Given, (12x)₃ = (123)_x
Since LHS has 3 as the base and RHS has ‘x’ base,
1 * 3*3 + 2 * 3 + x * 1 = 1 * x*x + 2 * x + 3
9 + 6 + x = x2 + 2x + 3
x2 + x - 12 = 0
x2 + 4x - 3x - 12 = 0
x( x + 4 ) - 3(x + 4) = 0
(x + 4)(x - 3) = 0
x = 3 and -4
But, both the values are infeasible.

Alternative explanation :
According to the rules of number systems , the numbers present in a number system should not be greater than the base of the number system.
According to LHS , (12x)3 tells us that the value of x should be less than 3.
According to RHS , (123)x tells us that the value of x should be greater than 3 as largest digit in 123 is 3.
Therefore, any combination is not possible.

The computer uses 8 digit mantissa and 2 digit exponent:

a = 0.052 We can represent the number in M*E So a= 0.052 = 0.52*10-1

mantissa = 0.52, exponent = −1.

b = 28E+11, We can represent the number in M*E So b = 28E+11= 0.28*1013

mantissa = 0.28, exponent = 13.

To add b+a, Small exponent number, a is shifted

to 13-(-1) = 14 places to right side

a = 0.0000000000000052E+13

From the given data computer uses only 8 digit mantissa, so digits beyond 8th position will be discarded.

So a = 0.00000000E+13 = 0.0 E+13

b + a = (0.28E + 13) + (0.0E + 13 )

= 0.28E + 13

Then b + a - b = (0.28E + 13) - (0.28E + 13)

= 0

Gray codes are less error-prone for mechanical devices that involve making and breaking electrical circuits because they only change in one bit position at a time.

So, they are considered as the minimum error code.

The range of integers that can be represented by n bit 2’s complement number system is -2^n-1 to (2^n-1 – 1).

→ The Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.
→ In other words, it measures the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other.
→ Given HexaDecimal numbers are 0xAA and 0x55.
Decimal equivalent of 0xAA is 170
Binary form of 0xAA is 1010 1010
Decimal equivalent of 0x55 is 85
Binary form of 0xAA is 0101 0101
The two numbers binary length length is 8.
→ If you observe all the bits of above two binary numbers, both numbers have different bits in all positions.
→ So according to definition , the number of positions at which the corresponding symbols are different which is 8.

(42.75)10 = (?)2