What is the unit digit in the expansion of $(57242{)}^{9\times 7\times 5\times 3\times 1}$?

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:

• $945÷4=236$ with a remainder of 1.

The unit digit of $2^{945}$ is equivalent to the unit digit of $2^{\text{remainder}}$. 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.