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9^3(2^8 + 2^9/2) =

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9^3(2^8 + 2^9/2) =  [#permalink]

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New post 02 Sep 2015, 22:24
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:17) correct 28% (01:47) wrong based on 231 sessions

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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 02 Sep 2015, 22:48
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Bunuel wrote:
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)


Kudos for a correct solution.


=9^3*2^8(2+1)/2
=3^2*3*2^8(3)/2
=3^6*2^7*3
=(3*2)^7
Answer C
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9^3(2^8 + 2^9/2) =  [#permalink]

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New post 03 Sep 2015, 00:12
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1
9^3(\(\frac{(2^8+2^9)}{2}\))
=9^3(2^-1)(2^8+2^9)
=9^3 * 2^7+9^3 * 2^8
=9^3*2^7(1+2)
=3^6*2^7(3)
=3^7 * 2^7
=(3*2)^7
=6^7

So,the Correct Answer is C
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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 03 Sep 2015, 00:27
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\(9^3*\frac{2^8+2^9}{2}\)
\((3^2)^3*\frac{2^8*(1+2)}{2}\)
\(3^6*\frac{(3*2^8)}{2}\)
\(3^6*3*2^7\)
\(3^7*2^7\)
\(6^7\)
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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 03 Sep 2015, 08:35
1
Bunuel wrote:
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)


Kudos for a correct solution.



\(9^{3}(\frac{2^{8} + 2^{9}}{2})\)

= \(3^{6}2^8(\frac{1+2}{2})\)

= \(3^{6}2^8(\frac{3}{2})\)

= \(3^7 2^7\)

= \(6^7\)

Answer:- C
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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 07 Sep 2015, 03:51
Bunuel wrote:
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

When working with exponents, you should try to find common (typically prime) bases and multiply. First to multiply, you can factor a 2^8 out of each additive term within the parentheses to get: \(9^{3} 2^{8} (\frac{(1+2)}{2})\). Then, perform the addition within the parentheses to get: \(9^{3}2^{8} (\frac{3}{2})\).

You can break down \(9^{3}\) to be \((3^{2})^3\), which is \(3^{6}\), and then you have common prime terms for each exponent: \(3^{6}2^{8} (\frac{3}{2})\).

Multiplying out the equation \(3^{6}2^{8} (\frac{3}{2})\), you get: \(3^7*3=3^8\) and \(\frac{2^{8}}{2} =2^7\), which can be simplified to \(3^7*2^7\).

This can be expressed as \((3*2)^{7} = 6^{7}\), and the correct answer is C.
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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 09 Jul 2017, 01:26
Bunuel wrote:
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)


Kudos for a correct solution.



\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

\(= 9^{3}(2^{7} + 2^{8})\)

\(= 9^{3} * 2^7 (2 + 1)\)

\(= 9^{3} * 2^7 * 3\)

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

\(= 3^{7} * 2^7\)

\(= 6^{7}\)

Hence, Answer is C
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Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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New post 30 Aug 2018, 18:03
Bunuel wrote:
\(9^{3}(\frac{2^{8} + 2^{9}}{2})=\)

(A) \(2^{7}3^{6}\)
(B) \(2^{6}3^{7}\)
(C) \(6^{7}\)
(D) \(2^{8}3^{7}\)
(E) \(12^{4}\)


We can re-express 9^3 as (3^2)^3, or 3^6, and we can re-express the fraction (2^8 + 2^9)/2 as two separate fractions. Thus, we have:

3^6[(2^8/2) + (2^9/2)]

3^6(2^7 + 2^8)

3^6[2^7(1 + 2)]

3^6[2^7(3)]

3^7 x 2^7 = 6^7

Answer: C
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Re: 9^3(2^8 + 2^9/2) =   [#permalink] 30 Aug 2018, 18:03
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