1 markMCQSection A

For any natural number n, $6^n$ ends with the digit : (a) 0 (b) 6 (c) 3 (d) 2

Real Numbers

Units digit of powers

Options

(A)0

(B)6

(D)2

Official Answer

(b) 6 — every power of 6 ends in 6, since $6\times6$ ends in 6 and $6^n = 2^n\times3^n$ never has a factor of 5 (so it can never end in 0).

units digitpowers of 66^n ends in 6no factor of 5cannot end in 0even numbercyclicity of units digit

Marking Scheme

11 mark: correct option (b) 6.
2Expected reasoning: $6\times6$ ends in 6, so every power of 6 ends in 6; equivalently $6^n = 2^n\times3^n$ has no factor of 5, so it can never end in 0.

Hint

Work out $6^1$ , $6^2$ , $6^3$ and look at the last digit each time — the pattern repeats.

Quick Oral Answer

Every power of 6 ends in 6 because $6\times6 = 36$ ends in 6, and since $6^n = 2^n\times3^n$ has no factor of 5 it can never end in 0.

Analysis & Explanation

This MCQ links the units-digit pattern of powers of 6 to the Fundamental Theorem of Arithmetic.

Concept

$6^n = 2^n \times 3^n$ ; since the factorisation never contains the prime 5, $6^n$ can never end in 0.
Successive powers: $6^1=6, 6^2=36, 6^3=216, 6^4=1296$ — the units digit stays 6 because $6\times6$ ends in 6.

Key points

The pattern is self-sustaining for every natural number n, so the units digit is always 6.
$6^n$ is always even, so it can never end in the odd digit 3.

Common mistakes

Thinking $6^n$ ends in 0 because it "looks like" 10's multiples — it never contains a factor of 5.
Guessing digit 2 by confusing it with powers of other even numbers.

Common Mistakes

1Choosing 0 by wrongly assuming any repeated multiplication eventually ends in 0 — that needs a factor of 5, which $6^n$ never has.
2Choosing an odd digit like 3 without noticing that $6^n$ is always even and must end in an even digit.
3Testing only $6^1 = 6$ and not confirming with $6^2 = 36, 6^3 = 216$ to see the pattern is stable.

Interesting Facts

6 is called an 'automorphic-like' digit for exponentiation because every power of 6 ends in 6; the digits 0, 1, 5 and 6 all share this 'self-repeating units digit' property.

For $6^n$ to end in 0 it would need a factor of $10 = 2\times5$ , but $6^n = 2^n\times3^n$ contains the prime 5 exactly zero times — a direct consequence of unique prime factorisation.

6 is also the smallest 'perfect number' ( $1+2+3 = 6$ ), a fact unrelated to this problem but a reason 6 appears often in number-theory questions.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why does every power of 6 end in the digit 6?

Because multiplying any number that ends in 6 by 6 gives a units digit of $6\times6 = 36$ , i.e. 6 again. So $6^1 = 6, 6^2 = 36, 6^3 = 216$ , and the last digit is permanently 6 for every natural number n.

Why can $6^n$ never end in 0?

A number ends in 0 only if it is divisible by $10 = 2\times5$ . But $6^n = 2^n\times3^n$ has no factor of 5 at all (by the uniqueness of prime factorisation), so it can never be a multiple of 10 and can never end in 0.

Which single-digit numbers have powers that always end in the same digit?

0, 1, 5 and 6. Powers of these digits always terminate in the same units digit — for example $5^2 = 25, 5^3 = 125$ (always 5), and $6^n$ always ends in 6.