Prime factorization — factor trees, ladder method: Worked examples

The scenario matrix

Before working examples, let's list every class of situation prime factorization throws at you. Each row is a distinct "shape" of problem; the last column names the example that lands on it.

Cell	The scenario	Why it's its own case	Example
A	Small even composite	Warm-up: keep halving	Ex 1
B	Perfect square	Every exponent turns out even	Ex 2
C	Number ending in 0 (two-tailed)	Has both $2$ and $5$ as factors	Ex 3
D	Large number, many factors	Ladder gets long; forecast tested	Ex 4
E	Prime disguised as composite	Ladder fails every small prime → number is itself prime	Ex 5
F	Odd number with a big prime hidden	Must test up to $n$	Ex 6
G	Power of a single prime	Only one prime, high exponent	Ex 7
H	Word problem (real-world)	Translate story → factorize → HCF/LCM	Ex 8
I	Exam twist: divisor-count backwards	Given $d (n)$ , reason about exponents	Ex 9

Ex 1 — Cell A · small even composite

Worked example Factorize

$72$ . Forecast: $72$ is even and looks "very halvable". Guess: how many times can you halve it before hitting an odd number? Write down your exponent of $2$ before reading on.

Ladder:

$72 \div 2 = 36$ . Why this step? $72$ is even → smallest prime $2$ fits. Keep going while even.
$36 \div 2 = 18$ , $18 \div 2 = 9$ . Why stop halving? $9$ is odd — $2$ no longer divides it.
$9 \div 3 = 3$ , $3 \div 3 = 1$ . Why $3$ now? Smallest prime dividing the odd $9$ is $3$ .
Reached $1$ → stop. $72 = 2^{3} \cdot 3^{2}$

Verify: $2^{3} \cdot 3^{2} = 8 \times 9 = 72$ . ✅ Halved three times, matching the $2^{3}$ .

Ex 2 — Cell B · a perfect square

Worked example Factorize

$196$ and show it is a perfect square from its factorization. Forecast: $196 = 1 4^{2}$ (since $14 \times 14 = 196$ ). If a number is a square, what must be true of every exponent in its factorization? Guess before checking.

Ladder:

$196 \div 2 = 98$ , $98 \div 2 = 49$ . Why? $196$ even, halve twice; $49$ is odd, stop halving.
$49 \div 7 = 7$ , $7 \div 7 = 1$ . Why $7$ ? $49$ isn't divisible by $3$ or $5$ ; next prime $7$ works ( $7^{2} = 49$ ). $196 = 2^{2} \cdot 7^{2}$

Verify: $2^{2} \cdot 7^{2} = (2 \cdot 7)^{2} = 1 4^{2} = 196$ . ✅ The lesson (see figure): every exponent is even ( $2$ and $2$ ). A number is a perfect square iff all exponents in its prime factorization are even — because a square splits into two identical halves.

Ex 3 — Cell C · ends in $0$ (has both tails)

Worked example Factorize

$150$ . Forecast: Ending in $0$ means divisible by $10 = 2 \times 5$ . So both $2$ and $5$ must appear. Guess the full factorization.

Ladder:

$150 \div 2 = 75$ . Why? $150$ even → one factor of $2$ . But $75$ is odd → only one $2$ .
$75 \div 3 = 25$ . Why $3$ ? $7 + 5 = 12$ is divisible by $3$ (divisibility rule), and $75$ is odd so skip $2$ .
$25 \div 5 = 5$ , $5 \div 5 = 1$ . Why $5$ ? $25 = 5^{2}$ . $150 = 2 \cdot 3 \cdot 5^{2}$

Verify: $2 \cdot 3 \cdot 5^{2} = 6 \times 25 = 150$ . ✅ One trailing zero ⇒ exactly one pair of $(2, 5)$ ; the extra $5$ is why there's a $5^{2}$ but only one $2$ .

Ex 4 — Cell D · large, many factors (forecast tested)

Worked example Factorize

$2520$ . Forecast: $2520$ is famously divisible by lots of small numbers. Before the ladder, forecast which of $2, 3, 5, 7$ appear.

Ladder:

Three halvings: $2520 \to 1260 \to 630 \to 315$ . Why stop at $315$ ? It's odd.
$315 \div 3 = 105$ , $105 \div 3 = 35$ . Why two $3$ s? $3 + 1 + 5 = 9$ divisible by $3$ ; then $1 + 0 + 5 = 6$ divisible by $3$ .
$35 \div 5 = 7$ . Why $5$ ? ends in $5$ . Then $7 \div 7 = 1$ . $2520 = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7$

Verify: $8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520$ . ✅ Bonus (divisor count): $d = (3 + 1) (2 + 1) (1 + 1) (1 + 1) = 4 \cdot 3 \cdot 2 \cdot 2 = 48$ divisors — that's why $2520$ feels "divisible by everything".

Ex 5 — Cell E · a prime disguised as composite

Ex 6 — Cell F · odd number, big prime hidden

Ex 7 — Cell G · a pure prime power

Ex 8 — Cell H · real-world word problem

Worked example Two lighthouses flash: one every

$60$ s, one every $84$ s. They flash together at midnight. (a) After how many seconds do they next flash together? (b) In a $30$ -minute window, how many simultaneous flashes (including midnight)? Forecast: "Next time together" screams LCM. Guess whether the answer is bigger or smaller than $84$ .

Factorize each. $60 = 2^{2} \cdot 3 \cdot 5$ , $84 = 2^{2} \cdot 3 \cdot 7$ . Why? LCM is built from prime powers.
LCM = take the max exponent of each prime: $2^{2} \cdot 3 \cdot 5 \cdot 7$ . Why max? The combined cycle must contain both individual cycles, so it needs the richer count of every prime. $LCM (60, 84) = 2^{2} \cdot 3 \cdot 5 \cdot 7 = 420 s = 7 min .$
(b) $30$ minutes $= 1800$ s. Simultaneous flashes occur at $0, 420, 840, 1260, 1680$ s. Count $= ⌊ 1800/420 ⌋ + 1 = 4 + 1 = 5$ . Why $+ 1$ ? The flash at $t = 0$ (midnight) counts too.

Verify: $420 = 60 \times 7 = 84 \times 5$ — a whole number of each cycle. ✅ And $HCF (60, 84) = 2^{2} \cdot 3 = 12$ , so $HCF \times LCM = 12 \times 420 = 5040 = 60 \times 84$ . ✅ ( $5$ flashes in the window.)

Ex 9 — Cell I · exam twist (reason backwards)

Degenerate & edge inputs (never skip these)

Recall Rapid self-test

Factorize $150$ ::: $2 \cdot 3 \cdot 5^{2}$ Factorize $2520$ ::: $2^{3} \cdot 3^{2} \cdot 5 \cdot 7$ Is $149$ prime? Why? ::: Yes — no prime $\leq 149 \approx 12$ divides it. $221 = ?$ ::: $13 \times 17$ A number with exactly 3 divisors is always... ::: the square of a prime, $p^{2}$ LCM of $60$ and $84$ ::: $420$

Related: HCF and LCM · Number of Divisors · Fundamental Theorem of Arithmetic · Divisibility Rules · Sieve of Eratosthenes · Simplifying Fractions