If \(S_n = 2^n\), what is the unit's digit of \(S_{67}?\)
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
[spoiler]OA=D[/spoiler]
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If \(S_n = 2^n\), what is the unit's digit of \(S_{67}?\)
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henilshaht
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The important thing is to observe the cycle of the unit digits with 2:
$$2^1 = 2$$
$$2^2 = 4$$
$$2^3 = 8$$
$$2^4 = 16$$
$$2^5 = 32$$
$$2^6 = 64$$
$$2^7 = 128$$
So there is a cycle for unit digits of 2,4, 8 and 6.
In this case, 16*4 = 64, so the unit digit for $$2^{67}$$ will be same $$2^3$$
$$2^1 = 2$$
$$2^2 = 4$$
$$2^3 = 8$$
$$2^4 = 16$$
$$2^5 = 32$$
$$2^6 = 64$$
$$2^7 = 128$$
So there is a cycle for unit digits of 2,4, 8 and 6.
In this case, 16*4 = 64, so the unit digit for $$2^{67}$$ will be same $$2^3$$
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Scott@TargetTestPrep
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S(67) = 2^67VJesus12 wrote:If \(S_n = 2^n\), what is the unit's digit of \(S_{67}?\)
(A) 2
(B) 4
(C) 6
(D) 8
(E) 0
[spoiler]OA=D[/spoiler]
Source: GMAT Club Tests
Since the units-digit pattern of powers of 2 is 2-4-8-6, we see that when 2 is raised to a multiple of 4, the units digit is 6. In other words, 2^4, 2^8.....2^(4k) all end in 6. Thus, 2^68 ends in 6, and so 2^67 ends in 8.
Answer: D
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