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# (3^2+1)(81^2+1)(9^2+1)(3^2−1)=

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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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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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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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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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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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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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This is an excellent question, which can be solved purely based on ONE algebraic identity. The only thing that makes it look difficult is the order in which the terms are given in the question.

The algebraic identity to be used in solving this question is $$a^2$$ – $$b^2$$ = (a+b) (a-b).

Refer to the figure below to understand how it can be broken down step by step using the algebraic identity:

Attachment:

3rd Sept 2019 - Reply 1.JPG [ 31.97 KiB | Viewed 4149 times ]

The correct answer option is C.
In such a question, observation is key. Observe the different numbers and try to see how they are related or how they can be tied in using a mathematical operation.
Hope this helps!
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