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Q22·CSAT · Prelims 2026

Quantitative Aptitude — Unit Digit Cyclicity

NumericalNumber SystemSeries & Pattern● Easy

Question

The digit in the unit place of the number 6^129 × 7^307 is

Options

a

2

Answer
b

4

c

8

d

6

Explanation

To find the final unit digit of the product 6^129 × 7^307, we analyze the unit digit behavior and power cyclicity for base 6 and base 7 independently.

Analyze Base 6: The number 6 has a cyclicity of 1. Any positive integer power of a number ending in 6 will always yield a unit digit of 6 (6^1=6, 6²=36, 6³=216). Thus, Unit Digit(6^129) = 6.
Analyze Base 7: The number 7 has a pattern cyclicity of 4 (7^1=7, 7²=49, 7³=343, 7^4=2401).
Divide the exponent 307 by 4 to find the remainder: 307 \div 4 = 76 with a remainder of 3.
This means the unit digit corresponds to the third step in the pattern sequence, which is 3 (7³ \rightarrow 3). Thus, Unit Digit(7^307) = 3.
Calculate the Final Product: Multiply the individual unit digits:

Final Unit Digit = 6 × 3 = 18 \rightarrow 8

Let let us re-verify the input text and options structure carefully. The expression is written as 6^129 × 7^307. The units output yields 6 × 3 = 18, giving a final digit of 8.

Unit digit analysis relies on using the remainder of the exponent divided by 4 to map values within a base number's repeating pattern sequence.

Answer: (c).

Question details

Year

2026

Paper

CSAT

Question

Q22

Section

Quantitative Aptitude

Sub-topic

Number System

Type

Series & Pattern

Difficulty

Easy

Source hint

Unit digits — pattern cyclicity algorithms

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