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hurryupgmat
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Post subject: 2^5+2^5+3^5+3^5+3^5= Posted: Fri Aug 08, 2008 1:01 pm |
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This problem is from GMAT Prep..
QUESTION : 2^5+2^5+3^5+3^5+3^5= [EDITOR: question edited so that it's a correct transcription of the original from gmatprep] Choices are:
a. 5^6
b. 13^5
c. 2^6+3^6
d. 2^7+3^8
e. 4^5+9^5
Correct Answer is C
Please explain how is it C & Y not E
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guest
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Post subject: Posted: Fri Aug 08, 2008 4:18 pm |
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(2^5 +2^5)+(3^5+3^5+3^5)= 2^5(1+1)+3^5(1+1+1)=2^6+3^6
[editor: also changed, so that it's a correct answer to the question posted above]
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hurryupgmat
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Post subject: Thanks a Million :) Posted: Sat Aug 09, 2008 12:57 am |
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akshata.udiavar
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Post subject: Re: 2^6+2^6+3^6+3^6+3^6= Posted: Wed Dec 09, 2009 12:22 pm |
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| Students |
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Posts: 2
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I believe the actual question is 2^5+2^5+3^5+3^5+3^5 => 2^5(1+1)+3^5(1+1+1) => 2^5*2+3^5*3 => 2^6+3^6
Thus C is the correct option.
Thanks
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RonPurewal
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Post subject: Re: 2^5+2^5+3^5+3^5+3^5= Posted: Sat Jan 09, 2010 4:24 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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by the way, guys - even if you have NO IDEA how to solve this problem, there's still no excuse for getting it wrong.
because...
you can easily solve this problem WELL within the time guidelines, simply by DOING OUT THE ARITHMETIC.
2^5 is 32; 3^5 is 243. (you should probably have these memorized, but you can calculate them in a matter of seconds if necessary.) so the prompt is 32 + 32 + 243 + 243 + 243, which works out to 793.
(a) wrong, since all powers of 5 end with the digit 5.
(b) this is ridiculously big. note that 13^2 = 169 (a value you should know by heart), so that 13^3 is already much bigger than 793. and 13^5 is two powers above that.
(c) = 64 + 729 = 793
(d) and especially (e) are both several orders of magnitude larger than 793.
done.
don't forget these sorts of lowbrow, scrappy ways to solve problems. remember that there are no style points; if you can solve the problem, then you can solve the problem. "ugly" solutions are worth the same points as "pretty" solutions.
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