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Math Expert V
Joined: 02 Sep 2009
Posts: 58453
9^3(2^8 + 2^9/2) =  [#permalink]

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Difficulty:   35% (medium)

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

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

(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.

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Director  P
Joined: 21 May 2013
Posts: 636
Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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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
Director  B
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9^3(2^8 + 2^9/2) =  [#permalink]

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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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$$9^3*\frac{2^8+2^9}{2}$$
$$(3^2)^3*\frac{2^8*(1+2)}{2}$$
$$3^6*\frac{(3*2^8)}{2}$$
$$3^6*3*2^7$$
$$3^7*2^7$$
$$6^7$$
Manager  Joined: 29 Jul 2015
Posts: 155
Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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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.

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

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

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

= $$3^7 2^7$$

= $$6^7$$

Math Expert V
Joined: 02 Sep 2009
Posts: 58453
Re: 9^3(2^8 + 2^9/2) =  [#permalink]

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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.

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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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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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

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: 9^3(2^8 + 2^9/2) =   [#permalink] 30 Aug 2018, 18:03
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