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

Question

Homi · Accepted Answer

Understand the Question

For any number to end with the digit 0, it must be divisible by $10$ , which means its prime factorisation must contain both 2 and 5 as factors ( $10 = 2 \times 5$ ).
We find the prime factorisation of $6^n$ and check whether the prime factor 5 is present.
By the Fundamental Theorem of Arithmetic, the prime factorisation of any number is unique, meaning no other prime factors can appear.

Step 1 · Prime Factorisation of $6^n$

Prime factorising the base $6$ $6 = 2 \times 3$

For any natural number $n$

\begin{aligned} 6^n &= (2 \times 3)^n \\ &= 2^n \times 3^n \end{aligned}

Step 2 · Check for Factor 5

For a number to end with the digit $0$ , it must be divisible by $10$ , requiring both $2$ and $5$ as prime factors.

The prime factorisation of $6^n$ contains only the prime numbers $2$ and $3$ .

By the Fundamental Theorem of Arithmetic, this prime factorisation is unique, so $5$ is not a prime factor of $6^n$ .

Therefore, $6^n$ is not divisible by $5$ and cannot end with the digit $0$ for any natural number $n$ .

Answer

$6^n$ cannot end with the digit $0$ for any natural number $n$ .

Common Mistakes

Overlooking the Uniqueness Theorem: Forgetting to mention the Fundamental Theorem of Arithmetic, which proves that no other prime factor (like $5$ ) can exist for $6^n$ .
Incomplete Condition for Ending in 0: Assuming that being divisible by $2$ alone is sufficient; a number ending in $0$ must have both $2$ and $5$ in its prime factorisation.

Question 5