Can the number $(15)^n$, $n$ being a natural number, end with the digit 0? Give reasons.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer
No, $(15)^n$ cannot end with the digit 0.
For a number to end with 0, it must be divisible by 10, i.e., its prime factorisation must contain both 2 and 5.
Now, $15^n = (3 \times 5)^n = 3^n \times 5^n$.
The prime factorisation of $15^n$ contains only the primes 3 and 5 — it does not contain 2 as a factor. By the uniqueness of the Fundamental Theorem of Arithmetic, 2 can never appear in this factorisation. Hence, $15^n$ can never end with the digit 0.
Source: Chapter 1, Section 1.2 – The Fundamental Theorem of Arithmetic
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Explanation
- The key idea: a number ends in 0 only if both 2 and 5 are factors (since $10 = 2 \times 5$).
- Show the prime factorisation of $15^n$ and note that 2 is absent.
- Invoke the uniqueness part of the Fundamental Theorem of Arithmetic — this is the formal justification examiners expect, not just an intuitive argument.
- This mirrors the textbook's Example 1 on $4^n$; follow the same reasoning pattern.