Check whether $6^{n}$ can end with the digit $0$ for any natural number $n$ .

medium 2 marks Class 10 · Mathematics · Real Numbers

Answer

No.

6^{n}

can never end with the digit

0

The correct answer is No. $6^{n}$ can never end with the digit $0$ . — Since $5$ never divides $6^{n}$ , neither does $10$ , and so $6^{n}$ cannot end in $0$ for any natural number $n$ .

Step-by-step solution

A number ends with the digit $0$ only if it is divisible by $10$ .
Since $10 = 2 \times 5$ , such a number must have both $2$ and $5$ in its prime factorisation.
Now factorise $6^{n}$ . Since $6 = 2 \times 3$ , we get $6^{n} = 2^{n} \times 3^{n}$ .
The only primes appearing here are $2$ and $3$ . There is no factor of $5$ .
By the Fundamental Theorem of Arithmetic the prime factorisation of a number is unique, so a factor of $5$ cannot appear from anywhere else.
Since $5$ never divides $6^{n}$ , neither does $10$ , and so $6^{n}$ cannot end in $0$ for any natural number $n$ .

Common mistake: Testing a few values such as

6

36

and

216

and concluding from those. Checking examples is not a proof; the argument has to rule out every

n

at once.

Concept tested: Fundamental Theorem of Arithmetic

Published 19 August 2026 · How this solution was prepared and checked

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Check whether 6n can end with the digit 0 for any natural number n.

Step-by-step solution

Check whether $6^{n}$ can end with the digit $0$ for any natural number $n$ .