GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2018, 10:14

# Saturday Quant Quiz:

Starts promptly at 10 AM PST - Join in to Have Fun & Win Prizes

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If a and b are positive integers, is a^4-b^4 divisible by 4?

Author Message
TAGS:

### Hide Tags

Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1314
Location: Malaysia
If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

Updated on: 23 Mar 2017, 00:23
3
15
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:01) correct 54% (01:46) wrong based on 229 sessions

### HideShow timer Statistics

If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Originally posted by hazelnut on 23 Mar 2017, 00:22.
Last edited by Bunuel on 23 Mar 2017, 00:23, edited 1 time in total.
Edited the tags.
Intern
Joined: 02 Feb 2017
Posts: 17
Location: India
Schools: ISB '20 (I)
GMAT 1: 740 Q50 V40
If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

24 Mar 2017, 06:24
12
4
$$a^4−b^4=(a+b)(a−b)(a^2+b^2)$$

1. $$a+b$$ is divisible by 4

Straightforward and sufficient.

2. Let $$a^2+b^2 = 4x+2$$ (since it gives 2 as remainder when divided by 4)

So, $$a^{4}−b^{4} = (4x+2)(a+b)(a-b)$$ which breaks down to ..$$2(2x+1)(a+b)(a-b)$$

Now, any number which has product of two even numbers will be divisible by 4.
We already have 2. So we need to know if we have another even number or not.

So, 2x+1 .. always odd
(a+b) and (a-b) both will be even if a & b both are odd or both are even.

Since $$a^2+b^2$$ gives a remainder of 2 when divided by 4, it means the sum sum is even.

Now, the sum will be even only if $$a^2$$ & $$b^2$$ both are even or both are odd.

That means, a & b both are even or both are odd (since square of odd numbers are odd and even numbers are even).

So, sufficient.

##### General Discussion
Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

23 Mar 2017, 02:50
2
6
ziyuen wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi

Very good queston.

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

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder $$r$$ when divided by $$c$$, then $$a^2$$ will have remainder $$r^2$$ when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> $$a^2 = r^2$$ (mod 4), $$b^2 = q^2$$ (mod 4)

We are given $$r^2 + q^2 = 2$$. There is only one possibility $$1^2 + 1^2 = 1 + 1 = 2$$ (We can't get $$0 + 2$$ or $$2 + 0$$ because that means that $$a$$ has a remainder $$\sqrt{2}$$ when divided by $$4$$ and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by $$4$$. Sufficient.

Senior Manager
Joined: 02 Apr 2014
Posts: 471
GMAT 1: 700 Q50 V34
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

19 Nov 2017, 11:29
1
$$(a^4 - b^4$$) % 4 = 0?

$$(a^2 + b ^2)(a - b)(a+b)$$ % 4 = 0?

Statement 1:
(a + b) is divisible by 4 => (a + b) % 4 = 0 => $$(a^2 + b ^2)(a - b)(a+b)$$ is divisible by 4

Sufficient

Statement 2:
$$(a^2 + b^2)$$ % 4 = 2, Let $$(a^2 + b^2)$$ = 4x + 2 (where x is any integer)

To Note, since remainder is 2, $$(a^2 + b^2)$$ must be even
=> either $$a^2 hence (a) and b^2 hence(b) both are Odd$$
$$or$$
$$a^2 hence (a) and b^2 hence(b) both are Even$$

Substituting $$(a^2 + b^2) = 4x + 2$$ in $$(a^2 + b ^2)(a - b)(a+b)$$
=> $$(4x + 2)(a^2 - b^2)$$
=> $$4x(a^2-b^2) + 2 (a^2 - b^2)$$

First term, 4x(a^2-b^2) is clearly divisble by 4
Second term, $$2 (a^2 - b^2)$$ , since we know a, b are either both odd or even, so that $$(a^2 - b^2)$$is even => $$2 * even$$ => clearly divisible by 4

Sufficient

Manager
Joined: 03 Mar 2018
Posts: 182
If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

28 Mar 2018, 05:47
vitaliyGMAT wrote:
ziyuen wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi

Very good queston.

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

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder $$r$$ when divided by $$c$$, then $$a^2$$ will have remainder $$r^2$$ when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> $$a^2 = r^2$$ (mod 4), $$b^2 = q^2$$ (mod 4)

We are given $$r^2 + q^2 = 2$$. There is only one possibility $$1^2 + 1^2 = 1 + 1 = 2$$ (We can't get $$0 + 2$$ or $$2 + 0$$ because that means that $$a$$ has a remainder $$\sqrt{2}$$ when divided by $$4$$ and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by $$4$$. Sufficient.

vitaliyGMAT
Although you have made a good analysis, Your initial expression $$a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)$$ is not correct.
Correct form of expression: $$a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)$$
_________________

Intern
Joined: 29 Mar 2015
Posts: 21
GMAT 1: 590 Q44 V23
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

26 Apr 2018, 19:41
itisSheldon wrote:
vitaliyGMAT wrote:
ziyuen wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi

Very good queston.

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

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder $$r$$ when divided by $$c$$, then $$a^2$$ will have remainder $$r^2$$ when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> $$a^2 = r^2$$ (mod 4), $$b^2 = q^2$$ (mod 4)

We are given $$r^2 + q^2 = 2$$. There is only one possibility $$1^2 + 1^2 = 1 + 1 = 2$$ (We can't get $$0 + 2$$ or $$2 + 0$$ because that means that $$a$$ has a remainder $$\sqrt{2}$$ when divided by $$4$$ and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by $$4$$. Sufficient.

vitaliyGMAT
Although you have made a good analysis, Your initial expression $$a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)$$ is not correct.
Correct form of expression: $$a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)$$

Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?
Manager
Joined: 17 May 2015
Posts: 239
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

26 Apr 2018, 21:58
2
2
hazelnut wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi,
Using the formula $$a^{2} - b^{2} = (a+b)(a-b)$$, we can simplify the given expression as follows:

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

(1) a+b is divisible by 4

Sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4.

=> $$a^{2} + b^{2} = 4k + 2 = 2(2k + 1)$$ (where k is an integer) --- (1)

=> $$a^{2} = 4k + 2 - b^{2}$$

$$a^{2} - b^{2}$$ can be written as $$4k + 2 - b^{2} - b^{2} = 4k + 2 - 2b^{2} = 2(2k + 1 - b^{2})$$ --- (2)

By using equation (1) and (2) we can write $$a^4 - b^4$$ as follows:

$$a^4 - b^4 = (a^{2} + b^{2})(a^{2} - b^{2}) = 2(2k + 1) \times 2(2k + 1 - b^{2}) = 4 (2k + 1)(2k + 1 - b^{2})$$

Hence, divisible by 4. Sufficient.

Thanks.
Senior DS Moderator
Joined: 27 Oct 2017
Posts: 888
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

27 Apr 2018, 14:09
Hi akara2500
As per Statement 2 :
When a^2 + b^2 is divisible by 4, remainder =2.
Hence a^2+ b^2 = 4k+2, thats is even.
Hence either both a,b are even or both a,b are odd.
If both are even, a =2p, b =2q. Hence a^4- b^4 = 16(p^4-q^4). Hence divisible by 4.
If both are odd, a=(2m+1), b=(2n+1)
Now a^2-b^2=4(m^2-n^2+m-n). Hence divisible by 4. Also a^4-b^4= (a^2-b^2)(a^2+b^2). Hence divisible by 4.
Statement 2 is also sufficient.

akara2500 wrote:
itisSheldon wrote:
vitaliyGMAT wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi

Very good queston.

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

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder $$r$$ when divided by $$c$$, then $$a^2$$ will have remainder $$r^2$$ when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> $$a^2 = r^2$$ (mod 4), $$b^2 = q^2$$ (mod 4)

We are given $$r^2 + q^2 = 2$$. There is only one possibility $$1^2 + 1^2 = 1 + 1 = 2$$ (We can't get $$0 + 2$$ or $$2 + 0$$ because that means that $$a$$ has a remainder $$\sqrt{2}$$ when divided by $$4$$ and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by $$4$$. Sufficient.

vitaliyGMAT
Although you have made a good analysis, Your initial expression $$a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)$$ is not correct.
Correct form of expression: $$a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)$$

Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?[/quote]
_________________
Manager
Joined: 01 Aug 2017
Posts: 188
Location: India
Schools: ISB '20 (S), IIMA (S)
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

06 May 2018, 21:22
$$a^4−b^4=(a+b)(a−b)(a^2+b^2)$$ ---------1

1. 1st statement is very simple and straight forward. And it is sufficient.

2. Let see how we will do for 2nd Statement.

Let $$a^2+b^2=4x+2$$ -----------2

where x is the quotient and 2 is the remainder.

Using 2 in 1 we get $$a^4−b^4=(4x+2)(a+b)(a−b)$$

Which breaks down to $$a^4−b^4=2(2x+1)(a+b)(a−b)$$

We have to find out if eq 1 is divisible by 4.
Product of any two even numbers will be divisible by 4.

We have one even number- 2. So we need to find out if any of the expression (2x+1), (a+b) & (a-b) will give us even number?

We know that 2x+1 is always odd. NOTE - it is general representation of ODD numbers.

In what scenario (a+b) & (a-b) will give us odd/even.
1) If a and b is odd. Then both (a+b) and (a-b) will be even.
2) If a and b is even. Then both (a+b) and (a-b) will be even.

But how can we decide if a & b is even or odd?

Since $$a^2+b^2$$ gives a remainder of 2 when divided by 4, it means the sum is even.

From here we have to backtrack with logic to determine if a & b is odd or even.

Again, sum will be even if both the numbers are odd or both the numbers are even.

There are two scenario both leading to common results:-
1) If the square of a number is odd. then the number itself will be odd.
2) Similarly, if square of the numbers are even then, the numbers itself will be even.

From scenario 1 :- Since sum of expression $$a^2+b^2$$ is even, let's assume both $$a^2$$ and $$b^2$$ are odd. Then it implies both a & b are odd. Also the expression (a+b) & (a-b) will be even.

From scenario 1 :- Since sum of expression $$a^2+b^2$$ is even, let's assume both $$a^2$$ and $$b^2$$ are even. Then it implies both a & b are even. Also the expression (a+b) & (a-b) will be even.

Hence the expression $$a^4−b^4$$ is divisible by 4. So, sufficient.

Hope this helps!!
Thanks
_________________

If it helps you please press Kudos!

Thank You
Sudhanshu

Intern
Joined: 11 Feb 2018
Posts: 30
Re: If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

### Show Tags

23 Jun 2018, 03:31
vitaliyGMAT wrote:
ziyuen wrote:
If a and b are positive integers, is $$a^4 - b^4$$ divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

Hi

Very good queston.

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

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when $$a^2 + b^2$$ is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder $$r$$ when divided by $$c$$, then $$a^2$$ will have remainder $$r^2$$ when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> $$a^2 = r^2$$ (mod 4), $$b^2 = q^2$$ (mod 4)

We are given $$r^2 + q^2 = 2$$. There is only one possibility $$1^2 + 1^2 = 1 + 1 = 2$$ (We can't get $$0 + 2$$ or $$2 + 0$$ because that means that $$a$$ has a remainder $$\sqrt{2}$$ when divided by $$4$$ and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by $$4$$. Sufficient.

Hi Bunuel
Can u plz help me understand the above highlighted part.
Thanks
Re: If a and b are positive integers, is a^4-b^4 divisible by 4? &nbs [#permalink] 23 Jun 2018, 03:31
Display posts from previous: Sort by