Change 432\(_{five}\)  to a number in base three.

A.

10100\(_{three}\)

B.

11100\(_{three}\)

C.

11101\(_{three}\)

D.

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\(_{three}\)