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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 28 Mar 2018, 05:57
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Numbers a and b are positive integers. If 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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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 28 Mar 2018, 11:15
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itisSheldon
Numbers a and b are positive integers. If 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
Show more

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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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 26 Apr 2018, 19:35
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itisSheldon
Numbers a and b are positive integers. If 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
Show more

Expert please explain more about the approach to this question!
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 23 Jun 2018, 06:40
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Numbers a and b are positive integers. If 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

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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Hi Bunuel
I think the Ans is A..Plz correct me below if i am wrong in solving it

Statement-1 is pretty straight forward
Statement-2


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^2-b^2)*( a^2+b^2) = (a^2-b^2)*(3K+2)
Solving this i get, (3a^2K)+(2a^2)-(3Kb^2)-(2b^2).....now only this term we need to check (2a^2-2b^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
Scenario-1 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 Statement-2 gives different answers hence no sufficient
So final answer is A

Plz correct me if i am wrong anywhere.Thanks
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 24 Jun 2018, 06:09
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Numbers a and b are positive integers. If 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

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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Thank 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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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 24 Jun 2018, 07:37
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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^4-b^4 is divided by 3 25 Jul 2018, 05:11
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Numbers a and b are positive integers. If 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

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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Hi 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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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 25 Jul 2018, 10:28
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amanvermagmat
itisSheldon
Numbers a and b are positive integers. If 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

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

Hi 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?
Show more


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).
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 03 Jun 2020, 14:23
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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.

Show more


is there more content on this specific rule of 3? e.g. does this apply to other numbers when raised to an even/odd power
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 08 Jan 2021, 04:49
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Numbers a and b are positive integers. If 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

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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Apart from the number '3', are there any other divisibility rules that we should know for other integers?

Could anyone or experts please enlighten?
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 11 Jan 2021, 01:31
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Numbers a and b are positive integers. If 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

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

Apart from the number '3', are there any other divisibility rules that we should know for other integers?

Could anyone or experts please enlighten?
amanvermagmat MathRevolution VeritasKarishma
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Hello,

One must also know divisibility patterns for 4, 7, and some prime numbers like 11, 13, and 17 at least.

Thanks
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 12 Jan 2021, 06:38
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Numbers a and b are positive integers. If 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
Show more

\(a^4 - b^4\)
\(= (a^2 + b^2) (a+b) (a-b)\)

A:
(a+b) is divisible by 3; then \(= (a^2 + b^2) (a+b) (a-b)\) is also divisible by 3.
Hence remainder is 0.
A is sufficient

B:
\((a^2 + b^2)\) when divided by 3 gives 2 as remainder.
Means \((a^2 + b^2)\) is of form 3Q +2;
Which gives \((a^2 + b^2)\) as 2,5,8,11,14,16,...;

Now it is stated that a and b is positive int. and square of int will always be greater than or equal to 0.
if a = 0 and b\neq{0} then also above expression (1) does not hold true.
For above expression simply says that the summation of sq of int. And sq of sum of int is only 2,8,16,...
and the remainder for\( (a^2 - b^2)\) is 0, as a will be equal to b.
hence B is also sufficient.
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 19 Aug 2023, 01:28
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Another way to approach (B), the 2nd statement.

Given a^4-b^4 = (a^2+b^2)(a^2-b^2)
We know from (B) that a^2+b^2 = (3n+2), then a^2-b^2 = (3n+2-2b^2)
Then a^4-b^4 = (a^2+b^2)(a^2-b^2) = (3n+2)(3n+2-2b^2) = (3n)(3n+2-2b^2) + (2)(3n+2-2b^2)
We can safely ignore the first term, as the factor of 3n will make the first term divisible by 3.
The second term can further be writteen as 6n+4-4b^2, and similarly, the term 6n can be ignored.

Now what we really need to worry about is 4-4b^2, or 4b^2-4, (doesnt really matter). It can be written as 4(b-1)(b+1).
The property of 3 is that among 3 consecutive numbers x, x+1, x+2, there's gotta be 1 number divisible by 3.

Case 1. either (b-1) or (b+1) is divisible by 3, then the answer is obvious.

Case 2. b is divisible by 3. However, can this be the case?

It would mean a^2 has to be 3m+2 for some m, since a^2+b^2 = 3n+2.
If you are not familiar with the property of squares mentioned above by others (which I didn't know), you can deduce from the following:
If a^2 = 3m+2, then a^2-1=3m+1, which means (a+1)(a-1) = 3m+1. However, we know that one of the 3 numbers a, (a+1), and (a-1) is divisible by 3. If (a+1)(a-1)=3m+1 is not divisible by 3, it means that a has to be divisible by 3. This contradicts with a^2 = 3m+2, which is not divisible by 3. Therefore, b is not divisible by 3, and this case is not valid.
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3 02 Sep 2023, 14:58
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amanvermagmat
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Numbers a and b are positive integers. If 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

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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The real question here is Whether that property is expected (or even useful) knowledge on GMAT?
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