Prime factorization — factor trees, ladder method: Question bank

Before we start, here is every word and symbol this page leans on. Each is spelled out in plain words and anchored to a picture, so nothing is used before it is earned.

Figure — Prime factorization — factor trees, ladder method

True or false — justify

TF1. "

1

is the smallest prime number."

False — a prime needs exactly two distinct divisors;

1

has only one (

1

itself), so it is a unit (a number dividing everything), not a prime. The smallest prime is

2

. See Fundamental Theorem of Arithmetic.

TF2. "

2

is the only even prime."

True — any other even number has

2

as a divisor besides

1

and itself, giving it three-plus divisors, so it can't be prime.

2

is even yet still has only

{1, 2}

TF3. "Every number greater than

1

is either prime or a product of primes."

True — that's exactly the existence half of FTA. If it isn't prime it splits into smaller factors, and repeating drives everything down to prime atoms.

TF4. "Two different factor trees for

72

can give genuinely different prime factorizations."

False — the shape of the tree differs, but the multiset of leaves is always

{2, 2, 2, 3, 3}

. Uniqueness (FTA) forbids two different bags.

TF5. "If a number is odd, it has no factor of

2

, so the ladder should skip straight to

3

True — odd means not divisible by

2

, so trying

2

(dividing out the smallest prime) wastes a step. But you must still test

3, 5, 7, \dots

in order, not guess.

TF6. "A number and its prime factorization always have the same number of divisors."

True — they are the same number written two ways; a rewrite can't change

d (n)

, the count of its divisors. The factored form just makes the count easy: Number of Divisors.

TF7. "

p^{2}

for a prime

p

always has exactly

3

divisors."

True — divisors are

1, p, p^{2}

, i.e.

d (p^{2}) = (2 + 1) = 3

choices of exponent from

0

2

. This holds for every prime, e.g.

49 = 7^{2}

has divisors

1, 7, 49

TF8. "Multiplying two primes always gives a number with exactly

4

divisors."

True when the primes are distinct (

pq

gives

1, p, q, pq

). If they're equal you get

p^{2}

with only

3

divisors, so the word "distinct" matters.

TF9. "The factorization

60 = 2 \cdot 2 \cdot 3 \cdot 5

is wrong because it isn't written with exponents."

False — it's fully correct, just uncompressed. Exponent form

2^{2} \cdot 3 \cdot 5

is only preferred because divisor, HCF and LCM formulas read exponents directly.

Spot the error

SE1. "

1

divides

6

, so

6 = 1 \cdot 2 \cdot 3

is a prime factorization."

Error:

1

is a unit, not prime, so it can't be a factor in a prime factorization. Allowing it would destroy uniqueness (

6 = 1 \cdot 2 \cdot 3 = 1 \cdot 1 \cdot 2 \cdot 3 = \dots

SE2. "Ladder for

60

: I divided by

6

then by

10

— quick!"

Error: ladder rungs must be primes (you divide out the smallest prime that fits).

6

and

10

are composite, so you get composite pieces, not atoms. Splitting by

6

is legal only in a tree, where you split the

6

again.

SE3. "

45

is odd, and I can't halve it, so

45

must be prime."

Error: odd

\neq =

prime. Test the next primes —

45 = 3^{2} \cdot 5

. Stopping at "can't divide by

2

" is the classic stop-too-early trap.

SE4. "

77

isn't even and isn't a multiple of

3

5

, so

77

is prime."

Error: you stopped before testing

7

. Since

7^{2} = 49 < 77

, keep going:

77 = 7 \times 11

. Only stop trialling once the prime squared exceeds the number.

SE5. "

36 = 2^{2} \cdot 3

because

36 = 4 \times 9

and

4 = 2^{2}

9 = 3^{2}

… so exponents

2

and

1

Error: the writer kept

3^{2}

's base but dropped its exponent.

9 = 3^{2}

, so

36 = 2^{2} \cdot 3^{2}

. Always carry both exponents from every branch.

SE6. "

100 = 1 0^{2}

, and

10

is a factor, so

100

's primes are

10

and

10

Error:

10

is not prime. Break it:

10 = 2 \times 5

, so

100 = (2 \cdot 5)^{2} = 2^{2} \cdot 5^{2}

. Never leave a composite on the leaf pile.

SE7. "For HCF (Highest Common Factor) I took the max exponent of each prime because it should be big."

Error: the HCF must divide both numbers, so it can't use more of a prime than the poorer number owns — that's the min exponent. Max exponents give the LCM (Lowest Common Multiple). See HCF and LCM.

SE8. "

d (12)

(the divisor count of

12

): since

12 = 2^{2} \cdot 3

, divisors

= 2 \times 1 = 2

Error: the formula adds

1

to each exponent (a prime can appear

0

up to

a_{i}

times). So

d (12) = (2 + 1) (1 + 1) = 6

, matching

1, 2, 3, 4, 6, 12

Why questions

WY1. Why does dividing by the smallest prime each time (the ladder rule) still land on the correct factorization?

Because FTA guarantees the final bag of primes is fixed; the order you extract them can't change the multiset, so "smallest first" is just a safe, no-guess route.

WY2. Why do we stop trial-dividing once the trial prime's square exceeds the remaining number?

If the leftover had a prime factor

p

with

p^{2} >

leftover, then leftover

= p \times (something < p)

, but that smaller factor would itself have a prime you'd already tested. So the leftover must itself be prime. See the Sieve of Eratosthenes for the same logic.

WY3. Why must

1

(a unit) be excluded from the primes for FTA's uniqueness to survive?

Because you could sprinkle any number of

1

's into a product without changing its value, giving infinitely many "factorizations." Excluding units pins the representation to exactly one.

WY4. Why does the divisor count

d (n)

multiply the

(a_{i} + 1)

terms instead of adding them?

Each prime's exponent is chosen independently from

0

a_{i}

. Independent choices multiply (like outfits from shirts × trousers), so the counts combine by product, not sum.

WY5. Why can HCF (Highest Common Factor)

\times

LCM (Lowest Common Multiple) equal the product of the two numbers?

For each prime,

min (a, b) + max (a, b) = a + b

. Summing exponents corresponds to multiplying, so combining HCF and LCM rebuilds exactly

n \times m

. See HCF and LCM.

WY6. Why does a prime factorization instantly tell you if a fraction is already in lowest terms?

Lowest terms means numerator and denominator share no prime; if their bags of primes are disjoint, their HCF

= 1

and nothing cancels. See Simplifying Fractions.

WY7. Why is the ladder method "safer under exam pressure" than the tree?

The ladder removes all choice — you always divide out the next smallest prime — so there's no risk of a lucky-but-wrong split or a lost branch. The tree's freedom is also its trap.

Edge cases

EC1. What is the prime factorization of a prime itself, say

13

Just

1 3^{1}

— a prime is a product of one prime (itself). Its bag is

{13}

; the ladder ends immediately with

13

dividing

13

, then quotient

1

EC2. Does

0

have a prime factorization?

No — FTA only speaks about integers

n > 1

0

is divisible by every prime, so no finite bag of primes multiplies to it; it's outside the theorem.

EC3. Does

1

have a prime factorization?

Yes, the empty product — a bag with no primes, whose product is defined as

1

. That's why

d (1) = 1

(only divisor is itself) and

1

is a unit, neither prime nor composite.

EC4. How many divisors does a prime

p

have, and why is that the boundary case of the formula?

Exactly

2

p = p^{1}

gives

d (p) = (1 + 1) = 2

, namely

1

and

p

. This is the smallest possible divisor count above

1

— the very definition of prime.

EC5. If two numbers share no prime factors, what are their HCF and LCM?

Their HCF (Highest Common Factor)

= 1

(empty overlap, min

= 0

for every prime), and their LCM (Lowest Common Multiple)

=

their product (take max, i.e. keep every prime from both). Such numbers are called coprime. See HCF and LCM.

EC6. Is

2^{0}

ever "part" of a factorization?

Only implicitly — a prime absent from a number has exponent

0

, and

p^{0} = 1

contributes nothing. We simply omit it, e.g.

5^{0}

is why

5

doesn't appear in

84 = 2^{2} \cdot 3 \cdot 7

EC7. Can a perfect square have an odd number of divisors, and what's special about it?

Yes — perfect squares are exactly the numbers with an odd

d (n)

, because every exponent

a_{i}

is even, so each

(a_{i} + 1)

is odd and their product is odd. Non-squares always have divisors in pairs, hence even counts.

EC8. How do we prime-factorize a negative integer like

- 60

Pull out the unit

- 1

first:

- 60 = (- 1) \cdot 2^{2} \cdot 3 \cdot 5

. The unit

- 1

carries the sign and is not a prime; the prime bag

{2, 2, 3, 5}

is identical to that of

+ 60

. FTA extends to integers "up to units," which is exactly the role

\pm 1

play.

Recall One-line summary of the traps

Never let $1$ , $0$ , or a composite sneak into a prime bag; the unit $\pm 1$ only carries a sign, never a factor; stop the ladder only at quotient $1$ or once the trial prime squared passes the leftover; add $1$ to exponents for $d (n)$ , take min for HCF and max for LCM. Every "different" factorization is the same multiset by FTA.