Question 31 markMCQSection A

  1. For any natural number n, 6n6^n ends with the digit : (a) 0 (b) 6 (c) 3 (d) 2

Real Numbers
Units digit of powers

Options

(A)0
(B)6
(C)3
(D)2

Official answer

(b) 6 — every power of 6 ends in 6, since 6×66\times6 ends in 6 and 6n=2n×3n6^n = 2^n\times3^n never has a factor of 5 (so it can never end in 0).

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

Marking scheme

  1. 1.1 mark: correct option (b) 6.
  2. 2.Expected reasoning: 6×66\times6 ends in 6, so every power of 6 ends in 6; equivalently 6n=2n×3n6^n = 2^n\times3^n has no factor of 5, so it can never end in 0.

Hint

Work out 616^1, 626^2, 636^3 and look at the last digit each time — the pattern repeats.

Quick oral answer

Every power of 6 ends in 6 because 6×6=366\times6 = 36 ends in 6, and since 6n=2n×3n6^n = 2^n\times3^n has no factor of 5 it can never end in 0.

Common mistakes

  1. 1.Choosing 0 by wrongly assuming any repeated multiplication eventually ends in 0 — that needs a factor of 5, which 6n6^n never has.
  2. 2.Choosing an odd digit like 3 without noticing that 6n6^n is always even and must end in an even digit.
  3. 3.Testing only 61=66^1 = 6 and not confirming with 62=36,63=2166^2 = 36, 6^3 = 216 to see the pattern is stable.

Analysis and explanation

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


Concept

  • 6n=2n×3n6^n = 2^n \times 3^n; since the factorisation never contains the prime 5, 6n6^n can never end in 0.
  • Successive powers: 61=6,62=36,63=216,64=12966^1=6, 6^2=36, 6^3=216, 6^4=1296 — the units digit stays 6 because 6×66\times6 ends in 6.

Key points

  • The pattern is self-sustaining for every natural number n, so the units digit is always 6.
  • 6n6^n is always even, so it can never end in the odd digit 3.

Common mistakes

  • Thinking 6n6^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.

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 6n6^n to end in 0 it would need a factor of 10=2×510 = 2\times5, but 6n=2n×3n6^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=61+2+3 = 6), a fact unrelated to this problem but a reason 6 appears often in number-theory questions.

Question details

Q3 of 53 questions

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

Spotted a mistake or something unclear?

Tell us and we will fix it.

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 6×6=366\times6 = 36, i.e. 6 again. So 61=6,62=36,63=2166^1 = 6, 6^2 = 36, 6^3 = 216, and the last digit is permanently 6 for every natural number n.

Why can 6n6^n never end in 0?

A number ends in 0 only if it is divisible by 10=2×510 = 2\times5. But 6n=2n×3n6^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 52=25,53=1255^2 = 25, 5^3 = 125 (always 5), and 6n6^n always ends in 6.