Explain why $6^{n}$ cannot end with the digit 0 for any natural number $n$ . Choose the correct reason.

Class 10Real NumbersNumber Systems

Step-by-Step Solution

A number ending in 0 must have both 2 and 5 as prime factors.

6^{n} = 2^{n} \times 3^{n}

contains no factor 5, so it can never end in 0.

Answer: B) Because

6^{n} = 2^{n} \times 3^{n}

has no factor of 5

**Lemma**: For positive integers a,

b

(with

b > 0

), there exist UNIQUE

q, r

with

a = b q + r

and 0 ≤

r < b

.

This is just formal long division —

q

is the quotient,

r

is the remainder.

**Algorithm to find HCF(a, b)** with

a > b

:
1. Apply lemma:

a = b q + r

.
2. If

r = 0

, HCF =

b

.
3. If

r

≠ 0, replace (a, b) with

(b

, r) and repeat.

**Example**: HCF(420, 130)

. - 420 = 130 \times 3 + 30 - 130 = 30 \times 4 + 10 - 30 = 10 \times 3 + 0

→ HCF = 10.

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