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(5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =

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(5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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New post 13 Nov 2014, 08:48
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A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

88% (01:12) correct 12% (02:56) wrong based on 80 sessions

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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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New post 13 Nov 2014, 09:35
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1
Bunuel wrote:

Tough and Tricky questions: Arithmetic.



\((5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =\)

A. \(5^{16} - 1\)
B. \(5^{16} + 1\)
C. \(5^{32} - 1\)
D. \(5^{128} - 1\)
E. \(5^{16}(5^{16} - 1)\)

Kudos for a correct solution.


Work shown below for the easy solution. As a general rule, it is always good to look at the patterns in the problem when it looks like there might be a lot of tedious math involved.

Skills you need to develop for this problem: understand exponent properties, multiplying equations (and factoring equations).

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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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New post 13 Nov 2014, 19:40
1
Answer = A = \(5^{16} - 1\)

Just add the powers of 5

\((5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) = 5^{(8+4+2+2)} - 1 = 5^{16} - 1\)
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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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New post 14 Nov 2014, 04:12
1
Bunuel wrote:

Tough and Tricky questions: Arithmetic.



\((5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =\)

A. \(5^{16} - 1\)
B. \(5^{16} + 1\)
C. \(5^{32} - 1\)
D. \(5^{128} - 1\)
E. \(5^{16}(5^{16} - 1)\)

Kudos for a correct solution.



The last two terms (5^2 + 1) (5^2 - 1) are of the form (a - b) (a + b). So it can be converted to the form (a^2 - b^2).

So now the 3rd term is ( (5^2)^2 - (1)^2) = (5^4 - 1)

Similarly 2nd and 3rd term becomes (5^8 - 1)

Now the first two becomes (5^16 - 1)

Hence Answer 'A'
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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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New post 14 Nov 2014, 08:24
Bunuel wrote:

Tough and Tricky questions: Arithmetic.



\((5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =\)

A. \(5^{16} - 1\)
B. \(5^{16} + 1\)
C. \(5^{32} - 1\)
D. \(5^{128} - 1\)
E. \(5^{16}(5^{16} - 1)\)

Kudos for a correct solution.


Official Solution:

\((5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =\)

A. \(5^{16} - 1\)
B. \(5^{16} + 1\)
C. \(5^{32} - 1\)
D. \(5^{128} - 1\)
E. \(5^{16}(5^{16} - 1)\)


The question asks us to simplify an expression.

We do not need to use FOIL to multiply the terms out. Instead, notice that \((5^2 + 1)(5^2 - 1)\) is in the form \((x + y)(x - y)\), which is the difference of two squares: \((x + y)(x - y) = x^2 - y^2\).

Since \((5^2 + 1)(5^2 - 1) = 5^4 - 1\), the original expression becomes: \((5^8 + 1)(5^4 + 1)(5^4 - 1)\).

Another difference of squares has appeared: \((5^4 + 1)(5^4 - 1)\).

We use the same method to get the final answer: \((5^8 + 1)(5^4 + 1)(5^4 - 1) = (5^8 + 1)(5^8 - 1) = (5^16 - 1)\).


Answer: A.
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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =   [#permalink] 14 Nov 2014, 08:24

(5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =

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