The digit in the unit place of the number $6^{129}\times 7^{307}$ is

The correct option is (c) - 8.

[as per provisional answerkey]

Solution

To find the unit digit of the product 6¹²⁹ × 7³⁰⁷, we must find the unit digit of each term individually and then multiply them.

Step 1: Find the unit digit of 6¹²⁹
The number 6 has a unique property: any positive integer power of 6 always ends in 6 (6¹=6, 6²=36, 6³=216, etc.).
Therefore, the unit digit of 6¹²⁹ is 6.

Step 2: Find the unit digit of 7³⁰⁷
The unit digits of powers of 7 follow a cycle of 4:
7¹ = 7
7² = 49 (unit digit 9)
7³ = 343 (unit digit 3)
7⁴ = 2401 (unit digit 1)
The cycle is [7, 9, 3, 1]. To find the position in the cycle, we divide the exponent by 4 and look at the remainder:
307 ÷ 4 = 76 with a remainder of 3.
The 3rd number in the cycle is 3.
Therefore, the unit digit of 7³⁰⁷ is 3.

Step 3: Multiply the results
Unit digit of (6¹²⁹ × 7³⁰⁷) = (Unit digit of 6¹²⁹) × (Unit digit of 7³⁰⁷)
= 6 × 3 = 18.
The unit digit of 18 is 8.

Why the other options are incorrect

• Option (a) - 2: This result would occur if the unit digit of 7³⁰⁷ was incorrectly calculated as 2, or if the product 6 × 7 was used without considering the exponents correctly.
• Option (b) - 4: This result might be reached if a student incorrectly identifies the cyclicity of 7 or makes a calculation error in the multiplication step (e.g., thinking 6 × 4 = 24).
• Option (d) - 6: This is the unit digit of the first term (6¹²⁹) only; it ignores the impact of the second term (7³⁰⁷) in the product.

Key Concept

Unit digit cyclicity, where the last digit of any power follows a repeating pattern (periodicity) of at most 4.