Factors and their count

Take any whole number — say $180$ — and ask a simple question: which numbers divide it exactly, with nothing left over? The answer is its factors, and learning to find them all, without missing one or counting one twice, is the very first skill needed before a Class 8 student can talk about perfect squares and perfect cubes. This topic opens the chapter "A Square and A Cube." Before recognising whether a number is a perfect square or a perfect cube, a student must be comfortable finding a number's prime factorisation — writing it as a product of primes only — because that factorisation is what later gets grouped into pairs (for squares) or triples (for cubes). Along the way there is a neat shortcut: once you know the prime factorisation, you can count how many factors a number has without listing a single one.

The core insight

Every number's prime factorisation is unique, and adding one to each prime's exponent and multiplying those results together counts its total number of factors without listing any of them.

The core idea

A factor of a number $n$ is a whole number that divides $n$ exactly, leaving remainder $0$ . Every number greater than $1$ has at least two factors: $1$ and itself.

A number greater than $1$ with exactly two factors is a prime number (e.g. $2, 3, 5, 7, 11, \dots$ ).
A number with more than two factors is a composite number (e.g. $4, 6, 8, 9, \dots$ ).
The number $1$ is neither prime nor composite — it has only one factor.

Prime factorisation writes a composite number as a product of primes only, found by repeated division by the smallest possible prime or by drawing a factor tree:

36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2

This form is unique for every number, except for the order in which the primes are written — a fact called the Fundamental Theorem of Arithmetic. Because it is unique, the exponents in the factorisation carry real information: grouping every prime into a pair (even exponent) signals a perfect square, and grouping every prime into a triple (exponent a multiple of $3$ ) signals a perfect cube — the very test the rest of the chapter builds on. The exponents also unlock a shortcut for counting factors without listing them: add $1$ to each exponent and multiply the results. This works because choosing a factor of $n$ is the same as choosing, independently, how many copies of each prime to include. Two habits make this reliable in an exam. First, always divide by the smallest prime that fits before moving to the next one — jumping straight to a large prime often skips a smaller factor still hiding inside the quotient. Second, write the final factorisation with exponents rather than as a long string of repeated primes; it is the exponent form that plugs directly into the counting formula and into the square/cube tests later in the chapter.

Figure

Factor tree for one hundred eighty splitting into two two three three five with the count formula three times three times two equals eighteen

Derivation ACounting the factors of a number from its prime factorisation

Show that if a number $n$ is written in prime-factorised form as $n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k}$ , then the total number of whole-number factors of $n$ , including $1$ and $n$ itself, equals the product $(a_1+1)(a_2+1)\cdots(a_k+1)$ , without ever having to list the factors individually.

Step 1. Every factor of $n$ must be built only from the primes $p_1, p_2, \dots, p_k$ that appear in $n$ 's factorisation, because any other prime dividing a factor would have to divide $n$ too, and $n$ has no such prime. So a general factor has the form $p_1^{b_1} \times p_2^{b_2} \times \cdots \times p_k^{b_k}$ .

Step 2. For the factor to actually divide $n$ , each exponent $b_i$ cannot exceed the exponent $a_i$ present in $n$ — you cannot pull out more copies of a prime than $n$ itself contains. So $b_i$ can be any whole number from $0$ up to $a_i$ .

Step 3. Count the choices for a single prime $p_i$ : the exponent $b_i$ can be $0, 1, 2, \dots, a_i$ , which is $a_i + 1$ possible values (remember to count $0$ itself as a choice — it means "leave this prime out entirely").

Step 4. Since the choice of exponent for each prime is made independently of the choices for the other primes, and each distinct combination of exponents gives a distinct factor, the total number of factors is the product of the individual counts: $(a_1+1) \times (a_2+1) \times \cdots \times (a_k+1)$ .

\text{number of factors of } n = (a_1+1)(a_2+1)\cdots(a_k+1)

where: $p_1, \dots, p_k$ are the distinct prime factors of $n$ , and $a_1, \dots, a_k$ are their respective exponents in the prime factorisation.

Worked example

Find the prime factorisation of $180$ , count its total number of factors, and check whether it is a perfect square.

Step 1. Divide $180$ repeatedly by the smallest prime that fits:

180 \div 2 = 90, \quad 90 \div 2 = 45, \quad 45 \div 3 = 15, \quad 15 \div 3 = 5, \quad 5 \div 5 = 1

Step 2. Collect the primes used: $180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$ .

Step 3. Apply the counting formula from the derivation above, adding $1$ to each exponent and multiplying:

(2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18

Step 4. Check the perfect-square condition — every exponent must be even. Here $2^2$ and $3^2$ are fine, but $5^1$ has an odd exponent, so $180$ is not a perfect square, even though it has a large, even total count of $18$ factors. $180$ has prime factorisation $2^2 \times 3^2 \times 5^1$ , exactly 18 factors, and is not a perfect square (listing them confirms it: $1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180$ ).

Competitive stretch

For $360 = 2^3 \times 3^2 \times 5^1$ , the same formula gives $(3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24$ factors — but since neither $3$ nor $2$ nor $1$ are all even, $360$ is neither a perfect square nor a perfect cube, despite having more factors than $180$ .

Common traps

Confusing "factors" with "multiples." Factors of $12$ are numbers that divide $12$ ( $1,2,3,4,6,12$ ); multiples of $12$ are numbers $12$ divides into ( $12, 24, 36, \dots$ ) — opposite directions entirely.
Forgetting the $+1$ in the counting formula. The number of factors is $(a+1)$ , not $a$ , because the exponent $0$ (leaving that prime out) is itself a valid choice.
Mixing up "factor count is even" with "is a perfect square." A large even total, like $180$ 's $18$ factors, says nothing about whether each individual exponent is even — that is a separate check, done exponent by exponent.
Stopping the factor tree too early, leaving a composite number like $4$ at the end of a branch instead of continuing to split it into primes ( $2 \times 2$ ), which would silently undercount the factors.

Ask the tutor: if you are stuck building a factor tree or want more practice counting factors for a tricky number, ask the AI tutor to walk through it with you step by step.

Why does adding 1 to each exponent give the factor count instead of just multiplying the exponents

Each prime's exponent in a factor can range from $0$ up to the exponent in $n$ itself — that is $a_i + 1$ separate choices, not $a_i$ , because $0$ is a valid choice too (it means that prime is left out of the factor). Multiplying the exponents directly would silently drop the "leave it out" option for every prime, undercounting every time.

Does a bigger number always have more factors than a smaller one

No. Factor count depends on the pattern of the prime factorisation, not the size of the number. A number like $2^{10} = 1024$ has only $11$ factors (since $10+1=11$ ), while a smaller number like $360 = 2^3\times3^2\times5$ has $24$ factors, because it spreads its prime power across more distinct primes, giving more ways to combine them.

If a number has an odd number of factors, does that mean anything special

Yes — a number has an odd total number of factors exactly when it is a perfect square. This happens because $(a_1+1)(a_2+1)\cdots$ is odd only when every single factor $(a_i+1)$ is odd, which means every $a_i$ is even, which is precisely the perfect-square condition. It is a quick sanity check, though the full test still needs each exponent examined individually.

Factors and their count

The core idea

Worked example

Competitive stretch

Common traps

Where this leads