Correct Option (A)

To determine the unit digit of $(57242)^{9 \times 7 \times 5 \times 3 \times 1}$ , the focus is on the unit digit of the base and the exponent.

The unit digit of the base is 2.

First, calculate the value of the exponent:

$9 \times 7 \times 5 \times 3 \times 1 = 945$

The problem is thus reduced to finding the unit digit of $(2)^{945}$ . The unit digits of powers of 2 follow a cyclical pattern:

$2^{1} = 2$
$2^{2} = 4$
$2^{3} = 8$
$2^{4} = 16$ (unit digit is 6)
$2^{5} = 32$ (unit digit is 2)

This pattern (2, 4, 8, 6) repeats every four powers. Therefore, the cyclicity of the unit digit 2 is 4.

To find the unit digit of $2^{945}$ , divide the exponent (945) by the cyclicity (4) and determine the remainder:

$945 \div 4 = 236$ with a remainder of 1.

The unit digit of $2^{945}$ is equivalent to the unit digit of $2^{remainder}$ . Since the remainder is 1, the unit digit is $2^{1} = 2$ .

Hence, the unit digit in the expansion of $(57242)^{9 \times 7 \times 5 \times 3 \times 1}$ is 2.

Incorrect Options:

Options B (4), C (6), and D (8) are incorrect. The systematic application of the cyclicity rule for the unit digit of 2, based on the calculated exponent of 945, unequivocally establishes 2 as the unit digit. Any other result would contradict the principles of unit digit cyclicity and exponentiation.