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Solution



Given:
• The value of the integer x is greater than 1

To find:
• The unit digit of the expression \(6^x\) – \(6^{(x-1)}\)

Approach and Working:
• If we observe the unit digit pattern of 6, we can see for any positive power of 6, the number will always end with 6 only
    o \(6^1\) = 6
    o \(6^2\) = 36
    o \(6^3\) = 216 etc.
• As x > 1, x-1 > 0
    o Hence, \(6^x\) and \(6^{(x-1)}\) will always end with the unit digit 6
• Therefore, \(6^x\) - \(6^{(x-1)}\) will end with 0
Hence, the correct answer is option A.

Answer: A
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Theory: To find the units digit of power of 6 we need to check the cyclicity in the units digit of powers of 6

\(6^1\) units’ digit is 6 [ 6 ]
\(6^2\) units’ digit is 3 [ 36 ]
\(6^3\) units’ digit is 3 [ 216 ]

=> Units digit of any positive integer power of 6 is 6
=> Units digit of both \(6^x\) and \(6^{(x−1)}\) will be 6
=> Units digit of \(6^x − 6^{(x−1)}\) = 6-6 = 0

So, Answer will be A
Hope it helps!

Watch the following video (from 2:48 mins) to learn how to find cyclicity of 3 and other numbers
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