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

Question

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Answer 1

If any digit has the last digit $10$ that means it divisible by $10 .$

The factor of $10 = 2 \times 5,$

So value of $6^{n}$ should be divisible by $2$ and $5 .$

Both $6^{n}$ is divisible by $2$ but not divisible by $5 .$

So, it can not end with $0 .$

Answer 2

If any digit has the last digit

10

that means it divisible by

10 .

The factor of

10 = 2 \times 5,

So value of

6^{n}

should be divisible by

2

and

5 .

Both

6^{n}

is divisible by

2

but not divisible by

5 .

So, it can not end with

0 .