If $S_n = 2^n$, what is the unit's digit of $S_{67}?$

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

by VJesus12 » Sun Dec 08, 2019 7:49 am

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

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Source: Beat The GMAT — Problem Solving | Sun Dec 08, 2019 7:49 am

henilshaht: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Mon Sep 30, 2019 5:04 pm

by henilshaht » Sun Dec 08, 2019 8:20 am

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 $$2^3$$

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8155; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Tue Dec 10, 2019 7:18 pm

VJesus12 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

S(67) = 2^67

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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