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(3^2+1)(81^2+1)(9^2+1)(3^2−1)=

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(3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 03:05
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A
B
C
D
E

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79% (01:53) correct 21% (02:13) wrong based on 218 sessions

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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 03:11
Bunuel wrote:
(3^2+1)(81^2+1)(9^2+1)(3^2−1)=

A. 81^4+1
B. 9^9−1
C. 3^16−1
D. 3^32−1
E. 3^96+1


(3^2+1)(81^2+1)(9^2+1)(3^2−1)=(81^2+1)(9^2+1)(9^2−1)=(81^2+1)(9^4-1)=(81^2+1)(81^2-1) = (81^4-1)
Answer is A
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 03:14
Senthil1981 wrote:
Bunuel wrote:
(3^2+1)(81^2+1)(9^2+1)(3^2−1)=

A. 81^4+1
B. 9^9−1
C. 3^16−1
D. 3^32−1
E. 3^96+1


(3^2+1)(81^2+1)(9^2+1)(3^2−1)=(81^2+1)(9^2+1)(9^2−1)=(81^2+1)(9^4-1)=(81^2+1)(81^2-1) = (81^4-1)
Answer is A


Notice that A is not 81^4-1 it's 81^4+1.
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 03:27
Bunuel wrote:
Senthil1981 wrote:
Bunuel wrote:
(3^2+1)(81^2+1)(9^2+1)(3^2−1)=

A. 81^4+1
B. 9^9−1
C. 3^16−1
D. 3^32−1
E. 3^96+1


(3^2+1)(81^2+1)(9^2+1)(3^2−1)=(81^2+1)(9^2+1)(9^2−1)=(81^2+1)(9^4-1)=(81^2+1)(81^2-1) = (81^4-1)
Answer is C A


Notice that A is not 81^4-1 it's 81^4+1.


Yup, just noticed. Thanks for highlighting.

so 81^4-1 = > (3^4)^4-1 => 3^16-1
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 03:34
Bunuel wrote:
(3^2+1)(81^2+1)(9^2+1)(3^2−1)=

A. 81^4+1
B. 9^9−1
C. 3^16−1
D. 3^32−1
E. 3^96+1


Option C

81^4-1= 3^16-1
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(3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post Updated on: 29 Aug 2016, 11:52
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Bunuel wrote:
\((3^2+1)(81^2+1)(9^2+1)(3^2−1)=\)

A. \(81^4+1\)

B. \(9^9−1\)

C. \(3^{16}−1\)

D. \(3^{32}−1\)

E. \(3^{96}+1\)


This question tests our ability to recognize a potential difference of squares.
That is: (a + b)(a - b) = a² - b²

(3² + 1)(81² + 1)(9² + 1)(3² − 1) = (3² + 1)(3² − 1)(9² + 1)(81² + 1)
= (9 + 1)(9 − 1)(9² + 1)(81² + 1)
= (9² − 1)(9² + 1)(81² + 1)
= (81 - 1)(81 + 1)(81² + 1)
= (81² - 1)(81² + 1)
= 81⁴ - 1
Check the answer choices......81⁴ - 1 is not among the answer choices. We'll have to REWRITE our answer...
81⁴ - 1 = (3⁴)⁴ - 1
= 3¹⁶ - 1

Answer:

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Originally posted by GMATPrepNow on 29 Aug 2016, 11:17.
Last edited by GMATPrepNow on 29 Aug 2016, 11:52, edited 1 time in total.
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(3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 29 Aug 2016, 11:47
Bunuel wrote:
\((3^2+1)(81^2+1)(9^2+1)(3^2−1)=\)

A. \(81^4+1\)

B. \(9^9−1\)

C. \(3^{16}−1\)

D. \(3^{32}−1\)

E. \(3^{96}+1\)


Property : a^2-b^2= (a+b)(a-b)

So, (3^2+1)(3^2−1) = 3^4-1

The expression becomes : (3^4-1)(3^8+1)(3^4+1)

Expression= (3^8+1)(3^8-1)
or, (3^16-1)
Answer C Simple as that!


Or through another route-
The expression becomes : (3^4-1)(81^2+1)(9^2+1)
Since (3^4-1) = (9^2-1)
Expression= (9^2-1)(81^2+1)(9^2+1)
or, (81^2+1)(81^2-1)
or, (81^4-1)
or, (3^16-1)

C again :lol:
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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New post 30 Aug 2016, 17:18
Most of us (including me) have solved this using the classical approach.

I have another possible approach but would need help from the experts. Notice that the first term is 10 so the entire expression will end in a 0. Now, let's use concept of units digits and check each options.

a. Unit digit = 1 + 1 = 2 -> reject
b. Unit digit = 9 - 1 = 8 -> reject
c. Unit digit = 1 - 1 = 0 -> hold
d. Unit digit = 1 - 1 = 0 -> hold
e. Unit digit = 1 + 1 = 2 -> reject

At this point, I'm not entirely sure how to pick between c and d. Can we eyeball the powers of 3 in the original expression to rule out option d since the power of 3 is too big and it is unlikely to have that many 3s in the expansion?

Experts, please suggest.
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=  [#permalink]

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