1101001
1011001
1001101
101101
11001
Correct answer is B
\(89_{10}\)
| 2 | 89 |
| 2 | 44 r 1 |
| 2 | 22 r 0 |
| 2 | 11 r 0 |
| 2 | 5 r 1 |
| 2 | 2 r 1 |
| 2 | 1 r 0 |
| 0 r 1 |
\(89_{10} = 1011001_{2}\)
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