Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 56251

Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
05 Dec 2015, 08:52
Question Stats:
29% (02:15) correct 71% (01:48) wrong based on 270 sessions
HideShow timer Statistics
Math Revolution and GMAT Club Contest Starts! QUESTION #6:For positive integers a and b, when a^4  b^4 is divided by 3, what is the remainder? (1) When a + b is divided by 3, the remainder is 0 (2) When a^2 + b^2 is divided by 3, the remainder is 2 Check conditions below: Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum We are happy to announce a Math Revolution and GMAT Club Contest For the following four (!) weekends we'll be publishing 4 FRESH math questions per weekend (2 on Saturday and 2 on Sunday). To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific. Then a week later, the forum moderator will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6months access to GMAT Club Tests. PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize: PS + DS course with 502 videos that is worth $299! All announcements and winnings are final and no whining GMAT Club reserves the rights to modify the terms of this offer at any time. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you!
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Intern
Joined: 29 Sep 2014
Posts: 15

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 13:54
Wow, I found this one super hard and got it wrong at first, but I've managed to prove that the answer is D, either statement is sufficient. I'm sure i've missed the obvious trick here, but this is how i've done it: For positive integers a and b, when \(a^4  b^4\) is divided by 3, what is the remainder? (1) When \(a + b\) is divided by 3, the remainder is 0 (2) When \(a^2 + b^2\) is divided by 3, the remainder is 2 Firstly note that a and b are positive integers therefore anything constructed from a, b and integer powers of a and b without any division will be an integer. Let's start by factorising the original statement to break it down into easier parts. \(a^4  b^4 = (a^2 + b^2)(a^2  b^2) = (a^2 + b^2)(a + b)(a  b)\) (1) tells us that \(a + b\) is divisible by 3 as the remainder is 0. Therefore any multiple of \((a + b)\) is divisible by 3. Since we have shown above that \(a^4  b^4\) is a multiple of \((a + b)\), and since we know that the other terms in the above equation are integers, we know that \(a^4  b^4\) is divisible by 3 and will therefore have a remainder of 0 when divided by 3. Therefore statement (1) is sufficient and the possible answers are A and D. (2) tells us that \(a^2 + b^2\) is not divisible by 3, and has a remainder of 2 when divided by 3. From the above we can see that \(a^4  b^4 = (a^2 + b^2)(a^2  b^2)\), and therefore we need to test whether \(a^2 + b^2\) having a remainder of 2 means that \((a^2  b^2)\) has a certain remainder when divided by 3. Let's consider \(a^2\) and \(b^2\) separately. If \(a^2 + b^2\) has a remainder of 2, then three options are possible: \(a^2\) has a remainder of 0 and \(b^2\) has a remainder of 2 \(a^2\) and \(b^2\) each have a remainder of 1 \(a^2\) has a remainder of 2 and \(b^2\) has a remainder of 0 Integers can only possibly have remainders 0, 1, or 2, when divided by 3, so each integer fits into one of 3 camps: \(x\), \(x+1\), or \(x+2\), where x is divisible by 3. Therefore every square number is either \(x^2\), \((x+1)^2 = x^2 + 2x + 1\), or \((x+2)^2 = x^2 + 4x + 4\). The only terms here not divisible by 3 are 1 and 4, both of which have a remainder 1. Therefore square numbers are either divisible by 3 or have remainder 1 when divided by 3. This let's us rule out the first and last possibility, so we know that if \(a^2 + b^2\) has a remainder of 2 when divided by 3, \(a^2\) and \(b^2\) each have a remainder of 1. If \(a^2\) and \(b^2\) each have a remainder of 1, then \((a^2  b^2)\) has a remainder 0 and is always divisible by 3. In that case \(a^4  b^4\) is divisible by 3 as it is a multiple of \((a^2  b^2)\). Therefore (2) is also sufficient to calculate the remainder of \(a^4  b^4\) when divided by 3. The answer to the question is D, either statement is sufficient on it's own.
_________________
Target GMAT 650. Please help by giving Kudos, it's free!




Board of Directors
Joined: 17 Jul 2014
Posts: 2539
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
05 Dec 2015, 11:58
I've been struggling to find the answer for this one...very tough and tricky, but if you try algebraic way, then one works and other one not always, and while picking numbers  different results...anyways...below my explanation. we can rewrite the initial statement: a^4b^4 = (a^2+b^2)(a^2b^2) (a^2b^2) can be rewritten as (a+b)(ab) or the whole: a^4b^4 = (a^2+b^2)(a+b)(ab) first statement: a+b is divisible by 3. well, in this case, the reminder will always yield 0, since a^4b^4 is a multiple of 3, because a+b is a multiple of 3. statement 1 is sufficient. statement 2: a^2+b^2, if divided by 3, will yield a remainder of 2. well, this doesn't tell much. What if a+b is a multiple of 3? what if ab is a multiple of 3? clearly 2 should't be sufficient, right? well, not exactly! if you try picking numbers, you can see that all the times, the remainder is 0. and this is what confuses me the most. Maybe I miss something? a=2 b=1 a^2 + b^2 = 5. 5/3 = 1 and remainder 2. a^4  b^4 = 16  1 = 15. 15/3 = 5 and remainder 0. a=4 and b=1 a^2 + b^2 = 16+1 = 17. 17/3 = 5 and remainder 2. a^4  b^4 = 2561 = 255. remainder again 0. and the pattern is the same for all numbers that are picked and that satisfy the condition of the statement 2. I really doubt this is a 600700 level question. Or maybe I am not in the mood for solving math today? :\
_________________



Retired Moderator
Joined: 22 Jun 2014
Posts: 1102
Location: India
Concentration: General Management, Technology
GPA: 2.49
WE: Information Technology (Computer Software)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
05 Dec 2015, 15:21
when a^4  b^4 is divided by 3, what is the remainder? using the formula (a^2  b^2) = (a + b)*(a  b), we can write a^4  b^4 as (a^2 + b^2)*(a^2  b^2) i.e. (a^2 + b^2)*(a + b)*(a  b) statement1: When a + b is divided by 3, the remainder is 0.it means a+b is a multiple of 3. when a multiple of 3 is multiplied some other number the resulting number would also be a multiple of 3. (a^2 + b^2)*(a + b)*(a  b) would be divisible by 3. hence remainder would always be 0. We get a definite answer hence statement1 is SUFFICIENT. statement2: When a^2 + b^2 is divided by 3, the remainder is 2It means Neither a nor b can be a multiple of 3. if a or b or both are multiple of 3 then statement2 will not hold. this can be explained as below: 1) if a & b both are multiple of 3 then remainder would be zero NOT 2 when a^2 + b^2 is divided by 3 2) Neither a nor b can be a multiple of 3. suppose a is multiple of 3 but b is not. In this case clearly a^2/3 will give remainder as 0 BUT b^2/3 will give remainder 0 or 1 NOT 2 because no perfect square ends in 2,8.we conclude that a and b can take any value which is NOT a multiple of 3 i.e. if n is a multiple of 3 then a or b or both can take either (n+1) or (n+2) or (n1) or (n2). but a or b can never taken n. For all such values of a & b either a+b or ab would be divisible by 3 i.e. remainder would be zero and a^2 + b^2/3 would give remainder of 2. this can be illustrated by following examples to bring more clarity: say n = 6 then a = (n+1) = 7 and b = (n+2) = 8 (notice that both 7 and 8 gives remainder of 1 when divided by 3, their squares also will give remainder as 1) a^2 + b^2 = (49 + 64)/3 = 113/3 gives remainder 2 because 79 and 64 both gives 1 as remainder. a+b/3 = 8+7 = 15/3 gives remainder 0 say n = 6 then a = (n+1) = 7 and b = (n2) = 4 (notice that both 7 and 4 gives remainder of 1 when divided by 3, their squares also will give remainder as 1) a^2 + b^2 = (49 + 16)/3 = 65/3 gives remainder 2 because 49 and 16 both gives 1 as remainder. ab/3 = 74= 3/3 gives remainder 0 say n = 6 then a = (n+1) = 7 and b = (n1) = 5 ( notice that gives remainder of 1 BUT 5 gives remainder of 2 when divided by 3, BUT THE FACT is that their squares will give remainder as 1 only) a^2 + b^2 = (49 + 25)/3 = 74/3 gives remainder 2 because 49 and 25 both gives 1 as remainder. a+b/3 = 7+5= 12/3 gives remainder 0 now we see When a^2 + b^2 is divided by 3, the remainder is 2 but (a^2 + b^2)*(a + b)*(a  b) would be divisible by 3 because either a+b or ab would be divisible by 3. hence remainder would always be 0. We get a definite answer hence statement1 is SUFFICIENT.Option (D) is the correct answer.
_________________



Intern
Joined: 09 Jul 2015
Posts: 49

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
05 Dec 2015, 17:35
The answer is A. Question is what is the reminder of a^4b^4 when divided by 3. a^4b^4, can be written as (a^2+b^2)(a^2b^2) or (a^2+b^2)(a+b)(ab). so when this is divided by 3, what is the reminder. 1) (a+b) / 3 > remainder is 0. This means (a^2+b^2)(a+b)(ab) is divisible by 3 itself, so the answer when a^4b^4 is divided by 3, the reminder is of course 0. (SUFFICIENT) 2) a^2+b^2 divided by 3 > remainder is 2. We will now need to see the different possibilities of what a^2 and b^2 will be, so we can find out the remainder when (a^2+b^2)(a^2b^2) is divided by 3. There are the following possibilities  when a^2+b^2 divided by 3, 1. (a^2)/3 yields the remainder of 2 and (b^2)/3 yields the remainder 0. In this case a^2 can be written as 3*x+2 and b^2 can be 3y 2. (a^2)/3 yields the remainder of 0 and (b^2)/3 yields the remainder 2. In this case a^2 can be written as 3*x and b^2 can be 3y+2 3. (a^2)/3 yields the remainder of 1 and (b^2)/3 yields the remainder 1. In this case a^2 can be written as 3*x+1 and b^2 can be 3y+1. Here is x and y are any +ve integers. Let us now substitute each of these values of of a^2 and b^2 into (a^2+b^2)(a^2b^2) 1.> a^2 = (3x+2), b^2 =3y. Simplifying (a^2+b^2)(a^2b^2) = 9(x^2y^2)+6(x+y)+6(xy)+4. In this case the remainder is 1. 2.> a^2 = (3x), b^2 =(3y+2). Simplifying (a^2+b^2)(a^2b^2) = 9(x^2y^2)6(x+y)+6(xy)4. In this case the remainder is 2. 3.> a^2 = (3x+1), b^2 =(3y+1). Simplifying (a^2+b^2)(a^2b^2) = 9(x^2y^2)+6(x+y). In this case the remainder is 0. Hence, option 2 is NOT SUFFICIENT. Hence answer is A.
_________________
Please kudos if you find this post helpful. I am trying to unlock the tests



Retired Moderator
Joined: 13 Apr 2015
Posts: 1676
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
05 Dec 2015, 21:50
a^4  b^4 = (a^2  b^2)(a^2 + b^2) = (a + b)(a  b)(a^2 + b^2) R = ? when (a + b)(a  b)(a^2 + b^2)/3
St1: R = 0 when (a + b)/3 > Implies that a^4  b^4 is divisible by 3 > Sufficient
St2: R = 2 when (a^2 + b^2)/3 This implies, a^2 + b^2 = 3k + 2 = 2, 5, 8, 11, .................... i.e. Sum of squares should be equal to 2, 5, 8, 11, .... Note that certain integers such as 11 cannot be considered here because 11 cannot be obtained as sum of the squares of two positive integers. Assume, a^2 + b^2 = 5 > a = 2; b = 1 > R = 0 when (a + b)/3 > So the entire expression is divisible by 3 Assume, a^2 + b^2 = 10001 > a = 100; b = 1 > R = 0 when (a  b)/3 > Expression is divisible by 3 Assume, a^2 + b^2 = 32 > a = 4; b = 4 > R = 0 when (ab)/3 > Expression is divisible by 3 Therefore, Statement 2 alone is sufficient.
Answer: D



Intern
Joined: 21 Nov 2014
Posts: 31
Location: Viet Nam

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 04:27
\(a^4  b^4=(a^2b^2)(a^2+b^2)=(ab)(a+b)(a^2+b^2)\) (1) is sufficient and we will have the remainder when \(a^4b^4\) is divided by 3 is 0. (2) When \(a^2 + b^2\) is divided by 3, the remainder is 2. For a random number A we have 3 cases: A = 3k, A = 3k+1, or A = 3k + 2. The first case: \(A^2=9k^2\) > The remainder when \(A^2\) is divided by 3 is 0. The second case: \(A^2=9k^2+6k+1\) > The remainder when \(A^2\) is divided by 3 is 1. The third case: \(A^2=9k^2 + 12k + 4\) > The remainder when \(A^2\) is divided by 3 is also 1. => From (2) both a & b are not divisible by 3 (if so the remainder of \(a^2 + b^2\) divided by 3 is either 1 or 0) We have 4 cases for a & b: a = 3u + 1 & b = 3v + 1 => ab = 3(uv) divisible by 3 and hence \(a^4b^4\) is divisible by 3. a = 3u + 2 & b = 3v + 2 => ab = 3(uv) divisible by 3 and hence \(a^4b^4\) is divisible by 3. a = 3u + 2 & b = 3v + 1 => a+b = 3(u+v+1) divisible by 3 and hence \(a^4b^4\) is divisible by 3. a = 3u + 1 & b = 3v + 2 => a+b = 3(u+v+1) divisible by 3 and hence \(a^4b^4\) is divisible by 3. => The remainder of \(a^4b^4\) when divided by 3 is 0. and hence (2) is sufficient. Answer: C
_________________
GMAT Group for Vietnamese:
https://www.facebook.com/groups/644070009087525/



Manager
Joined: 02 Nov 2014
Posts: 185
GMAT Date: 08042015

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 09:20
QUESTION #1:
For positive integers a and b, when a^4  b^4 is divided by 3, what is the remainder?
(1) When a + b is divided by 3, the remainder is 0 (2) When a^2 + b^2 is divided by 3, the remainder is 2
Soln: We know, \(x^2 y^2 = (x+y)*(xy)\), so, \(a^4  b^4 = (a^2 + b^2)*(a^2  b^2) = (a^2 + b^2)*(a+b)*(ab)\)
1. Since \((a+b)\) is divisible by 3. We can conclude that \(a^4  b^4 = (a^2 + b^2)*(a+b)*(ab)\) will also be divisible by 3. So, remainder must be 0. SUFFICIENT.
2. \(a^2 + b^2\) is divided by 3, the remainder is 2. Lets put some values of a and b and check the value of \(a^2  b^2.\) If \(a = 2, b = 1,\) so \(a^2 + b^2 = 5 = 3*1+2.\) We have \(a^2  b^2 = 3\) ...divisible by 3. So, the remainder when \(a^4  b^4\) is divided by 3 will be 0. YES. If \(a = 4, b = 2,\) so \(a^2 + b^2 = 20 = 3*6+2.\) We have \(a^2  b^2 = 12\) ...divisible by 3. So, the remainder when \(a^4  b^4\) is divided by 3 will be 0. YES. If \(a = 2, b = 5,\) so \(a^2 + b^2 = 29 = 3*9+2.\) We have \(a^2  b^2 = 21\) ...divisible by 3. So, the remainder when \(a^4  b^4\) is divided by 3 will be 0. YES. If \(a = 2, b = 10,\) so \(a^2 + b^2 = 104 = 3*34+2.\) We have \(a^2  b^2 = 96\)...divisible by 3. So, the remainder when \(a^4  b^4\) is divided by 3 will be 0. YES. If \(a = 10, b = 4,\) so \(a^2 + b^2 = 116 = 3*38+2.\) We have \(a^2  b^2 = 84\) ...divisible by 3. So, the remainder when \(a^4  b^4\) is divided by 3 will be 0. YES.
We can conclude for every a and b, \(a^4  b^4\) will be divided by 3 or remainder will be 0. SUFFICIENT.
Answer: D.



Intern
Joined: 10 Sep 2015
Posts: 32

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 12:04
a^4b^4 = (a^2+b^2) (a^2b^2)
= (a+b) (ab) ( a^2+b^2)
1) suff .. remainder is 0
2) can't say NS ..
A



Intern
Joined: 27 Oct 2015
Posts: 25

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 17:30
When this function is simplified, we get (a^2 + b^2)(a + b)(a  b)
From 1st statement, (a+b) is a multiple of 3. Hence the product will be a multiple of 3 and the remainder will be 0 Sufficient. This eliminates B, C, E
From 2nd statement, there are multiple options and hence B is eliminated



Intern
Joined: 21 Jul 2015
Posts: 34

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 19:48
\(a^4b^4 = (a^2b^2)*(a^2+b^2) = (a+b)*(ab)*(a^2+b^2)\)
Statement (1): When a + b is divided by 3, the remainder is 0 since \((a+b)\) is a factor of 3, \(a^4b^4\) when divided by 3, the remainder will be 0 Sufficient
Statement (2): When \((a^2 + b^2)\) is divided by 3, the remainder is 2 lets choose a = 2 and b = 1 \((a^2 + b^2) = 5\) when divided by 3, the remainder is 2 \(a^4b^4 = 15\) when divided by 3, the remainder is 0 Sufficent
Answer (D)



Manager
Joined: 07 May 2015
Posts: 175
Location: India
GPA: 3

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
06 Dec 2015, 20:33
Statement 1 : (a+ b)/ 3 leaves no remainder let a = 5 , b = 2 now (a^4  b^4)/3 plugging in values: Remainder= 0
Similarily for a = 2, b = 1 Remainder=0
Thus Sufficient
Statement 2: (a^2+b^2) / 3 leaves remainder of 2 let a = 2 and b = 1..this satisfies the condition now (a^4  b^4)/3 plugging in values: Remainder= 0
Simlarily values a = 5 , b = 2 satisfies condition now (a^4  b^4)/3 plugging in values: Remainder= 0
Sufficient
Ans: D



Retired Moderator
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 304

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
07 Dec 2015, 07:03
\((a^4b^4)\)(ab)(a+b)\((a^2+b^2)\)
While (ab) and (a+b) will be divisible by 3, it will be \((a^2+b^2)\) that will give us remainder if it will.
St 1: When a + b is divided by 3, the remainder is 0
2 cases will give us this: 1. Both a and b are divisible by 3 2. Both a and b are not divisible by 3
While case 1 will not give us a remainder with \((a^2+b^2)\), case 2 will give us.
Insufficient
St 2: When a^2 + b^2 is divided by 3, the remainder is 2
Exactly what we want.
Sufficient
Answer: B



Intern
Joined: 23 Sep 2011
Posts: 46
Location: Singapore
Concentration: Finance, Entrepreneurship
GPA: 3.44
WE: Information Technology (Investment Banking)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
07 Dec 2015, 07:44
OA: D
Solution a^4  b^4 = (ab)(a+b)(a^2+b^2) Hence, clearly (1) is sufficient.
Now consider (2). Notice that n^2 for any integer n when divided by 3 leaves a remainder of either 0 or 1. Therefore, a^2 + b^2 divided by 3 leaves remainder 2 => (1) a^2 divided by 3 leaves remainder 1 AND (2) b^2 divided by 3 leaves remainder 1. (no other combos are possible). Hence, a^2  b^2 is divisible by 3 => a^4  b^4 is divisible by 3. Sufficient.



Manager
Status: In the realms of Chaos & Night
Joined: 13 Sep 2015
Posts: 149

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
07 Dec 2015, 09:31
a^4  b^4 = (a+b)(ab)(a^2+b^2) (1) When a + b is divided by 3, the remainder is 0 = a^4  b^4 is divisible by 3 (2) When a^2 + b^2 is divided by 3, the remainder is 2 = a^4  b^4 is divisible by 3 with a remainder of 2 Both are sufficient : D
_________________
Good luck========================================================================================= "If a street performer makes you stop walking, you owe him a buck" "If this post helps you on your GMAT journey, drop a +1 Kudo ""Thursdays with Ron  Consolidated Verbal Master List  Updated"



Intern
Joined: 29 Aug 2013
Posts: 39
Location: Bangladesh
GPA: 3.76
WE: Supply Chain Management (Transportation)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
07 Dec 2015, 12:08
QUESTION #1: For positive integers a and b, when a^4  b^4 is divided by 3, what is the remainder? (1) When a + b is divided by 3, the remainder is 0 (2) When a^2 + b^2 is divided by 3, the remainder is 2 Solution: both the statement taken together is not sufficient to produce the remainder of the question asked. Answer: E
_________________
Appreciate Kudos if the post seems worthwhile!



Senior Manager
Joined: 23 Sep 2015
Posts: 371
Location: France
GMAT 1: 690 Q47 V38 GMAT 2: 700 Q48 V38
WE: Real Estate (Mutual Funds and Brokerage)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
08 Dec 2015, 04:14
\(a^4  b^4 = (a^2 + a ^2)(a – b)(a + b)\) 1) If a + b is divisible by 3 then the product above is divisible by 3 then the remainder is 0 sufficient 2) Possible values for \(a^2 + b^2\) are: 8 = \((2^2 + 2^2\)) ==> Then we have 2 + 2 = 4 not divisible by 3, but 2 – 2 is divisible by 3 so the remainder is 0 17 = \((4^2 + 1^2)\) ==> Then we have 4 + 1 = 5 not divisible by 3, but 4 – 1 is divisible by 3 so remainder is 0 20 = \((4^2 + 2^2)\) ==> Then we have 4 + 2 = 6 which divisible by 3 so remainder is 0 26 = \((5^2 + 1^2)\) ==> Then we have 5 + 1 = 6 which divisible by 3 so remainder is 0 29 = \((5^2 + 2^2)\) ==> Then we have 5 + 2 = 7 not divisible by 3, but 5 – 2 is divisible by 3 so remainder is 0 We can see a pattern for a + b> 4, when a²+b² is odd, a – b is divisible by 3 and when a²+b² is even, a + b is divisible by 3 Statement 2 is also sufficient, the remainder is zero. Each statement is sufficient answer D
_________________



Intern
Status: Current Student
Joined: 27 Mar 2014
Posts: 26
Location: Bangladesh
Concentration: Entrepreneurship, Finance

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
Updated on: 13 Dec 2015, 01:27
Simplifying the term a^4b^4 (a^2+b^2)(a+b)(ab)
statement 1: When (a+b) is divided by 3, remainder is 0. If any term in the form a*b*c is divisible by 3, then the whole term is divisible by 3. Statement 1 is sufficient because when a^4b^4 is divided by 3, it will have a remainder of 0.
Statement 2: [(nonmultiple of 3)^2+(nonmultiple of 3)^2] must have a remainder of 2 when divided by 3. And any two nonmultiple of 3 must be divisible by 3 upon summation or subtraction. In the question stem, we see both summation and subtraction of a and b. Hence, sufficient.
{for example: a=nonmultiple of 3=14 and b=17. Now, (a^2+b^2)=485, which has a remainder of 2. Here, (ab) is a multiple of 3 and hence remainder of a^4b^4 is 0}
answer is D.
**[(multiple of 3)^2+(nonmultiple of 3)^2] must have a remainder of 1 when divided by 3.
Originally posted by zmtalha on 08 Dec 2015, 11:38.
Last edited by zmtalha on 13 Dec 2015, 01:27, edited 1 time in total.



Intern
Joined: 16 Apr 2015
Posts: 39
Concentration: Operations, Strategy

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
08 Dec 2015, 21:39
the answer is A. as it is enough to answer the question



Intern
Joined: 21 Jan 2013
Posts: 39
Concentration: General Management, Leadership
GPA: 3.82
WE: Engineering (Computer Software)

Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
Show Tags
09 Dec 2015, 10:01
a^4  b^4 can be rewritten as (ab)(a+b)(a^2 + b^2) . Now if any of these parts are divisible by 3 , the remainder will be zero. we need to know whether one of the parts is divisible by 3  so the remainder will be zero. or the remainders of each part or combined intermediate parts when they're divided by 3 (e.g., reminder when a^2  b^2 is divisble by 3 AND reminder a^2 + b^2 is divisible by 3) OR ( (ab)/3 and (a+b)/3 and (a^2+b^2)/3) . Option A says that one of the parts (=a+b) is divisible by 3 . so a^4  b^4 is divisible by 3 with a reminder of 0  A is sufficient. Option B says a^2 + b^2 is NOT divisible by 3 and leaves a reminder of 2. It is possible that one of a+b or ab is divisible by 3  so the entire a^4b^4 leaves a reminder 0 It is also possible that neither a+b nor ab is divisible by 3  so a^4  b^4 leaves a nonzero reminder  B insuff. Answer A
_________________

Consider +1 Kudos if you find my post useful




Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
[#permalink]
09 Dec 2015, 10:01



Go to page
1 2
Next
[ 28 posts ]



