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If $n$ is a natural number, then $6^n$ can never end with the digit
A0
Bnone
C6
Dall of these are possible
Answer & Solution
Correct answer: A. 0
$6^n = 2^n \cdot 3^n$ has no factor of 5, so it can never end in 0 (which requires both 2 and 5 as factors). Every power of 6 actually ends in 6.
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