10100\(_{three}\)
11100\(_{three}\)
11101\(_{three}\)
10110\(_{three}\)
Correct answer is B
Convert from base 5 to base 10
432\(_{five}\) = (4 x 5\(^2\)) + (3 x 5\(^1\)) + (2 x 5\(^0\))
= (4 x 25) + (3 x 5) + (2 x 1)
= 100 + 15 + 2
= 117\(_{ten}\)
Then convert from base 10 to base 3
| 3 | 117 |
| 3 | 39 r 0 |
| 3 | 13 r 0 |
| 3 | 4 r 1 |
| 3 | 1 r 1 |
| 0 r 1 |
Selecting the remainders from bottom to top:
117\(_{ten}\) = 11100\(_{three}\)
Hence; 432\(_{five}\) = 11100\(_{three}\)
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