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If x is an integer greater than 1, what is the units digit of 6^x − 6^
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Bunuel
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If x is an integer greater than 1, what is the units digit of 6^x − 6^ [#permalink] New post  22 May 2018, 03:29
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If x is an integer greater than 1, what is the units digit of \(6^x − 6^{(x−1)}\)?

A. 0
B. 1
C. 3
D. 6
R. It cannot be determined.
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If x is an integer greater than 1, what is the units digit of 6^x − 6^ [#permalink] New post  22 May 2018, 03:49
Bunuel wrote:
If x is an integer greater than 1, what is the units digit of \(6^x − 6^{(x−1)}\)?

A. 0
B. 1
C. 3
D. 6
R. It cannot be determined.


It does not matter how many times 6 is multiplied by 6. The unit's digit is always 6 (unless x is 0 or -ve).
Hence , Units digit of both \(6^x\) & \(6^{(x−1)}\) is \(6\).
Hence after subtraction the units digit is 0.
Hence , I would go for option A.
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If x is an integer greater than 1, what is the units digit of 6^x − 6^ [#permalink] New post  22 May 2018, 03:56
Bunuel wrote:
If x is an integer greater than 1, what is the units digit of \(6^x − 6^{(x−1)}\)?

A. 0
B. 1
C. 3
D. 6
R. It cannot be determined.


If \(x=1\), then the units digit of \(6^x − 6^{(x−1)}\) is \(6-1=5\)

For all other cases \(...6-...6 = ...0\)

Answer: A
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Re: If x is an integer greater than 1, what is the units digit of 6^x − 6^ [#permalink] New post  22 May 2018, 04:15
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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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If x is an integer greater than 1, what is the units digit of 6^x 6^ [#permalink] New post  23 May 2022, 10:29
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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!

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If x is an integer greater than 1, what is the units digit of 6^x 6^ [#permalink]
23 May 2022, 10:29
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