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# (3^2+1)(81^2+1)(9^2+1)(3^2−1)=

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Math Expert
Joined: 02 Sep 2009
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29 Aug 2016, 04:05
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Difficulty:

25% (medium)

Question Stats:

80% (01:51) correct 20% (02:04) wrong based on 249 sessions

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$$(3^2+1)(81^2+1)(9^2+1)(3^2−1)=$$

A. $$81^4+1$$

B. $$9^9−1$$

C. $$3^{16}−1$$

D. $$3^{32}−1$$

E. $$3^{96}+1$$

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29 Aug 2016, 04:11
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)
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29 Aug 2016, 04:14
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)

Notice that A is not 81^4-1 it's 81^4+1.
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29 Aug 2016, 04:27
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)

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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29 Aug 2016, 04:34
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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Updated on: 29 Aug 2016, 12:52
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Top Contributor
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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

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Originally posted by GMATPrepNow on 29 Aug 2016, 12:17.
Last edited by GMATPrepNow on 29 Aug 2016, 12:52, edited 1 time in total.
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29 Aug 2016, 12:47
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)

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
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30 Aug 2016, 11:36
Nice Question Bunuel
Here is my approach =>
using a+b * a-b = a^2-b^2
=> (3^2+1) *(3^2-1)(9^2+1)(81^2+1)
=> (9^2-1)*(9^1+1)(81^1+1)
=> (81^2-1)(81^2+1)
=>81^4-1
=>(3^4)^4-1
=>3^16-1
SMASH THAT C
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30 Aug 2016, 18:18
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?

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09 Aug 2018, 07:07
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Re: (3^2+1)(81^2+1)(9^2+1)(3^2−1)=   [#permalink] 09 Aug 2018, 07:07
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