Factors and multiples — all factors of a number, factor pairs: Worked examples

Two symbols we lean on, defined in plain words before we use them:

The scenario matrix

Every factor problem you will ever meet lands in one of these cells. The rest of the page fills each cell with a fully worked example.

#	Case class	What's special about it	Example
A	Small even composite	ordinary pairs, quick scan	Ex 1 ( $n = 24$ )
B	Perfect square	middle pair $(n, n)$ counted once → odd factor count	Ex 2 ( $n = 36$ )
C	Prime	only $(1, n)$ — the degenerate minimum	Ex 3 ( $n = 17$ )
D	The two edge inputs $n = 1$ and $n = 0$	limiting / degenerate: fewest and "infinite" factors	Ex 4
E	Odd composite (no even factors)	tests that you don't assume $2$ is a factor	Ex 5 ( $n = 45$ )
F	Larger number, $n$ not whole	shows exactly where to stop the scan	Ex 6 ( $n = 90$ )
G	Real-world word problem	translate objects → factor pairs	Ex 7 (72 chairs)
H	Exam twist — reverse question	given the factor count, find $n$	Ex 8

Cases A–H together touch every corner: even/odd, prime/composite, perfect square, the degenerate inputs $0$ and $1$ , a big number where the stopping point matters, a word problem, and a backwards exam question.

Case A — Small even composite

Worked example Example 1 — All factors of

$24$ (cell A) List every factor of $24$ and every factor pair.

Forecast: Before reading on, guess: how many factors do you think $24$ has? Even? Odd? Where should the scan stop?

Step 1 — Find the stopping point. $24 \approx 4.9$ , so $⌊ 24 ⌋ = 4$ . We scan $d = 1, 2, 3, 4$ . Why this step? Every factor pair has a member $\leq n$ (if both were bigger their product would exceed $n$ ). So scanning to $4$ catches every pair — the big partners appear automatically.

Step 2 — Test each $d$ and record both partners.

$d$	$24 \div d$	exact?	pair
1	24	✅	$(1, 24)$
2	12	✅	$(2, 12)$
3	8	✅	$(3, 8)$
4	6	✅	$(4, 6)$

Why this step? Each tick hands us two factors at once — the buddy $n / d$ comes free.

Step 3 — Collect the factors in order. $1, 2, 3, 4, 6, 8, 12, 24$ — eight factors.

Verify: Multiply each pair back: $1 \times 24 = 2 \times 12 = 3 \times 8 = 4 \times 6 = 24$ . ✅ Count is even (8), which is what we expect for a number that is not a perfect square — every factor has a distinct buddy.

Figure — Factors and multiples — all factors of a number, factor pairs

The figure above shows the factor-pair rectangles of $24$ : each pair $(a, b)$ is a rectangle of $a$ rows and $b$ columns, all built from the same $24$ tiles. Notice how $(1, 24)$ and $(4, 6)$ are the same tiles rearranged — that's why partners come together.

Case B — Perfect square

Worked example Example 2 — All factors of

$36$ (cell B) List all factors of $36$ .

Forecast: $36 = 6 \times 6$ . Do you expect an even or odd number of factors this time? Why might it differ from $24$ ?

Step 1 — Stopping point. $36 = 6$ exactly, so $⌊ 36 ⌋ = 6$ . Scan $d = 1 \dots 6$ . Why this step? Same $n$ logic — but here $n$ is itself a whole number, which is the whole point of this case.

Step 2 — Test each $d$ .

$d$	$36 \div d$	pair
1	36	$(1, 36)$
2	18	$(2, 18)$
3	12	$(3, 12)$
4	9	$(4, 9)$
6	6	$(6, 6)$ ← same number

(We skip $d = 5$ : $36 \div 5 = 7.2$ , not exact.) Why this step? At $d = 6$ the two partners collide: $6$ and $36/6 = 6$ are the same. A factor paired with itself.

Step 3 — Write the middle factor once. Factors: $1, 2, 3, 4, 6, 9, 12, 18, 36$ — nine. Why this step? $6$ is one factor, not two. That single un-paired factor is exactly why the total is odd.

Verify: $1 \times 36 = 2 \times 18 = 3 \times 12 = 4 \times 9 = 6 \times 6 = 36$ . ✅ Nine factors, odd — the signature of a perfect square.

Recall Why

only perfect squares have an odd factor count Factors march in pairs $(d, n / d)$ . The only way one gets left un-partnered is $d = n / d$ , i.e. $d = n$ — which needs $n$ to be a whole number. So an odd factor count $⟺$ $n$ is a perfect square.

Case C — Prime (the degenerate minimum)

Case D — The edge inputs $n = 1$ and $n = 0$

Worked example Example 4 — Factors of

$1$ and of $0$ (cell D, degenerate) How many factors does $1$ have? How many does $0$ have?

Forecast: Guess before scanning. One of these has the fewest factors of any number; the other has infinitely many. Which is which?

Part (i) — $n = 1$ . Only whole number $d \geq 1$ with $d ∣ 1$ is $d = 1$ itself ( $1 = 1 \times 1$ ). Why? Any $d > 1$ gives $1 \div d < 1$ , never a whole number. So $1$ has exactly one factor. (Note: $1$ is neither prime nor composite — it fails the "exactly two factors" test.)

Part (ii) — $n = 0$ . Ask: which $d$ satisfy $0 = d \times k$ for some whole $k$ ? Take $k = 0$ and any $d$ : $d \times 0 = 0$ . ✅ Why? Every whole number divides $0$ (remainder is always $0$ ). So $0$ has infinitely many factors — the opposite extreme. This is why $n$ scanning and factor lists are only ever used for $n \geq 1$ .

Verify: $1$ has $1$ factor (odd count → consistent, since $1 = 1^{2}$ is a perfect square). $0$ : pick any $d$ , say $d = 7$ ; $0 \div 7 = 0$ exact. ✅ Infinitely many factors confirmed by counterexample-proof: no largest $d$ exists.

Case E — Odd composite (don't assume $2$ divides)

Worked example Example 5 — All factors of

$45$ (cell E) List every factor of $45$ .

Forecast: $45$ is odd. Will $2$ or $4$ ever show up as a factor? Guess the count.

Step 1 — Stopping point. $45 \approx 6.7$ , so $⌊ 45 ⌋ = 6$ . Scan $d = 1 \dots 6$ .

Step 2 — Test each.

$d$	$45 \div d$	exact?	pair
1	45	✅	$(1, 45)$
2	22.5	✗	— (odd number, no even factor)
3	15	✅	$(3, 15)$
4	11.25	✗	—
5	9	✅	$(5, 9)$
6	7.5	✗	—

Why this step? An odd $n$ can never have an even factor: an even factor would force $n$ to be even. This is a mini divisibility rule in action — so $d = 2, 4, 6$ are guaranteed misses, but we still list them to show why they fail.

Step 3 — Collect. Factors: $1, 3, 5, 9, 15, 45$ — six factors (even count → not a perfect square, correct since $45$ isn't one).

Verify: $1 \times 45 = 3 \times 15 = 5 \times 9 = 45$ . ✅ No even factor anywhere — matches "odd number ⇒ all factors odd."

Case F — Larger number, $n$ not a whole number

Worked example Example 6 — All factors of

$90$ (cell F) List every factor of $90$ , and show exactly where the scan stops.

Forecast: $90$ is between which two whole numbers? That gap is where beginners over- or under-scan.

Step 1 — Pin the stopping point carefully. $9^{2} = 81 \leq 90 < 100 = 1 0^{2}$ , so $9 < 90 < 10$ , giving $⌊ 90 ⌋ = 9$ . Scan $d = 1 \dots 9$ . Why this step? Stopping at $8$ would miss the pair whose small member is $9$ ; scanning to $45$ would waste effort. The floor of $n$ is the precise cut.

Step 2 — Test each.

$d$	$90 \div d$	exact?	pair
1	90	✅	$(1, 90)$
2	45	✅	$(2, 45)$
3	30	✅	$(3, 30)$
5	18	✅	$(5, 18)$
6	15	✅	$(6, 15)$
9	10	✅	$(9, 10)$

(Misses: $d = 4$ gives $22.5$ , $d = 7$ gives $\approx 12.86$ , $d = 8$ gives $11.25$ .) Why this step? $d = 9$ is the very last check, and it does divide — its partner $10$ is just above $90$ , exactly the border case that a too-early stop would drop.

Step 3 — Collect in order. $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90$ — twelve factors.

Verify: $1 \times 90 = 2 \times 45 = 3 \times 30 = 5 \times 18 = 6 \times 15 = 9 \times 10 = 90$ . ✅ Twelve factors (even → not a perfect square, correct).

The figure plots each divisor $d$ on a number line against its partner $n / d$ . Watch the two dots of a pair mirror across the vertical $90$ line (dashed): every left dot ( $\leq 90$ ) has a right partner, which is precisely why we only ever scan the left half.

Case G — Real-world word problem

Worked example Example 7 — Arranging

$72$ chairs (cell G, word problem) A hall manager has $72$ identical chairs and wants to arrange them in a perfect rectangle (equal rows). Every arrangement must use all chairs. How many different rectangular arrangements are possible, and what are they?

Forecast: Is " $8$ rows of $9$ " the same arrangement as " $9$ rows of $8$ "? Decide first — it changes the count.

Step 1 — Translate to factors. "Rectangle of $r$ rows and $c$ columns using all chairs" means $r \times c = 72$ . So $(r, c)$ is a factor pair of $72$ . Why this step? A rectangle of objects is a factor pair — this is the tiling picture from the parent note made real.

Step 2 — List factor pairs of $72$ . $72 \approx 8.49$ , scan $d = 1 \dots 8$ : $(1, 72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9)$ . (Misses: $5, 7$ .) Why this step? Six factor pairs found by the $n$ rule.

Step 3 — Count arrangements (interpret the question).

If " $8 \times 9$ " and " $9 \times 8$ " count as different orientations: each of the $6$ pairs gives $2$ arrangements ⇒ $12$ arrangements.
If rotations count as the same shape: exactly $6$ arrangements. Why this step? Word problems hinge on interpretation; we state both. $72$ has $12$ factors total, so the ordered count $=$ factor count $= 12$ .

Verify: Factors of $72$ : $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$ — that's $12$ , matching "ordered arrangements $= 12$ " and "distinct rectangle shapes $= 6$ ." Units check: chairs $\times$ chairs-per-arrangement $= 72$ chairs. ✅

Case H — Exam twist (reverse question)

Worked example Example 8 — Backwards: find

$n$ from its factor count (cell H, exam twist) A whole number $n$ between $30$ and $50$ has an odd number of factors. Find $n$ . Then, of all such $n$ , which has exactly three factors?

Forecast: What kind of number is forced to have an odd factor count? And what kind has exactly three?

Step 1 — Use the odd-count fact. Odd factor count $⟺$ $n$ is a perfect square (from Case B). Why this step? The parent note proved only perfect squares leave one factor un-paired.

Step 2 — List perfect squares in range. Between $30$ and $50$ : $36 = 6^{2}$ and $49 = 7^{2}$ . Why this step? $5^{2} = 25$ is below $30$ , $8^{2} = 64$ is above $50$ — only two qualify.

Step 3 — Find the one with exactly three factors.

$36$ : factors $1, 2, 3, 4, 6, 9, 12, 18, 36$ — nine factors.
$49$ : $49 = 7$ ; test $d = 1 \dots 7$ : only $1, 7, 49$ divide ( $7$ is prime, so no factor between $1$ and $7$ ). Factors $1, 7, 49$ — three. Why this step? A number with exactly three factors must be the square of a prime $p^{2}$ (factors $1, p, p^{2}$ ). Here $49 = 7^{2}$ fits; $36 = 6^{2}$ doesn't because $6$ isn't prime. This links straight to Prime factorisation.

Verify: $49 = 7^{2}$ , factors ${1, 7, 49}$ — count $3$ , odd, and $49$ is a perfect square. ✅ $36$ has $9$ factors, also odd, also in range — both satisfy "odd count," only $49$ satisfies "exactly three."

Wrap-up: the matrix, filled

Recall One line per case (hide and recall)

A even composite $24$ → 8 factors ::: even count, four distinct pairs. B perfect square $36$ → 9 factors ::: odd count, middle pair $(6, 6)$ once. C prime $17$ → 2 factors ::: degenerate minimum, only $(1, 17)$ . D edge $n = 1$ → 1 factor; $n = 0$ → infinitely many ::: the two extremes. E odd composite $45$ → 6 factors ::: no even factor ever appears. F big number $90$ → 12 factors ::: scan stops precisely at $⌊ 90 ⌋ = 9$ . G word problem 72 chairs → 6 shapes / 12 ordered ::: rectangle = factor pair. H reverse: odd count in 30–50 → 36 and 49; exactly three factors → 49 ::: three factors = square of a prime.

Connections

Prime numbers and composite numbers — cells C and H turn on factor counts.
Prime factorisation — "exactly three factors $= p^{2}$ " comes from here.
Perfect squares — the odd-count signature in cells B, D, H.
HCF and LCM — built from the factor lists you now generate flawlessly.
Divisibility rules — the "odd number has no even factor" shortcut in cell E.
Fractions in lowest terms — cancelling shared factors of a pair.