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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 118

Kudos [?]: 1899 , given: 397

Three digits have been removed from each of the following nu [#permalink]
Expert's post 00:00

Question Stats: 71% (01:05) correct 28% (01:31) wrong based on 14 sessions
Three digits have been removed from each of the following numbers. If n = 25, which of the numbers is equal to $$3 * 2^{n - 1}$$ ?

A. 47, __ __6, __23
B. 47, __ __6, __32
C. 49, __ __2, __64
D. 49, __ __2, __36
E. 50, __ __1, __48

Drill 2
Question: 12
Page: 551
[Reveal] Spoiler: OA

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Sandy
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Last edited by Carcass on 28 Mar 2018, 12:47, edited 1 time in total.
Edited by Carcass Manager  Joined: 15 Jan 2018
Posts: 147
GMAT 1: Q V
Followers: 3

Kudos [?]: 185  , given: 0

Re: Three digits have been removed from each of the following nu [#permalink]
2
KUDOS
They've deleted part of each number in the answer choices, and although it is possible (but painful) to figure out what digits are missing, it's not necessary. In fact, all we need is the last digit.

It's useful to know that any number, when multiplied by itself over and over and over, will form a pattern with the units digits of each successive term. For example, here are the first 8 powers of 2:
2
4
8
16
32
64
128
256
etc

Notice that the last digits form a pattern: 2, 4, 8, 6... etc. So, if we plug n into 3*2^(n-1), we get 3*2^24. Let's figure out what the units digit of 2^24 is. Since we know that every 4th term ends in 6, and we know that 2 to the 24th power will go through exactly 6 cycles of the pattern, and will therefore end with a 6. Only one of the answers ends in 6, but that's a trap.

We still need to deal with that 3. Any number you can think of that ends in 6, if multiplied by 3, will produce a number that ends in 8. Try it! So, since 2^24 ends in a 6, if we multiply it by 3, then the resulting huge number, even with a bunch of digits missing, must end in 8. Thus the answer is E.
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 118

Kudos [?]: 1899 , given: 397

Re: Three digits have been removed from each of the following nu [#permalink]
Expert's post
Explanation

The equation $$3 \times 2^{n–1}$$ follows a pattern.

When n = 1, the result is 3. When n = 2, the result is 6.

When n = 3, the result is 12. When n = 4, the result is 24. When n = 5, the result is 48. When n = 6, the result is 96. Beginning with the second term, the final digit in each result follows the pattern: 6, 2, 4, 8, 6, 2, 4, 8, etc.

The 25th term will thus end in the same digit as all the other kth terms, where k is one greater than a multiple of 4.

Thus, the (4 + 1)th, (8+1)th, (12+1)th.…(24+1)th terms all have a final digit of 8, and the only answer in which that is true is choice (E).
_________________

Sandy
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Try our free Online GRE Test Re: Three digits have been removed from each of the following nu   [#permalink] 06 Apr 2018, 15:12
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