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# If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which

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Joined: 02 Sep 2009
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If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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14 Sep 2015, 21:57
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20
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Difficulty:

65% (hard)

Question Stats:

64% (03:11) correct 36% (02:20) wrong based on 172 sessions

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If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which both a and b are integers, which of the following could be the value of b?

A. 5
B. 10
C. 15
D. 20
E. 25

Kudos for a correct solution.

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Posts: 51258
Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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20 Sep 2015, 20:20
3
3
Bunuel wrote:
If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which both a and b are integers, which of the following could be the value of b?

A. 5
B. 10
C. 15
D. 20
E. 25

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

One first natural move is to factor $$5^2$$ out of the numeric expression:

$$5^2 + 5^4 + 5^6 = 5^2(1 + 5^2 + 5^4)$$

Meanwhile, $$(a + b)(a – b)$$ becomes $$a^2 – b^2$$.

So, somehow we have to turn $$5^2(1 + 5^2 + 5^4)$$ into a difference of squares. It must come down to the stuff in the parentheses.

Flip around those terms, so that we have the larger powers first: $$5^4 + 5^2 + 1$$.

This should look almost like a special product: $$(x + y)^2 = x^2 + 2xy + y^2$$, if $$x = 5^2$$ and $$y = 1$$. In fact, write that out:

$$(5^2+1)^2 = 5^4 + 2(5^2) + 1$$

The expression we actually have, $$5^4 + 5^2 + 1$$, is very close. Add $$5^2$$ and subtract it as well:

$$5^4 + 5^2 + 1$$
$$= 5^4 + 5^2 + 1 + 5^2 – 5^2$$
$$= (5^4 + 5^2 + 1 + 5^2) – 5^2$$
$$= (5^4 + 2(5^2) + 1) – 5^2$$
$$= (5^2 + 1)^2 – 5^2$$
$$= 26^2 – 5^2$$

Almost there. Remember, we had 5^2 as well:

$$5^2 + 5^4 + 5^6 = 5^2(1 + 5^2 + 5^4) = 5^2(26^2 – 5^2)$$
$$= (26*5)^2 – (5*5)^2$$
$$= 130^2 – 25^2$$.

Oher differences of squares can be equal to the original expression, but this is the only one that fits an answer choice: (E) 25.

There are other ways to solve this problem as well, such as backsolving.
$$5^2 + 5^4 + 5^6 = a^2 – b^2$$
$$5^2 + 5^4 + 5^6 + b^2 = a^2$$

Try different b’s from the answer choices, and see which one, when added to the original expression, gives you a perfect square.

Testing (E):

$$5^2+ 5^4 + 5^6 + 25^2$$
$$= 5^2 + 5^4 + 5^6 + 5^4$$
$$= 5^2 + 2×5^4 + 5^6$$
$$= (5 + 5^3)^2$$

This is the only answer that fits.

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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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14 Sep 2015, 22:18
Bunuel wrote:
If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which both a and b are integers, which of the following could be the value of b?

A. 5
B. 10
C. 15
D. 20
E. 25

Kudos for a correct solution.

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

5^2+ 5^4 + 5^6 = 5^2 (1+5^2+5^4)
if the above expression is solved even then every term of the expression will remain a multiple of 5^2 which is out of parenthesis
hence, b must be a multiple of 5^2 i.e. 25

One possible explanation is here,
5^2+ 5^4 + 5^6 = 5^2 (1+5^2+5^4) = 5^2 (1+2*5^2+5^4 - 5^2) = 5^2 [(1+5^2)^2 - (5)^2] = 5^2 [(1+5^2+5)(1+5^2-5)]
i.e. (5+5^4+5^3)(5+5^4-5^3) = (a + b)(ab)
i.e. b = 5^3
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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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15 Sep 2015, 02:12
1
Bunuel wrote:
If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which both a and b are integers, which of the following could be the value of b?

A. 5
B. 10
C. 15
D. 20
E. 25

Kudos for a correct solution.

Solution: $$5^2[1 + 2(5^2) + 5^4 - 5^2] = 5^2[(1 + 5^2)^2 - 5^2] = (5(1 + 5^2))^2 - 5^4 = a^2 - b^2 ==> b^2 = 5^4 ==> b= 25$$

Option E
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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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15 Sep 2015, 03:23
1
(a+b) (a-b) = a²-b²
5²+5^4+5^6 = 5² (1+5²+5^4) = 5² (1+5²+5^4+5²-5²) =5² [(1+5²)²-5²] = (5(1+5))² - (5²)²
→ a²-b² = (5(1+5))² - (5²)² → b² = (5²)² → b = 25
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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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01 Dec 2018, 09:19

GMATbuster's Weekly Quant Quiz#11 Ques #6

If the quantity $$5^2 + 5^4 + 5^6$$ is written as (a+b)(a-b), in which both a and b are integers, which of the following could be the value of b?

A)5
B)10
C)15
D)20
E)25
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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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03 Dec 2018, 07:32
1
5^2 + 5^4 + 5^6 = (a-b) (a+ b)
5^2( 1 + 25 + 625) = a^2 - b^2
5^2 ( 641 ) + b^2 = a^2
Note = it is given that a is integer so sqareroot of left hand side will also be integer.
Note 2= to make calc easy 26^ 2 is 676 and we are less of 25 ( 676-641) . So the b^2 should be such that 5^2 ( 641 + 25)

Please do note that we took 5^2 common .

b^2 = 25*25
b=25

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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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03 Dec 2018, 20:57
Bunuel wrote:
If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which both a and b are integers, which of the following could be the value of b?

A. 5
B. 10
C. 15
D. 20
E. 25

Kudos for a correct solution.

5^2 + 5^4 + 5^6
= 5^2 (1+5^2+5^4)
= 5^2 (1+2.5^2+5^4 - 5^2)
= 5^2 ((1+5^2)^2 - 5^2)
= [(5.(1+5^2))^2 - 25^2]

Value of b = 25
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Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

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03 Dec 2018, 22:18
Using Prime Factorization we need to convert this expression into products of two integers as indicated in the question.
5^2+5^4+5^6 = 5^2(1+5^2+5^4) = 25(1+25+625) = 25*651 = (5*5)*(3*217) = 5*5*3*7*31 = 105*155 = (130+25) * (130-25)
Re: If the quantity 5^2 + 5^4 + 5^6 is written as (a + b)(a – b), in which &nbs [#permalink] 03 Dec 2018, 22:18
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