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Why is there no natural number $n$ for which $4^n$ ends with the digit 0?
ABecause $4^n$ is always odd.
BBecause $4^n$ is never divisible by 2.
CBecause the prime factorisation of $4^n$ contains only 2s and no factor 5.
DBecause $4^n$ is always less than 10.
Answer & Solution
Correct answer: C. Because the prime factorisation of $4^n$ contains only 2s and no factor 5.
A number ending in 0 must be divisible by 10, so it must contain both 2 and 5 as prime factors. But $4^n=(2^2)^n=2^{2n}$, so its prime factorisation contains only the prime 2. Hence no $4^n$ can end with 0.
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