Mathematics — Standard (041) — CBSE Class 10 board question

If $6^n$ ends with digit 0, it must be divisible by 10, so its prime factorisation must contain both 2 and 5 as factors.

Now, $6^n = (2 \times 3)^n = 2^n \times 3^n$.

The only prime factors of $6^n$ are 2 and 3. The prime 5 does not appear in its factorisation.

By the uniqueness of the Fundamental Theorem of Arithmetic, 5 can never be a factor of $6^n$. Hence, $6^n$ cannot end with digit 0 for any natural number $n$.

Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic

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• The key idea: a number ends in 0 only if it is divisible by both 2 and 5 (since $10 = 2 \times 5$).
• Show that the prime factorisation of $6^n$ contains only 2 and 3, never 5.
• Invoke the uniqueness part of the Fundamental Theorem of Arithmetic — this is what examiners want to see explicitly stated.
• Don't forget to write the conclusion clearly. Many students lose the final half-mark by not stating it.