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Numbers a and b are positive integers. If a^4b^4 is divided by 3
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28 Mar 2018, 04:57
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Numbers a and b are positive integers. If a^4b^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: Numbers a and b are positive integers. If a^4b^4 is divided by 3
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28 Mar 2018, 10:15
itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 There is a certain property of positive integers when they are divided by 3. If a positive integer is divisible by 3, then any of its higher powers will also be divisible by 3. Eg, 6 is divisible by 3  then 6^n will also be divisible by 3, where 'n' is a positive integer. If a positive integer is NOT divisible by 3, then any of its EVEN powers will always give a remainder of '1' when divided by 3. Eg, 4 is not divisible by 3. And every even power of 4 (4^2, 4^4, 4^6,.... ) will always give a remainder of '1' when divided by 3. Another example, take 5. Every even power of 5 (5^2, 5^4, 5^6,... ) will also give a remainder '1' when divided by 3. Also, a^4  b^4 = (a^2  b^2)(a^2 + b^2). And it can further be broken down as (a  b)(a + b)(a^2 + b^2). Statement 1:a+b is divisible by 3. And since a+b is a factor of a^4  b^4, this means a^4  b^4 will also be divisible by 3. Thus remainder is '0'. Sufficient. Statement 2:a^2 + b^2 gives remainder '2' when divided by 3. Now as explained above, each of a^2 and b^2 can give a remainder of either '0' or '1' when divided by 3. But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. So this means each of a^2 and b^2 gives a remainder of '1' when divided by 3. Now, if a^2 gives remainder '1' when divided by 3, same will be with a^4 if b^2 gives remainder '1' when divided by 3, same will be with b^4 Thus a^4  b^4 will give a remainder of 1  1 = '0' when divided by 3. This is also sufficient. Thus D answer




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Re: Numbers a and b are positive integers. If a^4b^4 is divided by 3
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26 Apr 2018, 18:35
itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 Expert please explain more about the approach to this question!



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Numbers a and b are positive integers. If a^4b^4 is divided by 3
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23 Jun 2018, 05:40
amanvermagmat wrote: itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 There is a certain property of positive integers when they are divided by 3. If a positive integer is divisible by 3, then any of its higher powers will also be divisible by 3. Eg, 6 is divisible by 3  then 6^n will also be divisible by 3, where 'n' is a positive integer. If a positive integer is NOT divisible by 3, then any of its EVEN powers will always give a remainder of '1' when divided by 3. Eg, 4 is not divisible by 3. And every even power of 4 (4^2, 4^4, 4^6,.... ) will always give a remainder of '1' when divided by 3. Another example, take 5. Every even power of 5 (5^2, 5^4, 5^6,... ) will also give a remainder '1' when divided by 3. Also, a^4  b^4 = (a^2  b^2)(a^2 + b^2). And it can further be broken down as (a  b)(a + b)(a^2 + b^2). Statement 1:a+b is divisible by 3. And since a+b is a factor of a^4  b^4, this means a^4  b^4 will also be divisible by 3. Thus remainder is '0'. Sufficient. Statement 2:a^2 + b^2 gives remainder '2' when divided by 3. Now as explained above, each of a^2 and b^2 can give a remainder of either '0' or '1' when divided by 3. But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. So this means each of a^2 and b^2 gives a remainder of '1' when divided by 3. Now, if a^2 gives remainder '1' when divided by 3, same will be with a^4 if b^2 gives remainder '1' when divided by 3, same will be with b^4 Thus a^4  b^4 will give a remainder of 1  1 = '0' when divided by 3. This is also sufficient. Thus D answerHi Bunuel I think the Ans is A..Plz correct me below if i am wrong in solving it Statement1 is pretty straight forward Statement2 we can write a^2+b^2= 3K+2; given K is a any positive integer Now after breaking the original question i can write (a^2b^2)*( a^2+b^2) = (a^2b^2)*(3K+2) Solving this i get, (3a^2K)+(2a^2)(3Kb^2)(2b^2).....now only this term we need to check (2a^22b^2)= 2(a^2 b^2) Now, (3K+2) can be both Even or Odd depending if K is even/odd. For eg if K=2, then (3K+2)= Even So, that means a^2+b^2 can be both Even or Odd Scenario1 If a^2+b^2 is Even then a^2 b^2 = Even..in this case 2(a^2 b^2) is not divisible by 3 But if a^2+b^2= Odd, then a^2 b^2 = Odd. In this case 2(a^2 b^2) can be divisible by 3..E.g if (a^2 b^2)= 3 or 9 or any multiple of 3...Then 2(a^2 b^2) is divisble by 3, giving remainder zero. Thus Statement2 gives different answers hence no sufficient So final answer is A Plz correct me if i am wrong anywhere.Thanks



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Re: Numbers a and b are positive integers. If a^4b^4 is divided by 3
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24 Jun 2018, 05:09
amanvermagmat wrote: itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 There is a certain property of positive integers when they are divided by 3. If a positive integer is divisible by 3, then any of its higher powers will also be divisible by 3. Eg, 6 is divisible by 3  then 6^n will also be divisible by 3, where 'n' is a positive integer. If a positive integer is NOT divisible by 3, then any of its EVEN powers will always give a remainder of '1' when divided by 3. Eg, 4 is not divisible by 3. And every even power of 4 (4^2, 4^4, 4^6,.... ) will always give a remainder of '1' when divided by 3. Another example, take 5. Every even power of 5 (5^2, 5^4, 5^6,... ) will also give a remainder '1' when divided by 3. Also, a^4  b^4 = (a^2  b^2)(a^2 + b^2). And it can further be broken down as (a  b)(a + b)(a^2 + b^2). Statement 1:a+b is divisible by 3. And since a+b is a factor of a^4  b^4, this means a^4  b^4 will also be divisible by 3. Thus remainder is '0'. Sufficient. Statement 2:a^2 + b^2 gives remainder '2' when divided by 3. Now as explained above, each of a^2 and b^2 can give a remainder of either '0' or '1' when divided by 3. But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. So this means each of a^2 and b^2 gives a remainder of '1' when divided by 3. Now, if a^2 gives remainder '1' when divided by 3, same will be with a^4 if b^2 gives remainder '1' when divided by 3, same will be with b^4 Thus a^4  b^4 will give a remainder of 1  1 = '0' when divided by 3. This is also sufficient. Thus D answerThank you for the detailed explanation. Is there another way to approach this problem, i.e. without applying the divisibility rule that you have stated?



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Re: Numbers a and b are positive integers. If a^4b^4 is divided by 3
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24 Jun 2018, 06:37
Statement 1 is sufficient.
Statement 2 needs some work: We are given that a,b are pos. integers. The remainder of a^2 + b^2 = 2 Let's think about this.
I'll write some integers, and their remainders when divided by 3 in the next row. Pay close attention.
Num: 1,2,3,4,5,6,7,8,9,10,11... Rem: 1,2,0,1,2,0,1,2,0,1,2,0,...
Going back to this: The remainder of a^2 + b^2 = 2. This can only be possible if each gives a remainder of 1. Okay, so that means Rem(a^2/3) = Rem(b^2/3) = 1 So, I can write a^2 = 3n + 1 Similarly, b^2 = 3m + 1 We need to find the Rem((a^4  b^4)/3) = Rem((a^2  b^2)*(a^2 + b^2)/3) = Rem((3m + 1  3n  1)*(3q + 2)/3) = 0
Answer D



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Numbers a and b are positive integers. If a^4b^4 is divided by 3
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25 Jul 2018, 04:11
amanvermagmat wrote: itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 There is a certain property of positive integers when they are divided by 3. If a positive integer is divisible by 3, then any of its higher powers will also be divisible by 3. Eg, 6 is divisible by 3  then 6^n will also be divisible by 3, where 'n' is a positive integer. If a positive integer is NOT divisible by 3, then any of its EVEN powers will always give a remainder of '1' when divided by 3. Eg, 4 is not divisible by 3. And every even power of 4 (4^2, 4^4, 4^6,.... ) will always give a remainder of '1' when divided by 3. Another example, take 5. Every even power of 5 (5^2, 5^4, 5^6,... ) will also give a remainder '1' when divided by 3. Also, a^4  b^4 = (a^2  b^2)(a^2 + b^2). And it can further be broken down as (a  b)(a + b)(a^2 + b^2). Statement 1:a+b is divisible by 3. And since a+b is a factor of a^4  b^4, this means a^4  b^4 will also be divisible by 3. Thus remainder is '0'. Sufficient. Statement 2:a^2 + b^2 gives remainder '2' when divided by 3. Now as explained above, each of a^2 and b^2 can give a remainder of either '0' or '1' when divided by 3. But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. So this means each of a^2 and b^2 gives a remainder of '1' when divided by 3. Now, if a^2 gives remainder '1' when divided by 3, same will be with a^4 if b^2 gives remainder '1' when divided by 3, same will be with b^4 Thus a^4  b^4 will give a remainder of 1  1 = '0' when divided by 3. This is also sufficient. Thus D answerHi amanvermagmat, Thanks a lot for the explanation. >> But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. Could you please explain this portion alone?



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Re: Numbers a and b are positive integers. If a^4b^4 is divided by 3
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25 Jul 2018, 09:28
KarthikvsGMAT wrote: amanvermagmat wrote: itisSheldon wrote: Numbers a and b are positive integers. If a^4b^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 There is a certain property of positive integers when they are divided by 3. If a positive integer is divisible by 3, then any of its higher powers will also be divisible by 3. Eg, 6 is divisible by 3  then 6^n will also be divisible by 3, where 'n' is a positive integer. If a positive integer is NOT divisible by 3, then any of its EVEN powers will always give a remainder of '1' when divided by 3. Eg, 4 is not divisible by 3. And every even power of 4 (4^2, 4^4, 4^6,.... ) will always give a remainder of '1' when divided by 3. Another example, take 5. Every even power of 5 (5^2, 5^4, 5^6,... ) will also give a remainder '1' when divided by 3. Also, a^4  b^4 = (a^2  b^2)(a^2 + b^2). And it can further be broken down as (a  b)(a + b)(a^2 + b^2). Statement 1:a+b is divisible by 3. And since a+b is a factor of a^4  b^4, this means a^4  b^4 will also be divisible by 3. Thus remainder is '0'. Sufficient. Statement 2:a^2 + b^2 gives remainder '2' when divided by 3. Now as explained above, each of a^2 and b^2 can give a remainder of either '0' or '1' when divided by 3. But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. So this means each of a^2 and b^2 gives a remainder of '1' when divided by 3. Now, if a^2 gives remainder '1' when divided by 3, same will be with a^4 if b^2 gives remainder '1' when divided by 3, same will be with b^4 Thus a^4  b^4 will give a remainder of 1  1 = '0' when divided by 3. This is also sufficient. Thus D answerHi amanvermagmat, Thanks a lot for the explanation. >> But none of them can give a remainder '0' because then a^2 + b^2 cannot give a combined remainder of '2'. Could you please explain this portion alone? Hello We are given in second statement that a^2 + b^2 gives a remainder of '2' when divided by 3. Now consider a^2, as I explained  even power of a positive integer 'a' can give a remainder of '0' (if a is divisible by 3) or a remainder of '1' (if a is not divisible by 3); when divided by 3. So from a^2, we can get either a '0' remainder or '1' remainder, when its divided by 3. Similarly from b^2, we can get either a '0' remainder or '1' remainder when its divided by 3. So when we add a^2 + b^2, we are either adding remainders 0 & 0, or we are adding 0 & 1, or we are adding 1 & 1. But in which case will we get a combined remainder of '2'? Only when each of a^2 and b^2 gives a remainder of 1 (1+1 will lead to 2).




Re: Numbers a and b are positive integers. If a^4b^4 is divided by 3 &nbs
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