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

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Math Expert
Joined: 02 Sep 2009
Posts: 64144
(5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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13 Nov 2014, 08:48
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Difficulty:

5% (low)

Question Stats:

88% (01:12) correct 12% (02:56) wrong based on 79 sessions

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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.

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Joined: 21 Jul 2014
Posts: 117
Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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13 Nov 2014, 09:35
1
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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Status: The Best Or Nothing
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Posts: 1709
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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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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Joined: 04 Jul 2014
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Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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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'
Math Expert
Joined: 02 Sep 2009
Posts: 64144
Re: (5^8 + 1)(5^4 + 1)(5^2 + 1)(5^2 - 1) =  [#permalink]

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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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