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5(2^4)(3^32) =

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Magoosh GMAT Instructor
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5(2^4)(3^32) =  [#permalink]

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New post 16 Jan 2015, 11:33
4
8
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

65% (01:50) correct 35% (02:24) wrong based on 158 sessions

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\(5(2^4)(3^{32})\) =
(A) \(5(6^{36})\)
(B) \((10^4)(3^{32})\)
(C) \((3^{34}) + (3^{32})\)
(D) \((3^{35}) - (3^{33})\)
(E) \((3^{36}) - (3^{32})\)


For a set of challenging GMAT problems on exponents and roots, as well as the OE for this problem, see:
http://magoosh.com/gmat/2014/challengin ... and-roots/

Mike :-)

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Re: 5(2^4)(3^32) =  [#permalink]

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New post 16 Jan 2015, 11:58
E.
Ques - 5 X (2^4=16) X 3^32 = 80 X 3^32
E = 3^32 x (3^4 - 1) = 80 x 3^32
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 16 Jan 2015, 13:11
1
IMO its E
firstly in now way we could use or modify 3^32
secondly a trick of using negative remainder.
so i used 2^4 with 5.. though after spending 1 min on it that its 80 and hence divisible by 3^4 as negative remainder of -1
so its 81-1... which is needed to be used
now here i got confused between D AND E
i did eliminated D simply because of its odd powers which didn't gave much sense to me as its a 2mins time constraint.
so hence i arrived at E
81 is 3^4 so no use of odd powers else result would be negative elsewise.

this is my first time i solved a question and gave a bit explanation too.

kudos will entice me to work hard...
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 17 Jan 2015, 10:09
Hello!
Another way out

5(2^4)(s^32)= 80(81^8)
as 3^4 is 81

80 can be written as 81-1
(81-1)(81^8)
(81^9)-81^8
3^36-3^32

Hence E
this is the simplest i could figured out

kindly share that binomial eq in imp for gmat?
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 19 Jan 2015, 01:38
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Answer = (E) \((3^{36}) - (3^{32})\)

\(5 * 2^4 * 3^{32}\)

\(= 80 * 3^{32}\)

\(= (81 - 1) * 3^{32}\)

\(= 3^{36} - 3^{32}\)
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 01 Mar 2015, 04:36
Yep I also did it similarly.

Starting with the question stem:
5 * 2^4 * 3^32
5 * 2^1 * 2^3 * 3^32
10 * 2^3 * 3^32
80 * 3^32

Then moving on to the answer choices, A and B are relatively easy to eliminate. Plus, the fact that we see additions and subtractions should alert as to the possibility of taking a common factor and creating a multiplication as an effect.

I started witd D which resulted in 3^33 * 8. Close, but not what I was looking for. So I moved to E, which results in:
3^36 - 3^32
3^32 (3^4 - 1)
3^32 (81 - 1)
3^32 * 80, which is what we are looking for. So, ANS E
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 12 Jul 2015, 11:19
1
\(5 * 2^4 * 3^{32}\)

\(= 10*8 * 3^{32}\)

\(= (3^2 + 1)*(3^2-1) * 3^{32}\)

\(= (3^4 - 1) * 3^{32}\)

\(= 3^{36} - 3^{32}\)[/quote]

--> Answer E
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Re: 5(2^4)(3^32) =  [#permalink]

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New post 12 Jan 2019, 07:51
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mikemcgarry wrote:
\(5(2^4)(3^{32})\) =
(A) \(5(6^{36})\)
(B) \((10^4)(3^{32})\)
(C) \((3^{34}) + (3^{32})\)
(D) \((3^{35}) - (3^{33})\)
(E) \((3^{36}) - (3^{32})\)



Quite an intriguing one.

\(5(2^4)(3^{32})\) = 80 * \((3^{32})\)

Now just browsing over the options you can get the value as E.
(A) \(5(6^{36})\)------------------------------------> 5* 2^{36} *3^36
(B) \((10^4)(3^{32})\)-----------------------------> wont give 80
(C) \((3^{34}) + (3^{32})\)-----------------------> wont give 80
(D) \((3^{35}) - (3^{33})\)------------------------> wont give 80
(E) \((3^{36}) - (3^{32})\)------------------------> took \(3^{32}\) common, giving \(3^4\)- 1 =80

Answer E.
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Re: 5(2^4)(3^32) =   [#permalink] 12 Jan 2019, 07:51
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