The digit in the unit place of the number 6¹²⁹ × 7³⁰⁷ is

According to the principles of number systems widely cited in quantitative aptitude references like R.S. Aggarwal's Quantitative Aptitude for Competitive Examinations, calculating the unit digit of an exponential expression relies on the mathematical concept of "cyclicity".

The unit digit cyclicity of 6 is 1. Any positive integer power of 6 always ends with the unit digit 6 (e.g., 6¹ = 6, 6² = 36). Therefore, the unit digit of 6¹²⁹ is unequivocally 6.

The unit digit cyclicity of 7 is 4. The unit digits for successive powers of 7 follow a repeating, four-step pattern: 7, 9, 3, 1 (for exponents yielding a remainder of 1, 2, 3, and 0 when divided by 4, respectively). To find the unit digit of 7³⁰⁷, we divide the exponent 307 by 4. This division yields a quotient of 76 with a remainder of 3 (307 = 4 × 76 + 3). A remainder of 3 dictates that the unit digit corresponds to the 3rd value in the cyclicity sequence, which is 3 (since 7³ = 343).

Multiplying the unit digits of the two components gives 6 × 3 = 18. The final digit in the unit place is 8. Thus, Option C is correct.

• Option A is incorrect: It incorrectly implies the product of the unit digits ends in 2. This would require the second term to end in 2 or 7 (e.g., 6 × 7 = 42), which contradicts the calculated unit digit of 3.
• Option B is incorrect: It assumes the final unit digit is 4, which would only occur if the unit digit of 7³⁰⁷ was 4 or 9 (e.g., 6 × 9 = 54).
• Option D is incorrect: It suggests the overall unit digit is 6. This would solely happen if the exponent of 7 was a perfect multiple of 4 (yielding a unit digit of 1, since 6 × 1 = 6), which is not the case for 307.

Concluding takeaway: Master the unit digit cyclicity theorem for CSAT Paper-II: digits 0, 1, 5, and 6 always repeat themselves (cyclicity of 1); 4 and 9 have a cyclicity of 2; and 2, 3, 7, and 8 have a cyclicity of 4.