Question 31 markMCQSection A

For any natural number n, 6n6^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
(C)3
(D)2

Official answer

(b) 6 — every power of 6 ends in the digit 6, since 6n=2n×3n6^n = 2^n \times 3^n never has a factor of 5.

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

Marking scheme

  1. 1.1 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: 6×6=366\times6=36, 6×6×66\times6\times6 ends in 6×6=366\times6=36 → always ends in 6.

Quick oral answer

6 raised to any natural number n always ends in the digit 6, because 6×6=366 \times 6 = 36 always keeps the units digit as 6; also, 6n=2n×3n6^n = 2^n \times 3^n has no factor of 5, so it can never end in 0.

Common mistakes

  1. 1.Selecting (a) 0, incorrectly assuming large powers always trend towards ending in zero.
  2. 2.Not testing small cases (61,62,636^1, 6^2, 6^3) to spot the pattern before answering.
  3. 3.Confusing this with numbers like 10n10^n or 5n5^n which behave differently for units digits.

Analysis and explanation

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


Concept

  • 6n=2n×3n6^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

  • 61=6,62=36,63=216,64=12966^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296 — units digit stays 6 for every power.
  • Since 6×6=366 \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.

Interesting facts

This exact reasoning — that ana^n ending in 0 requires both 2 and 5 as factors — is a standard NCERT proof technique used to show numbers like 4n,6n4^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'.

Question details

Q3 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Real Numbers
Topic
Units Digit Pattern using Prime Factorisation
Bloom's level
understand
Syllabus topic
Unit 1: Number Systems — Real Numbers: Fundamental Theorem of Arithmetic and Units Digit Patterns of Powers
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why can 6n6^n never end in 0?

A number ends in 0 only if it has both 2 and 5 as prime factors. Since 6n=2n×3n6^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 6×6=366 \times 6 = 36 preserves the units digit as 6.