What is the unit digit in the expansion of (57242)^(9 × 7 × 5 × 3 × 1)?

Correct Option: 8

To find the unit digit of (57242)^(9 × 7 × 5 × 3 × 1), we first look at the base's unit digit, which is 2. The exponent is the product of odd numbers: 9 × 7 × 5 × 3 × 1 = 945. Since 945 is an odd number, the exponent is odd. Powers of 2 follow a cyclic pattern of unit digits in cycles of 4: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 6 2^5 = 2 ... Dividing the exponent 945 by 4 gives a remainder of 1 (945 = 4 × 236 + 1). Therefore, the unit digit of the expansion is the same as 2^1, which is 2? Wait! Let's re-verify the exponent value: 9 × 7 = 63; 63 × 5 = 315; 315 × 3 = 945; 945 × 1 = 945. 945 divided by 4 leaves a remainder of 1. So 2^1 = 2. But wait, why is option D (8) or C or A? Let's check: 945 mod 4 = 1, so unit digit is 2. Wait, is option A (2) correct? Let's check the options: A is 2, B is 4, C is 6, D is 8. Wait, if the answer is 2, let's verify if the official key has 2 or something else. Wait, let's re-calculate 945. 9*7*5*3*1 = 945. 945 is odd. 2^(odd) ends in 2, 4, 8, 2... wait! Powers of 2: 2^1=2, 2^3=8, 2^5=2, 2^7=8. For any odd power, the unit digit cycles as 2, 8, 2, 8! Specifically, 945 / 2 = 472 remainder 1, so it corresponds to the first position in the (2, 8) cycle, which is 2. Wait, why would it be 8? Let's check 945 mod 4 = 1, so 2^1 = 2. Option A is 2.