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

It is currently 23 Jun 2018, 23:52

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If a, b, c are different positive integers, and a^2+b^2 = c^

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

5 KUDOS received
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 620
Premium Member
If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]

Show Tags

New post 03 Aug 2013, 06:56
5
6
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

39% (01:51) correct 61% (02:05) wrong based on 114 sessions

HideShow timer Statistics

If a, b, c are different positive integers, and a^2 + b^2 = c^2, then what is the value of (c - b)^2?

(1) a is a prime.

(2) b^2 is a multiple of 4.

Source: My own

_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

3 KUDOS received
Director
Director
User avatar
Joined: 14 Dec 2012
Posts: 810
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
GMAT ToolKit User
Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]

Show Tags

New post Updated on: 03 Aug 2013, 08:21
3
mau5 wrote:
If a,b,c are different positive integers,and \(a^2+b^2 = c^2\), then what is the value of \((c-b)^2\)?

I.a is a prime.

II.\(b^2\) is a multiple of 4.


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

statement1:
\(a\) is prime
therefore a = 1*a = (a cant have any other factor)
\(a^2= a*a = (c-b)(c+b)\) ==>since a is prime \(c-b\) should be \(1\) and \(c+b\) should be some perfect square.(as prime number factors are 1 and prime no.itself)
hence clearly \(c-b=1\)...hence sufficient

statement 2:
\(b^2\) is a multiple of 4

let \(b=4 c=5 a=3\)....therefore \(c-b = 1\)
another case : let\(b=8 c=10 a=6\)...therefore \(c-b=2\)
hence insufficient

hence it is A
_________________

When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...



GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabulary-list-for-gmat-reading-comprehension-155228.html
learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment
: http://www.youtube.com/watch?v=APt9ITygGss


Originally posted by blueseas on 03 Aug 2013, 08:15.
Last edited by blueseas on 03 Aug 2013, 08:21, edited 1 time in total.
3 KUDOS received
VP
VP
User avatar
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1099
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
GMAT ToolKit User
Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]

Show Tags

New post 03 Aug 2013, 08:16
3
1
If a,b,c are different positive integers,and \(a^2+b^2 = c^2\), then what is the value of \((c-b)^2\)?
\(a^2=c^2-b^2=(c-b)(c+b)\). Because \(c\neq{b},c-b\neq{0}\) we get \(c+b=\frac{a^2}{c-b}\).

I.a is a prime.
If a is prime, then \(a^2\) has three factors: 1,a and \(a^2\). Because c and b are integers, their sum must be an integer so \(c+b=integer=\frac{a^2}{c-b}\). That is true if c-b is a factor of \(a^2\):1, a or \(a^2\). If \(c-b=a^2\) then c+b=1 but this cannot be true as c and b are positive integers.
So we are left with \(c-b=a\) or \(c-b=1\). In the first case \(c-b=a\), substituting in the above equation we find that \((c-b)^2+b^2=c^2\) or \(2b(b-c)=0\), so b=0 (because \(c\neq{b}\)) but this contradicts the statement (b is a positive integer)
We are left with \(c-b=1\), so \((c-b)^2=1\).
Sufficient

II.\(b^2\) is a multiple of 4.
\(a^2+b^2 = c^2,3^2+4^2=5^2\) and \((c-b)^2=1\)
\(a^2+b^2 = c^2,(3*2)^2+(4*2)^2=(5*2)^2,=6^2+8^2=10^2\) and \((c-b)^2=4\). Not sufficient.
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Expert Post
1 KUDOS received
EMPOWERgmat Instructor
User avatar
D
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11823
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]

Show Tags

New post 25 Mar 2015, 12:10
1
1
Hi All,

While the math involved in this question is certainly algebra-based, the patterns are rooted more in Geometry (specifically, the Pythagorean Theorem). You can use that knowledge to your advantage here.

We're told that A, B and C are DIFFERENT positive integers and that A^2 + B^2 = C^2. We're asked for the value of (C-B)^2.

Since the variables are INTEGERS, and if you recognize the Pythagorean Theorem, you might choose to jot down a few examples....

eg
3/4/5
5/12/13
And the multiples of those Triplets....

Fact 1: A is PRIME.

Right away, we have a couple of TESTs that we can run....

IF...
A = 3, B = 4, C = 5
The answer to the question is .... (C-B)^2 = (5-4)^2 = 1

IF...
A = 5, B = 12, C = 13
The answer to the question is .... (C-B)^2 = (13-12)^2 = 1

Since we only have two TESTs so far, I'd look to also TEST a rarer Pythagorean Triplet....

IF....
A = 7, B = 24, C = 25
The answer to the question is... (C-B)^2 = (25-24)^2 = 1

At this point, a pattern clearly exists. The answer to the question is ALWAYS 1.
Fact 1 is SUFFICIENT

Fact 2: B^2 is a multiple of 4

IF...
A = 3, B = 4, C = 5
The answer to the question is .... (C-B)^2 = (5-4)^2 = 1

IF...
A = 6, B = 8, C = 10
The answer to the question is .... (C-B)^2 = (10-8)^2 = 4
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Manager
Manager
avatar
B
Joined: 04 May 2014
Posts: 162
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: If a, b, c are different positive integers, and a^2+b^2 = c^ [#permalink]

Show Tags

New post 09 Aug 2017, 18:49
a, b & c =+ve distinct integers
a²+b²=c²
or a²=c²-b²=(c-b)(c+b)
1) a is prime
plugin
a=3
a²=9
9=(c-b)(c+b)
now 9 is the product of (c-b)(c+b)
so it can be 1X9 or 9x1 or 3x3
c-b will be always less than c+b (sum of 2 positive distinct no will always be greater than their difference)
hence c-b can only be 1 and (c-b)²=(1)²-sufficient.
2) b² is multiple of 4
multiple values are possible
b²=4, 8, 16.....
b²=(c-a)(c+a)=1x4 or1x8 or 2x4 we can stop here-Not sufficient
Re: If a, b, c are different positive integers, and a^2+b^2 = c^   [#permalink] 09 Aug 2017, 18:49
Display posts from previous: Sort by

If a, b, c are different positive integers, and a^2+b^2 = c^

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.