by $3$ , so the tally for $3$ is exactly $2$ : $45 = 3^2\times5$ .

Step 4. Combine both tallies: $360 = 2^3\times3^2\times5$ . Since $5$ divides itself once and stops, its tally is $1$ , written as the bare prime with an understood exponent of $1$ . In general, for any prime $p$ dividing $N$ , its exponent $a$ in $N=p^a\times(\text{rest})$ is exactly the count of consecutive successful divisions of $N$ by $p$ — which is precisely what the division method tracks step by step.

Worked example

Find the prime factorisation of $1000$ , write it in exponential notation, and use the result to confirm $1000=10^3$ .

Step 1. Divide $1000$ by the smallest prime, $2$ , repeatedly: $1000\div2=500$ , $500\div2=250$ , $250\div2=125$ . Three successful divisions, and $125$ is odd, so the tally for $2$ is $3$ .

Step 2. Move to the next prime that fits, $5$ (since $3$ does not divide $125$ : its digit sum is $8$ , not a multiple of $3$ ): $125\div5=25$ , $25\div5=5$ , $5\div5=1$ . Three successful divisions, so the tally for $5$ is also $3$ .

Step 3. Collect the tallies into exponential form: $1000 = 2^3\times5^3$ .

Step 4. Regroup the same six factors differently — pair each $2$ with a $5$ instead of keeping all the $2$ s together — to see a second useful pattern: $2^3\times5^3=(2\times5)\times(2\times5)\times(2\times5)=10\times10\times10=10^3$ . Answer: $1000=2^3\times5^3=10^3$ .

Competitive stretch

Without fully factorising, how many prime factors (counted with repetition) does $2^3\times5^3$ have in total? Add the exponents directly: $3+3=6$ prime factors in total, matching the six leaves $2,2,2,5,5,5$ found by the tree or division method — a useful shortcut for checking a factorisation's length before writing every factor out.

Common traps

Stopping the factor tree too early, leaving a composite number like $4$ or $9$ at the end of a branch instead of splitting until every leaf is prime.
Dropping a prime that appears only once, e.g. writing $2^3\times3^2$ for $360$ and forgetting the lone $5$ — a bare prime always means an exponent of $1$ .
Miscounting repeats when converting to exponential form, such as writing $2^2\times3^2$ for $2\times2\times2\times3\times3$ , which actually has three $2$ s, not two. Tally each prime carefully before writing its exponent.
Dividing by a non-prime number (like $4$ or $6$ ) during the division method, which can hide the true count of a prime factor — always test primes in increasing order.

Ask the tutor: If a factor tree branch or a division step ever feels stuck, ask the AI tutor to check your work one division at a time.

Why does it not matter whether I use a factor tree or repeated division?

Both methods are just two different bookkeeping styles for the same underlying fact: the Fundamental Theorem of Arithmetic guarantees a composite number has exactly one set of prime factors. A factor tree lets you split in any order you like — starting with $360=36\times10$ or $360=8\times45$ both work — while the division method forces a fixed order, smallest prime first. Whichever path you take, once every leaf or quotient is prime, you collect the identical multiset of primes. Use whichever method is faster for the number in front of you; a tree feels natural for numbers with obvious factor pairs, division is more mechanical for large numbers.

Do I always have to start dividing by 2?

No, but it is the safest habit. Starting with the smallest prime that actually divides the number and working upward in order — $2$ , then $3$ , then $5$ , then $7$ — guarantees you never skip a factor or repeat one unnecessarily. If a number is odd, $2$ simply will not divide it evenly, and you move straight to $3$ ; the method self-corrects. Jumping around (trying $7$ before checking $2$ or $3$ ) still gives a correct final answer eventually, but it is far more likely to cause a missed or double-counted factor, especially under exam time pressure.

How is prime factorisation different from just listing all the factors of a number?

Listing all factors of $360$ gives every number that divides it evenly — $1, 2, 3, 4, 5, 6, 8, \ldots, 360$ — including composite ones like $4$ and $6$ . Prime factorisation is narrower and stricter: it lists only the prime building blocks, with repeats, whose product recreates the number, namely $2,2,2,3,3,5$ for $360$ . Every other factor of $360$ , including $4$ , $6$ , and $360$ itself, can be built by multiplying some combination of those six primes together — which is exactly why the prime factorisation is called the number's fundamental "recipe."

Prime factorisation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads