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 Pattern using Prime Factorisation

Options

(A)0

(B)6

(D)2

Official Answer

(b) 6 — every power of 6 ends in the digit 6, since $6^n = 2^n \times 3^n$ never has a factor of 5.

units digitpower of 6prime factorisation6 to the power nnever ends in 0Fundamental Theorem of Arithmetic

Marking Scheme

11 mark for correctly selecting option (b) 6, based on units digit pattern of powers of 6.

Hint

Multiply the units digit of 6 by itself repeatedly: $6\times6=36$ , $6\times6\times6$ ends in $6\times6=36$ → always ends in 6.

Quick Oral Answer

6 raised to any natural number n always ends in the digit 6, because $6 \times 6 = 36$ always keeps the units digit as 6; also, $6^n = 2^n \times 3^n$ has no factor of 5, so it can never end in 0.

Analysis & Explanation

Tests the units-digit pattern of powers of 6 and links it to prime factorisation reasoning.

Concept

$6^n = 2^n \times 3^n$ , which never contains a factor of 5, so 6^n can never end in 0; the units digit pattern of powers of 6 is always 6.

Key working

$6^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296$ — units digit stays 6 for every power.
Since $6 \times 6 = 36$ , any number ending in 6 multiplied by 6 again ends in 6.

Common mistakes (परीक्षा में सावधानी)

Guessing 0 by wrongly pattern-matching with powers of 10.
Confusing the base's prime factors (2 and 3) with the units digit of the power itself.

Real-world/exam link

This is the same reasoning NCERT uses to prove 6^n can never end in 0 — a favourite short-answer proof question.

Common Mistakes

1Selecting (a) 0, incorrectly assuming large powers always trend towards ending in zero.
2Not testing small cases ( $6^1, 6^2, 6^3$ ) to spot the pattern before answering.
3Confusing this with numbers like $10^n$ or $5^n$ which behave differently for units digits.

Interesting Facts

This exact reasoning — that $a^n$ ending in 0 requires both 2 and 5 as factors — is a standard NCERT proof technique used to show numbers like $4^n, 6^n$ never end in 0.

The units digit cycles of powers (period 1, 2, or 4 depending on the base) are a classical topic in number theory called 'multiplicative order modulo 10'.

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Frequently Asked Questions

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

A number ends in 0 only if it has both 2 and 5 as prime factors. Since $6^n = 2^n \times 3^n$ , it has no factor of 5, so it can never end in 0, regardless of how large n is.

Does this pattern apply to other numbers ending in 6?

Yes, any number ending in 6 raised to a natural number power will always end in 6, since $6 \times 6 = 36$ preserves the units digit as 6.