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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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11 Dec 2015, 10:09
A. Statement 1 alone is sufficient a^4b^4 can be written as (a^2+b^2)*(a^2b^2) which can further be written as (a^2+b^2)*(a+b)*(ab)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2: a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2b^2 when divided by 3, therefore it is not sufficient.



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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11 Dec 2015, 10:11
A. Statement 1 alone is sufficient a^4b^4 can be written as (a^2+b^2)*(a^2b^2) which can further be written as (a^2+b^2)*(a+b)*(ab)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2: a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2b^2 when divided by 3, therefore it is not sufficient.



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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11 Dec 2015, 22:49
a^4  b^4 is divided by 3
a^4b^4= (a+b)(ab)(a^2 + b^2) 1) When a + b is divided by 3, the remainder is 0
Since a+b is a factor, therefore a^4  b^4 divided by 3 will also give a remainder of 0 and factor. Hence choice 1 is sufficient.
(2) When a^2 + b^2 is divided by 3, the remainder is 2 When we square a number that gives a remainder of 1 when divided by 3, it gives remainder of 1 again. For example 4 gives a remainder of 1 when divided by 3, and 16 also gives a remainder of 1 when divided by 3 When we square a number that gives a remainder of 2 when divided by 3, it gives remainder of 1. For example 5 gives a remainder of 1 when divided by 3, and 25 also gives a remainder of 1 when divided by 3.
So point to note is When we square a number that gives a remainder of 2 when divided by 3, it always gives remainder of 1 Therefore point to infer from the statement, a^2 + b^2 is divided by 3, the remainder is 2, is both a and b are not multiples of 3. They must be numbers that either leave a remainder of 1 or 2 for a^2+b^2 to leave a remainder of 2 when divided by 3.
If we select numbers that either leave 1 or 2 as remainder when divided by 3 for the equation a^2 + b^2 , then the individual numbers added up or subtracted, a+b or ab always turn out to be a factor of 3 and therefore always leave remainder of 0. This can be tried with any numbers with the above conditions and will hold true.
Hence this choice is also sufficient.
As each statement alone is sufficient , therefore answer is choice D.



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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12 Dec 2015, 23:49
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
ANSWER:
\(a^4  b^4 = (a^2 + b^2)(a^2  b^2) = (a^2 + b^2)(a+b)(ab)\) (*)
(1) \((a+b)\) is divisible by 3 > (*) is always divisible by 3 > sufficient
(2) \((a^2 + b^2) = 3m +2\) > \(a^2 = 3q + 1\) and \(b^2 =3p +1\)> \(a^2  b^2 = 3 (qp) +1 1 = 3(qp)\) > (*) is divisible by 3 > sufficient
Answer: D



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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14 Dec 2015, 08:09
Bunuel wrote: 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! MATH REVOLUTION OFFICIAL SOLUTION:If we alter the original condition and question, it becomes \(a^4b^4=3t\)? (t is an integer). If we factor the question, the result is (ab)(a+b)(a^2+b^2)=3t?, which is question. Then, if we look at 1), ab=3n (n is an integer) becomes \(a^4b^4=3n(a+b)(a^2+b^2)\), which is surely divisible by 3. This is a “yes” and sufficient. If we look at 2), it looks like it will be \(a^2+b^2=3p+2\) (p is a positive integer) and not sufficient. Mistake Type 4(B) states “If answer choice A and B seem too attractive and obvious, consider answer choice D”. This Mistake Type commonly applies to a case similar to 1) and also to integer question, which is one of key questions. So, if we directly substitute 1 and 1 to a and b in 2), it yields (a,b)=(1,1) ← ab=multiples of 3. This is a “yes”. If we substitute 1 and 2, it yields (a,b)=(1,2) ← a+b=multiples of 3. This is also a “yes”. Thus, in any case, a4b4 is always a multiple of 3. This is a “yes” and sufficient. The correct answer choice is D. In case of GMAT math DS problems, students have to be extra cautious about Mistake Type 4(B). Remember, “If answer choice A and B seem too attractive and obvious, consider answer choice D”. It is nearly impossible to solve a problem in 2 to 3 minutes during the exam. So students have to approach problems with a complete sense of logic based on Mistake Types. Also, please remember that if one condition seems easy when the other seems rather challenging, the answer will be D. This type of question decides a score of 50 or a perfect 51.
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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19 Dec 2015, 06:21
Bunuel wrote: Bunuel wrote: 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! MATH REVOLUTION OFFICIAL SOLUTION:If we alter the original condition and question, it becomes \(a^4b^4=3t\)? (t is an integer). If we factor the question, the result is (ab)(a+b)(a^2+b^2)=3t?, which is question. Then, if we look at 1), ab=3n (n is an integer) becomes \(a^4b^4=3n(a+b)(a^2+b^2)\), which is surely divisible by 3. This is a “yes” and sufficient. If we look at 2), it looks like it will be \(a^2+b^2=3p+2\) (p is a positive integer) and not sufficient. Mistake Type 4(B) states “If answer choice A and B seem too attractive and obvious, consider answer choice D”. This Mistake Type commonly applies to a case similar to 1) and also to integer question, which is one of key questions. So, if we directly substitute 1 and 1 to a and b in 2), it yields (a,b)=(1,1) ← ab=multiples of 3. This is a “yes”. If we substitute 1 and 2, it yields (a,b)=(1,2) ← a+b=multiples of 3. This is also a “yes”. Thus, in any case, a4b4 is always a multiple of 3. This is a “yes” and sufficient. The correct answer choice is D. In case of GMAT math DS problems, students have to be extra cautious about Mistake Type 4(B). Remember, “If answer choice A and B seem too attractive and obvious, consider answer choice D”. It is nearly impossible to solve a problem in 2 to 3 minutes during the exam. So students have to approach problems with a complete sense of logic based on Mistake Types. Also, please remember that if one condition seems easy when the other seems rather challenging, the answer will be D. This type of question decides a score of 50 or a perfect 51. Thanks Bunuel. For option 2, I have an alternate approach to this. Please let me know, if this can be solved in this manner? \((a^2+b^2)=3p+2\). It can also be written as \(((ab)^2+2ab)=3p+2\). So this means (ab) is multiple of 3. Hence \(a^4b^4\) is divisible to 3. Please suggest if this solution is ok or it can be solved something similar to this?



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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15 Mar 2018, 09:00
I don't completely understand the solution posted by Math Revolution. Could Bunuel or chetan2u please post their explanation/solution? Thanks! 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



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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,
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27 Nov 2018, 11:18
Hi Victor, I find plugging in works best with these questions. I got D 1) a+b / 3 = Rem0 This basically means that when you add the numbers up they add up to 3 or a multiple of 3. Take 3 for easy calculation. Plug in a few ridiculous numbers in there. Try a = 1, b = 2 Try a = +7, b = 4 Try a = 1, b = 2 Looks sufficient to me. PS If you plug those into the original equation they work as well. Try 7^4  (4)^4 / 3 = 2401  254 / 3 = 2145 / 3 (just add the individual numbers up and you'll see that its divisible). 2) a^2 + b^2 / 3 = Rem2 Same idea. Plug in the same set of numbers or different ones. Doesn't matter really. You'll find this to be true as well. Let's try with our ridiculous example. 7^2 + (4)^2 / 3 = 49+16/3 = 65/3 = Rem 2 (63 closest) Try same idea with bunch of numbers if you have the time. Looks good to me. Answer = D mvictor wrote: 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? :\ Posted from my mobile device




Re: Math Revolution and GMAT Club Contest! For positive integers a and b, &nbs
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