GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2018, 12:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49858
If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

14 Sep 2015, 22:57
2
12
00:00

Difficulty:

75% (hard)

Question Stats:

58% (02:43) correct 42% (02:40) wrong based on 128 sessions

### HideShow timer Statistics

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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49858
If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

20 Sep 2015, 21:20
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.

_________________
##### General Discussion
CEO
Joined: 08 Jul 2010
Posts: 2529
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

14 Sep 2015, 23:18
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
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Manager
Joined: 10 Aug 2015
Posts: 103
Re: If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

15 Sep 2015, 03:12
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
Intern
Joined: 03 Feb 2014
Posts: 38
Location: United States
Concentration: Entrepreneurship, General Management
WE: General Management (Other)
Re: If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

15 Sep 2015, 04:23
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
_________________

--Shailendra

Non-Human User
Joined: 09 Sep 2013
Posts: 8392
Re: If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which  [#permalink]

### Show Tags

18 Apr 2018, 23:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If the quantity 5^2+ 5^4 + 5^6 is written as (a + b)(a – b), in which &nbs [#permalink] 18 Apr 2018, 23:40
Display posts from previous: Sort by