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chetan2u
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chetan2u
What is the units digit of \((2x)^8\)?
(1) The unit's digit of \(x^2\) is 9.
(2) The unit's digit of \(x^3\) is 7.


IMO D

\((2x)^8 = 2^8*x^8\)

The unit digits of \(2^n\) cycle as 2, 4, 6, 8, 0
The 8th power will have unit digit as 6.

Statement 1:
Unit digit of \(x^2\) is 9.
This means the unit digits of \(x^n\) cycle as 3, 9, 7, 1.
Therefore, the 8th power will have unit digit as 1. Multiplying 8th power of x to 8th power of 2 will give unit digit as 6.
Sufficient.

Statement 2:
Unit digit of \(x^3\) is 7.
From this again we can derive the unit digits of \(x^n\) cycle as 3, 9, 7, 1.
And again the unit digit of product of 8th power of x and 2 is 6.
Sufficient.

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chetan2u
What is the units digit of \((2x)^8\)?
(1) The unit's digit of \(x^2\) is 9.
(2) The unit's digit of \(x^3\) is 7.

From 1:- x can be 7 or 3 but unit digit of \(x^8\) will always be one(\(x^2\) 4 times or check for both 7 and 3).
so we can find single solution.
From 3:- x can only be 7. so sufficient

Option D
Excellent

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