The number (2217)8 is equivalent to
Octal to binary (3 bits per digit): 2 = 010, 2 = 010, 1 = 001, 7 = 111, so \(2217_8 = 010\,010\,001\,111\).
Regroup into 4 bits: 0100 1000 1111 = \(48F_{16}\) (= 048F).
Check: \(2217_8 = 1167_{10}\), and \(48F_{16} = 1024 + 128 + 15 = 1167\).
Answer: (C) (048F)₁₆
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