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Math Revolution and GMAT Club Contest! For positive integers a and b,

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Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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05 Dec 2015, 08:52
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Math Revolution and GMAT Club Contest Starts!

QUESTION #6:

For positive integers a and b, when 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

Check conditions below:

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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05 Dec 2015, 11:58
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I've been struggling to find the answer for this one...very tough and tricky, but if you try algebraic way, then one works and other one not always, and while picking numbers - different results...anyways...below my explanation.

we can rewrite the initial statement:
a^4-b^4 = (a^2+b^2)(a^2-b^2)
(a^2-b^2) can be rewritten as (a+b)(a-b)

or the whole:
a^4-b^4 = (a^2+b^2)(a+b)(a-b)

first statement:
a+b is divisible by 3.

well, in this case, the reminder will always yield 0, since a^4-b^4 is a multiple of 3, because a+b is a multiple of 3.

statement 1 is sufficient.

statement 2:
a^2+b^2, if divided by 3, will yield a remainder of 2.

well, this doesn't tell much. What if a+b is a multiple of 3?
what if a-b is a multiple of 3?

clearly 2 should't be sufficient, right?
well, not exactly!

if you try picking numbers, you can see that all the times, the remainder is 0. and this is what confuses me the most. Maybe I miss something?

a=2 b=1
a^2 + b^2 = 5. 5/3 = 1 and remainder 2.
a^4 - b^4 = 16 - 1 = 15. 15/3 = 5 and remainder 0.

a=4 and b=1
a^2 + b^2 = 16+1 = 17. 17/3 = 5 and remainder 2.
a^4 - b^4 = 256-1 = 255. remainder again 0.

and the pattern is the same for all numbers that are picked and that satisfy the condition of the statement 2.

I really doubt this is a 600-700 level question. Or maybe I am not in the mood for solving math today? :\
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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05 Dec 2015, 15:21
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when a^4 - b^4 is divided by 3, what is the remainder?

using the formula (a^2 - b^2) = (a + b)*(a - b), we can write a^4 - b^4 as (a^2 + b^2)*(a^2 - b^2) i.e. (a^2 + b^2)*(a + b)*(a - b)

statement-1: When a + b is divided by 3, the remainder is 0.

it means a+b is a multiple of 3. when a multiple of 3 is multiplied some other number the resulting number would also be a multiple of 3.

(a^2 + b^2)*(a + b)*(a - b) would be divisible by 3. hence remainder would always be 0.

We get a definite answer hence statement-1 is SUFFICIENT.

statement-2: When a^2 + b^2 is divided by 3, the remainder is 2

It means Neither a nor b can be a multiple of 3. if a or b or both are multiple of 3 then statement-2 will not hold. this can be explained as below:

1) if a & b both are multiple of 3 then remainder would be zero NOT 2 when a^2 + b^2 is divided by 3

2) Neither a nor b can be a multiple of 3.

suppose a is multiple of 3 but b is not. In this case clearly a^2/3 will give remainder as 0 BUT b^2/3 will give remainder 0 or 1 NOT 2 because no perfect square ends in 2,8.

we conclude that a and b can take any value which is NOT a multiple of 3 i.e. if n is a multiple of 3 then a or b or both can take either (n+1) or (n+2) or (n-1) or (n-2). but a or b can never taken n.

For all such values of a & b either a+b or a-b would be divisible by 3 i.e. remainder would be zero and a^2 + b^2/3 would give remainder of 2. this can be illustrated by following examples to bring more clarity:

say n = 6
then a = (n+1) = 7 and b = (n+2) = 8 (notice that both 7 and 8 gives remainder of 1 when divided by 3, their squares also will give remainder as 1)
a^2 + b^2 = (49 + 64)/3 = 113/3 gives remainder 2 because 79 and 64 both gives 1 as remainder.
a+b/3 = 8+7 = 15/3 gives remainder 0

say n = 6
then a = (n+1) = 7 and b = (n-2) = 4 (notice that both 7 and 4 gives remainder of 1 when divided by 3, their squares also will give remainder as 1)
a^2 + b^2 = (49 + 16)/3 = 65/3 gives remainder 2 because 49 and 16 both gives 1 as remainder.
a-b/3 = 7-4= 3/3 gives remainder 0

say n = 6
then a = (n+1) = 7 and b = (n-1) = 5 (notice that gives remainder of 1 BUT 5 gives remainder of 2 when divided by 3, BUT THE FACT is that their squares will give remainder as 1 only)
a^2 + b^2 = (49 + 25)/3 = 74/3 gives remainder 2 because 49 and 25 both gives 1 as remainder.
a+b/3 = 7+5= 12/3 gives remainder 0

now we see When a^2 + b^2 is divided by 3, the remainder is 2 but (a^2 + b^2)*(a + b)*(a - b) would be divisible by 3 because either a+b or a-b would be divisible by 3. hence remainder would always be 0.

We get a definite answer hence statement-1 is SUFFICIENT.

Option (D) is the correct answer.
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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05 Dec 2015, 17:35

Question is what is the reminder of a^4-b^4 when divided by 3.

a^4-b^4, can be written as (a^2+b^2)(a^2-b^2) or (a^2+b^2)(a+b)(a-b). so when this is divided by 3, what is the reminder.

1) (a+b) / 3 -> remainder is 0. This means (a^2+b^2)(a+b)(a-b) is divisible by 3 itself, so the answer when a^4-b^4 is divided by 3, the reminder is of course 0. (SUFFICIENT)

2) a^2+b^2 divided by 3 -> remainder is 2. We will now need to see the different possibilities of what a^2 and b^2 will be, so we can find out the remainder when (a^2+b^2)(a^2-b^2) is divided by 3.

There are the following possibilities - when a^2+b^2 divided by 3,

1. (a^2)/3 yields the remainder of 2 and (b^2)/3 yields the remainder 0. In this case a^2 can be written as 3*x+2 and b^2 can be 3y
2. (a^2)/3 yields the remainder of 0 and (b^2)/3 yields the remainder 2. In this case a^2 can be written as 3*x and b^2 can be 3y+2
3. (a^2)/3 yields the remainder of 1 and (b^2)/3 yields the remainder 1. In this case a^2 can be written as 3*x+1 and b^2 can be 3y+1.

Here is x and y are any +ve integers. Let us now substitute each of these values of of a^2 and b^2 into (a^2+b^2)(a^2-b^2)

1.-> a^2 = (3x+2), b^2 =3y. Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)+6(x+y)+6(x-y)+4. In this case the remainder is 1.
2.-> a^2 = (3x), b^2 =(3y+2). Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)-6(x+y)+6(x-y)-4. In this case the remainder is 2.
3.-> a^2 = (3x+1), b^2 =(3y+1). Simplifying (a^2+b^2)(a^2-b^2) = 9(x^2-y^2)+6(x+y). In this case the remainder is 0.

Hence, option 2 is NOT SUFFICIENT.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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05 Dec 2015, 21:50
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a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (a + b)(a - b)(a^2 + b^2)
R = ? when (a + b)(a - b)(a^2 + b^2)/3

St1: R = 0 when (a + b)/3 --> Implies that a^4 - b^4 is divisible by 3 --> Sufficient

St2: R = 2 when (a^2 + b^2)/3
This implies,
a^2 + b^2 = 3k + 2 = 2, 5, 8, 11, ....................
i.e. Sum of squares should be equal to 2, 5, 8, 11, .... Note that certain integers such as 11 cannot be considered here because 11 cannot be obtained as sum of the squares of two positive integers.
Assume, a^2 + b^2 = 5 --> a = 2; b = 1 --> R = 0 when (a + b)/3 --> So the entire expression is divisible by 3
Assume, a^2 + b^2 = 10001 --> a = 100; b = 1 --> R = 0 when (a - b)/3 --> Expression is divisible by 3
Assume, a^2 + b^2 = 32 --> a = 4; b = 4 --> R = 0 when (a-b)/3 --> Expression is divisible by 3
Therefore, Statement 2 alone is sufficient.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 04:27
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$$a^4 - b^4=(a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a^2+b^2)$$

(1) is sufficient and we will have the remainder when $$a^4-b^4$$ is divided by 3 is 0.
(2) When $$a^2 + b^2$$ is divided by 3, the remainder is 2.
For a random number A we have 3 cases: A = 3k, A = 3k+1, or A = 3k + 2.
The first case: $$A^2=9k^2$$ -> The remainder when $$A^2$$ is divided by 3 is 0.
The second case: $$A^2=9k^2+6k+1$$ -> The remainder when $$A^2$$ is divided by 3 is 1.
The third case: $$A^2=9k^2 + 12k + 4$$ -> The remainder when $$A^2$$ is divided by 3 is also 1.
=> From (2) both a & b are not divisible by 3 (if so the remainder of $$a^2 + b^2$$ divided by 3 is either 1 or 0)
We have 4 cases for a & b:
a = 3u + 1 & b = 3v + 1 => a-b = 3(u-v) divisible by 3 and hence $$a^4-b^4$$ is divisible by 3.
a = 3u + 2 & b = 3v + 2 => a-b = 3(u-v) divisible by 3 and hence $$a^4-b^4$$ is divisible by 3.
a = 3u + 2 & b = 3v + 1 => a+b = 3(u+v+1) divisible by 3 and hence $$a^4-b^4$$ is divisible by 3.
a = 3u + 1 & b = 3v + 2 => a+b = 3(u+v+1) divisible by 3 and hence $$a^4-b^4$$ is divisible by 3.
=> The remainder of $$a^4-b^4$$ when divided by 3 is 0. and hence (2) is sufficient.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 09:20
2
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QUESTION #1:

For positive integers a and b, when 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

Soln: We know, $$x^2 -y^2 = (x+y)*(x-y)$$, so, $$a^4 - b^4 = (a^2 + b^2)*(a^2 - b^2) = (a^2 + b^2)*(a+b)*(a-b)$$

1. Since $$(a+b)$$ is divisible by 3. We can conclude that $$a^4 - b^4 = (a^2 + b^2)*(a+b)*(a-b)$$ will also be divisible by 3. So, remainder must be 0. SUFFICIENT.

2. $$a^2 + b^2$$ is divided by 3, the remainder is 2. Lets put some values of a and b and check the value of $$a^2 - b^2.$$
If $$a = 2, b = 1,$$ so $$a^2 + b^2 = 5 = 3*1+2.$$ We have $$a^2 - b^2 = 3$$ ...divisible by 3. So, the remainder when $$a^4 - b^4$$ is divided by 3 will be 0. YES.
If $$a = 4, b = 2,$$ so $$a^2 + b^2 = 20 = 3*6+2.$$ We have $$a^2 - b^2 = 12$$ ...divisible by 3. So, the remainder when $$a^4 - b^4$$ is divided by 3 will be 0. YES.
If $$a = 2, b = 5,$$ so $$a^2 + b^2 = 29 = 3*9+2.$$ We have $$a^2 - b^2 = -21$$ ...divisible by 3. So, the remainder when $$a^4 - b^4$$ is divided by 3 will be 0. YES.
If $$a = 2, b = 10,$$ so $$a^2 + b^2 = 104 = 3*34+2.$$ We have $$a^2 - b^2 = -96$$...divisible by 3. So, the remainder when $$a^4 - b^4$$ is divided by 3 will be 0. YES.
If $$a = 10, b = 4,$$ so $$a^2 + b^2 = 116 = 3*38+2.$$ We have $$a^2 - b^2 = 84$$ ...divisible by 3. So, the remainder when $$a^4 - b^4$$ is divided by 3 will be 0. YES.

We can conclude for every a and b, $$a^4 - b^4$$ will be divided by 3 or remainder will be 0. SUFFICIENT.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 12:04
a^4-b^4 = (a^2+b^2) (a^2-b^2)

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

1) suff .. remainder is 0

2) can't say NS ..

A
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 13:54
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Wow, I found this one super hard and got it wrong at first, but I've managed to prove that the answer is D, either statement is sufficient. I'm sure i've missed the obvious trick here, but this is how i've done it:

For positive integers a and b, when $$a^4 - b^4$$ 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

Firstly note that a and b are positive integers therefore anything constructed from a, b and integer powers of a and b without any division will be an integer. Let's start by factorising the original statement to break it down into easier parts.

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

(1) tells us that $$a + b$$ is divisible by 3 as the remainder is 0. Therefore any multiple of $$(a + b)$$ is divisible by 3. Since we have shown above that $$a^4 - b^4$$ is a multiple of $$(a + b)$$, and since we know that the other terms in the above equation are integers, we know that $$a^4 - b^4$$ is divisible by 3 and will therefore have a remainder of 0 when divided by 3. Therefore statement (1) is sufficient and the possible answers are A and D.

(2) tells us that $$a^2 + b^2$$ is not divisible by 3, and has a remainder of 2 when divided by 3. From the above we can see that $$a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$$, and therefore we need to test whether $$a^2 + b^2$$ having a remainder of 2 means that $$(a^2 - b^2)$$ has a certain remainder when divided by 3.

Let's consider $$a^2$$ and $$b^2$$ separately. If $$a^2 + b^2$$ has a remainder of 2, then three options are possible:

$$a^2$$ has a remainder of 0 and $$b^2$$ has a remainder of 2
$$a^2$$ and $$b^2$$ each have a remainder of 1
$$a^2$$ has a remainder of 2 and $$b^2$$ has a remainder of 0

Integers can only possibly have remainders 0, 1, or 2, when divided by 3, so each integer fits into one of 3 camps: $$x$$, $$x+1$$, or $$x+2$$, where x is divisible by 3. Therefore every square number is either $$x^2$$, $$(x+1)^2 = x^2 + 2x + 1$$, or $$(x+2)^2 = x^2 + 4x + 4$$. The only terms here not divisible by 3 are 1 and 4, both of which have a remainder 1. Therefore square numbers are either divisible by 3 or have remainder 1 when divided by 3. This let's us rule out the first and last possibility, so we know that if $$a^2 + b^2$$ has a remainder of 2 when divided by 3, $$a^2$$ and $$b^2$$ each have a remainder of 1.

If $$a^2$$ and $$b^2$$ each have a remainder of 1, then $$(a^2 - b^2)$$ has a remainder 0 and is always divisible by 3. In that case $$a^4 - b^4$$ is divisible by 3 as it is a multiple of $$(a^2 - b^2)$$. Therefore (2) is also sufficient to calculate the remainder of $$a^4 - b^4$$ when divided by 3. The answer to the question is D, either statement is sufficient on it's own.
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 17:30
When this function is simplified, we get
(a^2 + b^2)(a + b)(a - b)

From 1st statement, (a+b) is a multiple of 3. Hence the product will be a multiple of 3 and the remainder will be 0
Sufficient. This eliminates B, C, E

From 2nd statement, there are multiple options and hence B is eliminated
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 19:48
1
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$$a^4-b^4 = (a^2-b^2)*(a^2+b^2) = (a+b)*(a-b)*(a^2+b^2)$$

Statement (1): When a + b is divided by 3, the remainder is 0
since $$(a+b)$$ is a factor of 3, $$a^4-b^4$$ when divided by 3, the remainder will be 0
Sufficient

Statement (2): When $$(a^2 + b^2)$$ is divided by 3, the remainder is 2
lets choose a = 2 and b = 1
$$(a^2 + b^2) = 5$$ when divided by 3, the remainder is 2
$$a^4-b^4 = 15$$ when divided by 3, the remainder is 0
Sufficent

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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06 Dec 2015, 20:33
1
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Statement 1 : (a+ b)/ 3 leaves no remainder
let a = 5 , b = 2
now (a^4 - b^4)/3
plugging in values: Remainder= 0

Similarily for a = 2, b = 1
Remainder=0

Thus Sufficient

Statement 2:
(a^2+b^2) / 3 leaves remainder of 2
let a = 2 and b = 1..this satisfies the condition
now (a^4 - b^4)/3
plugging in values: Remainder= 0

Simlarily values a = 5 , b = 2 satisfies condition
now (a^4 - b^4)/3
plugging in values: Remainder= 0

Sufficient

Ans: D
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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07 Dec 2015, 07:03
$$(a^4-b^4)$$(a-b)(a+b)$$(a^2+b^2)$$

While (a-b) and (a+b) will be divisible by 3, it will be $$(a^2+b^2)$$ that will give us remainder if it will.

St 1: When a + b is divided by 3, the remainder is 0

2 cases will give us this:
1. Both a and b are divisible by 3
2. Both a and b are not divisible by 3

While case 1 will not give us a remainder with $$(a^2+b^2)$$, case 2 will give us.

Insufficient

St 2: When a^2 + b^2 is divided by 3, the remainder is 2

Exactly what we want.

Sufficient

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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07 Dec 2015, 07:44
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OA: D

Solution
a^4 - b^4 = (a-b)(a+b)(a^2+b^2)
Hence, clearly (1) is sufficient.

Now consider (2). Notice that n^2 for any integer n when divided by 3 leaves a remainder of either 0 or 1.
Therefore,
a^2 + b^2 divided by 3 leaves remainder 2 => (1) a^2 divided by 3 leaves remainder 1 AND (2) b^2 divided by 3 leaves remainder 1. (no other combos are possible). Hence, a^2 - b^2 is divisible by 3 => a^4 - b^4 is divisible by 3. Sufficient.
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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07 Dec 2015, 09:31
1
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a^4 - b^4 = (a+b)(a-b)(a^2+b^2)

(1) When a + b is divided by 3, the remainder is 0 = a^4 - b^4 is divisible by 3
(2) When a^2 + b^2 is divided by 3, the remainder is 2 = a^4 - b^4 is divisible by 3 with a remainder of 2

Both are sufficient : D
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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07 Dec 2015, 12:08
QUESTION #1:

For positive integers a and b, when 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

Solution: both the statement taken together is not sufficient to produce the remainder of the question asked.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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08 Dec 2015, 04:14
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$$a^4 - b^4 = (a^2 + a ^2)(a – b)(a + b)$$

1) If a + b is divisible by 3 then the product above is divisible by 3 then the remainder is 0 sufficient

2) Possible values for $$a^2 + b^2$$ are:
8 = $$(2^2 + 2^2$$) ==> Then we have 2 + 2 = 4 not divisible by 3, but 2 – 2 is divisible by 3 so the remainder is 0
17 = $$(4^2 + 1^2)$$ ==> Then we have 4 + 1 = 5 not divisible by 3, but 4 – 1 is divisible by 3 so remainder is 0
20 = $$(4^2 + 2^2)$$ ==> Then we have 4 + 2 = 6 which divisible by 3 so remainder is 0
26 = $$(5^2 + 1^2)$$ ==> Then we have 5 + 1 = 6 which divisible by 3 so remainder is 0
29 = $$(5^2 + 2^2)$$ ==> Then we have 5 + 2 = 7 not divisible by 3, but 5 – 2 is divisible by 3 so remainder is 0

We can see a pattern for a + b> 4, when a²+b² is odd, a – b is divisible by 3 and when a²+b² is even, a + b is divisible by 3 Statement 2 is also sufficient, the remainder is zero.

Each statement is sufficient answer D
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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Updated on: 13 Dec 2015, 01:27
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Simplifying the term a^4-b^4
(a^2+b^2)(a+b)(a-b)

statement 1: When (a+b) is divided by 3, remainder is 0. If any term in the form a*b*c is divisible by 3, then the whole term is divisible by 3. Statement 1 is sufficient because when a^4-b^4 is divided by 3, it will have a remainder of 0.

Statement 2: [(non-multiple of 3)^2+(non-multiple of 3)^2] must have a remainder of 2 when divided by 3. And any two non-multiple of 3 must be divisible by 3 upon summation or subtraction. In the question stem, we see both summation and subtraction of a and b. Hence, sufficient.

{for example: a=non-multiple of 3=14 and b=17. Now, (a^2+b^2)=485, which has a remainder of 2. Here, (a-b) is a multiple of 3 and hence remainder of a^4-b^4 is 0}

**[(multiple of 3)^2+(non-multiple of 3)^2] must have a remainder of 1 when divided by 3.

Originally posted by zmtalha on 08 Dec 2015, 11:38.
Last edited by zmtalha on 13 Dec 2015, 01:27, edited 1 time in total.
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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08 Dec 2015, 21:39
the answer is A. as it is enough to answer the question
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Re: Math Revolution and GMAT Club Contest! For positive integers a and b, [#permalink]

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09 Dec 2015, 10:01
a^4 - b^4 can be rewritten as (a-b)(a+b)(a^2 + b^2) .
Now if any of these parts are divisible by 3 , the remainder will be zero. we need to know whether one of the parts is divisible by 3 - so the remainder will be zero.
or the remainders of each part or combined intermediate parts when they're divided by 3 (e.g., reminder when a^2 - b^2 is divisble by 3 AND reminder a^2 + b^2 is divisible by 3) OR ( (a-b)/3 and (a+b)/3 and (a^2+b^2)/3) .

Option A says that one of the parts (=a+b) is divisible by 3 . so a^4 - b^4 is divisible by 3 with a reminder of 0 - A is sufficient.
Option B says a^2 + b^2 is NOT divisible by 3 and leaves a reminder of 2.
It is possible that one of a+b or a-b is divisible by 3 - so the entire a^4-b^4 leaves a reminder 0
It is also possible that neither a+b nor a-b is divisible by 3 - so a^4 - b^4 leaves a non-zero reminder - B insuff.

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Re: Math Revolution and GMAT Club Contest! For positive integers a and b,   [#permalink] 09 Dec 2015, 10:01

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