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Bunuel
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(2^3^2 - (2^3)^2)/28 = 03 Aug 2018, 03:50
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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:

35% (medium)

Question Stats:

60% (01:16) correct 40% (00:52) wrong based on 128 sessions

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\(\frac{{2^3}^{2} - {(2^3)}^{2}}{28}\) =


A. 0

B. \(\frac{4}{7}\)

C. 8

D. 12

E. 16
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E
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LevanKhukhunashvili
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GMAT 1: 640 Q48 V28
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(2^3^2 - (2^3)^2)/28 = 03 Aug 2018, 03:55
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As math experts say begin from the top
\(\frac{2^9-2^9}{28}\)=0

IMO
Ans: A
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sudarshan22
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(2^3^2 - (2^3)^2)/28 = Updated on: 03 Aug 2018, 15:58
Originally posted by sudarshan22 on 03 Aug 2018, 04:04.
Last edited by sudarshan22 on 03 Aug 2018, 15:58, edited 2 times in total.
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(2^3^2 - (2^3)^2) / 28
(2^9 - 2^6 ) / 28
2^6 * (2^3 - 1) / 4 * 7
2^6 * 7 / 4 * 7
64 / 4
16

Hence, E.
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(2^3^2 - (2^3)^2)/28 = 03 Aug 2018, 15:51
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Bunuel
\(\frac{{2^3}^{2} - {(2^3)}^{2}}{28}\) =


A. 0

B. \(\frac{4}{7}\)

C. 8

D. 12

E. 16
Show more
Careful*: \({2^3}^{2}=(2)^{(3^2)}=2^9\)
On the other hand, \({(2^3)}^{2}=(2)^{(3*2)}=2^6\)

\(\frac{{2^3}^{2} - {(2^3)}^{2}}{28}\)

\(\frac{2^{9}-2^6}{2^2*7}\)

\(\frac{2^{6}(2^3-1)}{2^2*7}=\frac{2^6*7}{2^2*7}=\)

\(\frac{2^6}{2^2}=2^{(6-2)}=2^4=16\)

Answer
E


*\(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\). So the first term in the numerator is \(2^9\). The converse of the initial statement is also true. The second term in the prompt has one exponent inside and one exponent outside the parentheses. That placement of parentheses and exponents mimics this term: \((a^m)^n\) -- and not this term: \(a^{(m^n)}\) . The second term in the numerator, in other words, is \(2^6\). See Bunuel , here - scroll down to "Exponents."
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(2^3^2 - (2^3)^2)/28 = 06 Aug 2018, 23:26
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sudarshan22
(2^3^2 - (2^3)^2) / 28
(2^9 - 2^6 ) / 28
2^6 * (2^3 - 1) / 4 * 7
2^6 * 7 / 4 * 7
64 / 4
16

Hence, E.
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Hello sudarshan22 Bunuel,
As per my knowledge both elements are same and answer should be A(0).
Could you please suggest any resource for my reference.
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(2^3^2 - (2^3)^2)/28 = 07 Aug 2018, 00:57
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jackspire
Hello sudarshan22 Bunuel,
As per my knowledge both elements are same and answer should be A(0).
Could you please suggest any resource for my reference.
Show more

jackspire
Answers to of your questions have been provided in the above post by generis, please check the post and all your doubts will be clear.
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(2^3^2 - (2^3)^2)/28 = 07 Aug 2018, 03:41
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sudarshan22
jackspire
Hello sudarshan22 Bunuel,
As per my knowledge both elements are same and answer should be A(0).
Could you please suggest any resource for my reference.

jackspire
Answers to of your questions have been provided in the above post by generis, please check the post and all your doubts will be clear.
Show more

sudarshan22,
Yup. Missed it.
Thanks.
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(2^3^2 - (2^3)^2)/28 = 07 Aug 2018, 03:58
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Bunuel
\(\frac{{2^3}^{2} - {(2^3)}^{2}}{28}\) =


A. 0

B. \(\frac{4}{7}\)

C. 8

D. 12

E. 16
Show more
OA:E
\(\frac{{2^3}^{2} - {(2^3)}^{2}}{28}=\frac{{2^9 - 2^3*2^3}}{28}=\frac{{2^9 - 2^6}}{28}=\frac{2^6(2^3-1)}{28}=\frac{2^6*7}{2^2*7}=2^4=16\)
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