Welcome! Today we're tackling Last-Digit Proofs Using Prime Factorisation — a powerful technique that settles questions about digits once and for all.

Here's the puzzle:

Can $4^{1000}$ end in the digit 0?

Checking a few values — 4, 16, 64, 256 — might suggest no, but that's not a proof. You haven't checked all values.

There's a way to settle the question for ALL $n$ at once:

Use the prime factorisation of the base
Apply the uniqueness guarantee of FTA

By the end of this section, you'll be able to:

Goal	What you'll do
Prove	Whether any power $a^{n}$ can or cannot end in 0
Understand	The general rule in terms of prime factors

1. Translating 'ends in 0' to a prime factor condition

Can $4^{1000}$ end in the digit 0?

Checking a few values:

$4^{1} = 4$
$4^{2} = 16$
$4^{3} = 64$
$4^{4} = 256$

...might suggest no, but that's not a proof. You haven't checked all values!

There's a way to settle this question definitively, and it starts with translating the digit condition into the language of prime factors.

📋 Given Info

Here's what we know:

A number ends in the digit 0 when it is divisible by 10.

Examples:

$30$ — ends in 0, divisible by 10
$140$ — ends in 0, divisible by 10
$1000$ — ends in 0, divisible by 10

✍️ Question

Your turn 🤔

If a number ends in the digit 0, what two prime factors must it have?

Explain the chain of reasoning from 'ends in 0' to the prime factor requirement.

Here's the chain of reasoning:

Step 1: 'Ends in 0What you see at the end of a number' means the number is divisible by 10What you can conclude about its factors.

This is our starting point — the connection between the last digit and divisibility.

The last digit tells you the remainderThe last digit tells you the remainder when dividing by 10 when dividing by 10. If the last digit is 0, the remainder is 0You immediately know the number divides exactly by 10, meaning exact divisibility.

Think about it: 30÷10No calculation needed divided by 10 gives remainder 0. 140 divided by 10 gives remainder 0. 1000 divided by 10 gives remainder 0. See the pattern?

✍️ MCQ

Choose one

What is the prime factorisation of

10

?

✓

Step 2: $10 = 2 \times 5$ . These are the prime factors of 10.

Step 3: For a number to be divisible by 10, it must be divisible by both 2 AND 5Need both 2 and 5 to end in zero.

You need both — being divisible by just 2 (like 8) or just 5 (like 15) is not enough.

✍️ Yes/No

Yes or No?

Does the number

140

have both

2

and

5

as prime factors?

✓

So the translation is:

Ends in 0Zero ending requires divisibility by 10 → divisible by 10Ten equals 2 times 5 in prime factors → has both 2 and 5 as prime factors.Need both primes present

✍️ MCQ

Choose one

The number

1000

ends in

0

. According to our chain of reasoning, which prime factors must

1000

have?

✓

This translation is the key moveConverts the problem into something solvable. It converts a question about digitsHard to prove for all cases (which feels hard to prove for all cases) into a question about prime factorsThese have definite answers (which FTA can handle definitivelyFTA gives you the definitive answer).

2. Proving 4 to the n cannot end in 0

You have the translation — ending in 0 means the number must be divisible by both 2 and 5.

Now let's put it to work on a specific base.

The argument combines the prime factorisation of the base with FTA uniqueness to give a complete proof.

📋 Given Info

Here's what you know:

Ending in 0 requires both 2 and 5 as prime factors
The base is 4
You need to prove that $4^n$ never ends in 0, for any natural number $n$

✍️ Question

Prove that $4^n$ can never end with the digit 0key, for any natural number $n$ .

Your proof must explicitly use the Fundamental Theorem of Arithmetic.

Here's the complete proof:

Step 1 — Factorise the base: $4 = 2^2$ .

Step 2 — Raise to the nth power: $4^{n} = (2^{2})^{n} =$ $2^{2n}$ Only powers of 2, no factor of 5.

There's no 5 anywhere!Missing the 5 needed for a trailing zero

✍️ Yes/No

Yes or No?

Based on the prime factorisation

4^n = 2^{2n}

, does

4^n

have

5

as a prime factor?

✓

Step 3 — List the prime factors: The only prime factor is 22 is the only prime factor present. The number 5 does not appearmissing!5 is absent, which matters for the proof.

Step 4 — Invoke FTA: By the Fundamental Theorem of ArithmeticFTA guarantees uniqueness of factorisation, the prime factorisation $2^{2 n}$ is the unique prime factorisationOnly one way to factor this number of $4^{n}$ . There is no other way to write this number as a product of primes. So 5 cannot appear in any factorisation — it genuinely does not divide $4^n$ 5 can never appear, so 4^n never ends in 0.

✍️ MCQ

Choose one

Why is the Fundamental Theorem of Arithmetic essential to this proof?

✓

Step 5 — Connect to the digit condition:

Ending in 0 requires divisibility by $10 = 2 \times 5$ , which requires BOTH 2 and 5 as prime factorsYou need the complete pair, not just one of them.

Since 5 is absentmissing!The Fundamental Theorem says there's no other way to factor it from the prime factorisation of $4^{n}$ , we have:

$4^{n}$ is not divisible by 10Since 5 is missing, it can never be divisible by 10, and therefore cannot end in 0That's why it can never end in 0.

✍️ Yes/No

Yes or No?

If a number's prime factorisation is

2^{100}

, can it ever end in 0?

✓

The FTA step (Step 4) is what separates a proofProof requires rigorous justification, not just observation from a pattern observationSeeing powers of 4 don't end in zero is just observation.

Without it, someone could say "maybe there's some huge $n$ Even for enormous powers, the factorisation stays the same where 5key factorIf 5 isn't there now, it can never appear sneaks into the factorisation."

FTA says noThere is exactly one way to write any number as primes — the factorisation is uniqueUniqueness means the factorisation is fixed forever, so what you see is all there isNo new primes can ever sneak in.

✍️ MCQ

Choose one

Why can't we just check a few values of

4^n

(like

4^1, 4^2, 4^3

) and conclude that

4^n

never ends in 0?

✓

3. Applying the technique to a different base

You proved it for base 4. Now let's see whether you can apply the same reasoning to a base where the answer is less obvious — and identify which prime is missing.

You've proved that $4^{n}$ cannot end in 0 because its factorisation only contains 2, with no 5.

Now consider a different base: 15new base.

✍️ Question

Your Turn 🧠

Does $15^n$ end in 0 for any natural number $n$ ?

Prove your answer using prime factorisation and FTA. Identify specifically which prime is present✓ and which is missing✗.

This is a tricky one because 15 has a factor of 5, so students often think "it must end in 0."The trap is assuming 5 means it ends in 0

But ending in 0 needs BOTH 2 and 5You need both factors, not just one.

Number	Prime Factorisation	Has 2?	Has 5?	Ends in 0?
$15^{n}$	$3^n \times 5^n$ Only 3 and 5, no factor of 2	❌ NoMissing!Missing the required factor of 2	✅ Yes	❌ NoWithout both 2 and 5, no zero ending

✍️ MCQ

Choose one

Which prime factor is missing from

15^n

that prevents it from ending in 0?

✓

Let's factorise: $15 = 3 \times 5$ . So $15^n = 3^n \times 5^n$ .

Prime factors present: 3 and 5.

Prime factor missing:Checking what's present and missing 2key!2 is missing — key for divisibility by 10.

By FTA uniquenessOnly one possible factorisation, 2 cannot appear in any factorisation of $15^{n}$ — the factorisation $3^{n} \times 5^{n}$ is the only one.

So $15^n$ will NEVER be divisible by 10, and therefore will NEVER end in 0.Never divisible by 10, never ends in zero

✍️ MCQ

Choose one

Compare

4^n

and

15^n

. Both fail to end in 0, but for opposite reasons. Which prime is missing in each case?

✓

Since $15^{n}$ lacks the factor 2Missing factor 2 means can't divide by 10, it is not divisible by 2, hence not divisible by $10 = 2 \times 5$ 10 needs both 2 and 5 in its factorisation, hence cannot end in 0proven!Check which prime factors are present or missing.

Let's verify with examples:

$15^{1} = 15$ (ends in 5)
$15^{2} = 225$ (ends in 5)
$15^{3} = 3375$ (ends in 5)

All end in 5, not 0.pattern

✍️ MCQ

Choose one

For

b^n

to end in 0 for some natural number

n

, the base

b

must have which prime factors?

✓

The pattern: Having a factor of 5 is not enough. You need both 2 AND 5Both primes are required for a zero ending. Miss either one, and the number cannot end in 0Missing either prime means no zero.

✍️ MCQ

Choose one

Which base would have powers that CAN end in 0:

6

,

10

, or

25

?

✓

4. The general rule: which bases produce trailing zeros

You've now seen bases that lack 5 (like $4$ ) and bases that lack 2 (like $15$ ) — neither can end in 0.

The final step is to state the complete rule:

For which bases does the power always end in 0?

📋 Given Info

Here's what you've established so far:

Base	Prime Factors	Ends in 0?
$4^{n}$	only 2	Never
$15^{n}$	3 and 5 (no 2)	Never
$10^{n} = 2^{n} \times 5^{n}$	2 and 5	Always

The key difference: $10^{n}$ has both 2 and 5key as prime factors.

✍️ Question

State the general rule:

For which values of the base $a$ does $a^{n}$ always end in 0 (for all $n \geq 1$ )?

Express your answer in terms of the prime factorisation of $a$ .

Also give:

One example of a base that always produces trailing zeros
One example of a base that never does

Let's build the general rule from what you've seen:

Ending in 0 requires divisibility by $10 = 2 \times 5$ To end in 0, you need divisibility by 10. So $a^{n}$ ends in 0 only if it has both 2 and 5 as prime factorsYou must have both 2 and 5, not just one.

Now, the prime factors of $a^n$ Powers only change how many times each prime appears are exactly the prime factors of $a$ (just with larger exponents). So:

$a^{n}$ has a factor of 2 if and only ifThe condition works both ways $a$ has a factor of 2
$a^{n}$ has a factor of 5 if and only if $a$ has a factor of 5

✍️ MCQ

Choose one

For which bases

a

does

a^n

always end in 0 (for all

n \geq 1

)?

✓

Therefore: $a^{n}$ always ends in 0 if and only if the prime factorisation of $a$ contains BOTH 2 and 5Need both primes present, not just one.

Why both? Because $10 = 2 \times 5$ This is why we need both primes. To guarantee a trailing zero, we need at least one factor of 10Trailing zero means divisible by 10 in $a^{n}$ , which requires both prime factors 2 and 5 to be present in $a$ itselfBoth 2 and 5 must be in the original number.

✍️ MCQ

Choose one

Which of the following bases will always produce a trailing zero when raised to any positive integer power? (A)

a = 20

(B)

a = 25

(C)

a = 8

(D)

a = 35

✓

Examples that always end in 0:

$10 = 2 \times 5$ → both presentThe pair of primes needed for trailing zeros → 10, 100, 1000, ...
$20 = 2^{2} \times 5$ → both presentBoth required primes are there → 20, 400, 8000, ...
$30 = 2 \times 3 \times 5$ → both presentExtra primes don't matter if 2 and 5 are both there → 30, 900, 27000, ...

✍️ Yes/No

Yes or No?

Does

50 = 2 \times 5^2

always produce a trailing zero when raised to any power

n \geq 1

?

✓

Examples that never end in 0Missing one prime means no zeros ever:

$4 = 2^{2}$ → has 2 but not 5Has 2 but the 5 is missing
$6 = 2 \times 3$ → has 2 but not 5One of the required pair is absent
$15 = 3 \times 5$ → has 5 but not 2Has 5 but missing the 2
$9 = 3^{2}$ → has neither 2 nor 5Missing both primes needed for factor of 10

✍️ MCQ

Choose one

Which of these bases will never produce a trailing zero when raised to any power?

✓

The deep insight: Last-digit questions are prime-factor questions in disguise.

The translation "ends in 0This phrase signals you need factors of 2 and 5 → divisible by 10 → has 2 and 5The two prime factors you must find" is the move that unlocks the proofkeyThis translation unlocks every proof.