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Find the units digit of 3^{2018} - 2^{2018}.

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Find the units digit of 3^{2018} - 2^{2018}.  [#permalink]

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New post 21 Nov 2018, 03:39
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A
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C
D
E

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Question Stats:

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[Math Revolution GMAT math practice question]

Find the units digit of \(3^{2018} - 2^{2018}.\)

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)

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Re: Find the units digit of 3^{2018} - 2^{2018}.  [#permalink]

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New post 21 Nov 2018, 03:45
3 has a cyclicity of 4. Then \(3^{2018}\) is the same as the remainder when 2018 is divided by 4, which is 2.

\(3^{2018} = 3^2\)

Similarly, 2 has a cyclicity of 4. then \(2^{2018} = 2^2\)

\(3^2-2^2\) = 9-4 = 5

C is the answer.
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Re: Find the units digit of 3^{2018} - 2^{2018}.  [#permalink]

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New post 23 Nov 2018, 00:54
=>

The units digit is the remainder when \(3^{2018} - 2^{2018}\) is divided by \(10\).

The remainders when powers of \(3\) are divided by \(10\) are
\(3^1: 3,\)
\(3^2: 9,\)
\(3^3: 7,\)
\(3^4: 1,\)
\(3^5: 3,\)

So, the units digits of \(3^n\) have period \(4\): they form the cycle \(3 -> 9 -> 7 -> 1.\)
Thus, \(3^n\) has the units digit of \(9\) if \(n\) has a remainder of \(2\) when it is divided by \(4\).
The remainder when \(2018\) is divided by \(4\) is \(2\), so the units digit of \(3^{2018}\) is \(9\).

The remainders when powers of \(2\) are divided by \(10\) are
\(2^1: 2,\)
\(2^2: 4,\)
\(2^3: 8,\)
\(2^4: 6,\)
\(2^5: 2,\)

So, the units digits of \(2^n\) have period \(4\): they form the cycle \(2 -> 4 -> 8 -> 6.\)
Thus, \(2^n\) has the units digit of \(4\) since \(n\) has a remainder of \(2\) when it is divided by \(4\).
The remainder when \(2018\) is divided by \(4\) is \(2\), so the units digit of \(2^{2018}\) is \(4\).

\(3^{2018} - 2^{2018}\) has remainder \(9 – 4 = 5\) when it is divided by \(10\).

Therefore, the answer is C.
Answer: C
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Re: Find the units digit of 3^{2018} - 2^{2018}.  [#permalink]

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Re: Find the units digit of 3^{2018} - 2^{2018}.   [#permalink] 23 Nov 2019, 04:28
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