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

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Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Sat Apr 21, 2018 6:41 am

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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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Source: Beat The GMAT — Data Sufficiency | Sat Apr 21, 2018 6:41 am

Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Sat Apr 21, 2018 10:26 am

Any 1?

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by [email protected] » Sat Apr 21, 2018 4:02 pm

Hi Anaira Mitch,

We're told that A and B are POSITIVE INTEGERS. We're asked for the remainder when (A^4 - B^4) is divided by 3. This question is 'quirky', but there are a couple of different ways to approach it. There are some 'shortcuts' if you recognize the Algebra and subtle Arithmetic patterns involved...

To start, it's 'weird' to ask about A^4 - B^4. That's a remarkably specific calculation, so it's possible that the question itself involves a pattern. We can rewrite A^4 - B^4 as....
A^4 - B^4 =
(A^2 + B^2)(A^2 - B^2) =
(A^2 + B^2)(A - B)(A+B)

Since A and B are INTEGERS, each of the three 'pieces' of the calculation will be an INTEGER. The question's asking "what is the remainder when (A^2 + B^2)(A - B)(A+B) is divided by 3?"

1) When (A+B) is divided by 3, the remainder is 0.

This tells us that (A+B) has to be a MULTIPLE of 3. Multiplying a multiple of 3 by any other integers (which is what the question is ultimately asking us to do) will ALWAYS lead to a multiple of 3. By extension, the remainder of that big calculation will ALWAYS be 0.
Fact 1 is SUFFICIENT

2) When (A^2 + B^2) is divided by 3, the remainder is 2

There's a subtle Arithmetic rule that impacts Fact 2: EVERY square of an integer is EITHER a multiple of 3 OR 'one more' than a multiple of 3. Consider 0, 1, 4, 9, 16, 25, 36, 49, etc.... They are all EITHER multiples of 3 OR 'one more' than a multiple of 3.

Fact 2 tells us that (A^2 + B^2) divided by 3 gives us a remainder of 2. This means that BOTH A^2 and B^2 have to be perfect squares that are NOT multiples of 3 (since we need to add two squares that each have a remainder of 1 to end up with a sum that has a remainder of 2). By extension, those two individual values for A and B are either "1 more" than a multiple of 3 OR "2 more" than a multiple of 3. There are just three possible pairings of numbers:
"1 more" and "1 more", in which case (A-B) will be a multiple of 3
"1 more" and "2 more", in which case (A+B) will be a multiple of 3
"2 more" and "2 more", in which case (A-B) will be a multiple of 3

Thus, since at least one piece of the question will be a multiple of 3, the entire calculation will be a multiple of 3 and the answer to the question is ALWAYS 0.
Fact 2 is SUFFICIENT

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Sun Apr 22, 2018 1:32 am

Thanks Rich!!

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by GMATGuruNY » Sun Apr 22, 2018 2:24 am

Anaira Mitch 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

aâ�´-bâ�´ = (aÂ²+bÂ²)(aÂ²-bÂ²) = (aÂ²+bÂ²)(a+b)(a-b).

Statement 1:
Since a+b is divisible by 3, (aÂ²+bÂ²)(a+b)(a-b) is divisible by 3.
Thus, dividing aâ�´-bâ�´ by 3 will yield a remainder of 0.
SUFFICIENT.

Statement 2:
In other words, aÂ²+bÂ² is 2 more than a multiple of 3:
aÂ²+bÂ² = 3k + 2, where k is a nonnegative integer.
Thus:
aÂ²+bÂ² = 2, 5, 8, 11, 14, 17, 20, 23...
Integer options for a and b such that aÂ²+bÂ² yields a value in the list above:
a=1 and b=1, with the result that aÂ²+bÂ² = 2 and a-b = 0.
a=2 and b=1, with the result that aÂ²+bÂ² = 5 and a+b = 3.
a=2 and b=2, with the result that aÂ²+bÂ² = 8 and a-b = 0.
a=4 and b=1, with the result that aÂ²+bÂ² = 17 and a-b= 3.
a=4 and b=2, with the result that aÂ²+bÂ² = 20 and a+b = 6.
As the values in blue illustrate, every case yields a factor for aâ�´-bâ�´ that is divisible by 3.
Thus -- in every case -- dividing aâ�´-bâ�´ by 3 will yield a remainder of 0.
SUFFICIENT.

The correct answer is D.

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Anaira Mitch: Master | Next Rank: 500 Posts; Posts: 235; Joined: Wed Oct 26, 2016 9:21 pmFollowed by: 5 members; Thanked: 3 times

by Anaira Mitch » Sun Apr 22, 2018 3:45 pm

Thanks Mitch!!

GMATGuruNY wrote:
Anaira Mitch 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
aâ�´-bâ�´ = (aÂ²+bÂ²)(aÂ²-bÂ²) = (aÂ²+bÂ²)(a+b)(a-b).

Statement 1:
Since a+b is divisible by 3, (aÂ²+bÂ²)(a+b)(a-b) is divisible by 3.
Thus, dividing aâ�´-bâ�´ by 3 will yield a remainder of 0.
SUFFICIENT.

Statement 2:
In other words, aÂ²+bÂ² is 2 more than a multiple of 3:
aÂ²+bÂ² = 3k + 2, where k is a nonnegative integer.
Thus:
aÂ²+bÂ² = 2, 5, 8, 11, 14, 17, 20, 23...
Integer options for a and b such that aÂ²+bÂ² yields a value in the list above:
a=1 and b=1, with the result that aÂ²+bÂ² = 2 and a-b = 0.
a=2 and b=1, with the result that aÂ²+bÂ² = 5 and a+b = 3.
a=2 and b=2, with the result that aÂ²+bÂ² = 8 and a-b = 0.
a=4 and b=1, with the result that aÂ²+bÂ² = 17 and a-b= 3.
a=4 and b=2, with the result that aÂ²+bÂ² = 20 and a+b = 6.
As the values in blue illustrate, every case yields a factor for aâ�´-bâ�´ that is divisible by 3.
Thus -- in every case -- dividing aâ�´-bâ�´ by 3 will yield a remainder of 0.
SUFFICIENT.

The correct answer is D.

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