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28 Mar 2023, 07:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
63% (02:43) correct
37%
(02:54)
wrong
based on 219
sessions
History
Date
Time
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Not Attempted Yet
If \(A = (1+\frac{1}{3})(1+\frac{1}{3^2})(1+\frac{1}{3^4})(1+\frac{1}{3^8}),\) what is \(1 - \frac{2}{3}A\)?
A. \(\frac{1}{3 } \)
B. \(\frac{2}{3^2 } \)
C. \(\frac{2}{3^4 }\)
D. \(\frac{2}{3^8} \)
E. \( \frac{1}{3^{16}}\)
A. \(\frac{1}{3 } \)
B. \(\frac{2}{3^2 } \)
C. \(\frac{2}{3^4 }\)
D. \(\frac{2}{3^8} \)
E. \( \frac{1}{3^{16}}\)
28 Mar 2023, 10:35
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\( A = (1 + \frac{1}{3})(1 + \frac{1}{3^2})(1 + \frac{1}{3^4})(1 + \frac{1}{3^8}) \)
\( A = (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
We can write \(\frac{2}{3} = \frac{(3 - 1)}{3}\). Also, remember \( (a+b)(a-b) = a^2 - b^2 \). We will apply this rule to simplify the problem.
=>\( \frac{2}{3}A = (\frac{3 - 1}{3}) (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^2-1}{3^2}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^4-1}{3^4}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^8-1}{3^8}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^{16}-1}{3^{16}}) \)
=>\( 1-\frac{2}{3}A = 1- \frac{3^{16}-1}{3^{16}} \)
=>\( 1-\frac{2}{3}A = \frac{3^{16} - 3^{16} + 1 }{ 3^{16}} \)
=>\( 1-\frac{2}{3}A = \frac{1 }{ 3^{16}} \)
=> Answer is E IMO
\( A = (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
We can write \(\frac{2}{3} = \frac{(3 - 1)}{3}\). Also, remember \( (a+b)(a-b) = a^2 - b^2 \). We will apply this rule to simplify the problem.
=>\( \frac{2}{3}A = (\frac{3 - 1}{3}) (\frac{3+1}{3}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^2-1}{3^2}) (\frac{3^2+1}{3^2}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^4-1}{3^4}) (\frac{3^4+1}{3^4}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^8-1}{3^8}) (\frac{3^8+1}{3^8}) \)
=>\( \frac{2}{3}A = (\frac{3^{16}-1}{3^{16}}) \)
=>\( 1-\frac{2}{3}A = 1- \frac{3^{16}-1}{3^{16}} \)
=>\( 1-\frac{2}{3}A = \frac{3^{16} - 3^{16} + 1 }{ 3^{16}} \)
=>\( 1-\frac{2}{3}A = \frac{1 }{ 3^{16}} \)
=> Answer is E IMO
General Discussion
23 Aug 2023, 22:56
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