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# 2^1 + 2^2 + 2^3 + 2^4 + 2^5

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Senior Manager
Joined: 20 May 2014
Posts: 282
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Kudos [?]: 49 [0], given: 220

2^1 + 2^2 + 2^3 + 2^4 + 2^5 [#permalink]  05 Nov 2017, 01:41
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
$$2^1 + 2^2 + 2^3 + 2^4 + 2^5=$$

(A) 2^14
(B) 2^15
(C) 2^9 + 2^5
(D) 2^7 – 2
(E) 2^6 – 2

Kudos for correct solution.
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 521
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Kudos [?]: 334 [1] , given: 66

Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5 [#permalink]  05 Nov 2017, 06:32
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KUDOS
I think there are many strategies to face this kind of questions, mine is the following: factorizing 2 repetitively from the sum of powers of 2, i.e.

[m]2+2^2+2^3+2^4+2^5 = 2(1+2+2^2+2^3+2^4) = 2(1+2(1+2+2^2+2^3)) = \dots = 2(1+2(1+2(1+2(1+2)))) = 62/[m].

Then, knowing the first powers of 2 it is easy to see that 2^6 = 64, so that 62 = 2^6-2.

Re: 2^1 + 2^2 + 2^3 + 2^4 + 2^5   [#permalink] 05 Nov 2017, 06:32
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