IAS 2023 — CSAT Maths

Correct Option (A)

To determine the unit digit of $(57242)^{9 \times 7 \times 5 \times 3 \times 1}$(57242)9×7×5×3×1, 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$9×7×5×3×1=945

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

• $2^1 = 2$21=2
• $2^2 = 4$22=4
• $2^3 = 8$23=8
• $2^4 = 16$24=16 (unit digit is 6)
• $2^5 = 32$25=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}$2945, divide the exponent (945) by the cyclicity (4) and determine the remainder:

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

The unit digit of $2^{945}$2945 is equivalent to the unit digit of $2^{\text{remainder}}$2remainder. Since the remainder is 1, the unit digit is $2^1 = 2$21=2.

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