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Manager  P
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Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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Question Stats: 34% (01:52) correct 66% (01:57) wrong based on 201 sessions

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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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Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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itisSheldon wrote:
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.

##### General Discussion
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GMAT 1: 590 Q44 V23 Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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1
itisSheldon wrote:
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

Intern  B
Joined: 11 Feb 2018
Posts: 35
Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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amanvermagmat wrote:
itisSheldon wrote:
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.

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

Plz correct me if i am wrong anywhere.Thanks
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Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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amanvermagmat wrote:
itisSheldon wrote:
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.

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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Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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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

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

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Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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amanvermagmat wrote:
itisSheldon wrote:
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.

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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Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3  [#permalink]

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KarthikvsGMAT wrote:
amanvermagmat wrote:
itisSheldon wrote:
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.

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?

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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