Convert $(12x)_3$ to decimal:
$1\cdot 3^2 + 2\cdot 3 + x = 9 + 6 + x = 15 + x$
Convert $(123)_x$ to decimal:
$1\cdot x^2 + 2\cdot x + 3$
Equate:
$15 + x = x^2 + 2x + 3$
Bring all to one side:
$x^2 + x - 12 = 0$
Factor:
$(x + 4)(x - 3) = 0$
So $x = 3$ or $x = -4$
But base $x$ must be > 3 (because digits 1,2,3 appear).
Allowed value = 3 (but digit '3' cannot appear in base 3).
So no valid base exists.
Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contains 64 sectors(0-63). Data storage capacity in each sector is 512 bytes. Data are organized cyclinderwise and the addressing format is . A file of size 42797KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is cylinder number of the last sector of the file, if it is stored in a contiguous manner?