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

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

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New post 28 Mar 2018, 11:15
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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.


Thus D answer
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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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New post 26 Apr 2018, 19:35
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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


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 [#permalink]

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New post 23 Jun 2018, 06:40
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.


Thus D answer


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

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New post 24 Jun 2018, 06:09
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.


Thus D answer


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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New post 24 Jun 2018, 07:37
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
Re: Numbers a and b are positive integers. If a^4-b^4 is divided by 3   [#permalink] 24 Jun 2018, 07:37
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