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

 It is currently 18 Oct 2019, 12:43

### 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

# a, x, and k are positive integers. k is odd. Is a divisible by 24?

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58453
a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

11 Dec 2017, 00:50
1
11
00:00

Difficulty:

95% (hard)

Question Stats:

20% (02:11) correct 80% (02:40) wrong based on 140 sessions

### HideShow timer Statistics

a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$

(2) $$x^{k^3-k} = x^a$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58453
a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

20 Dec 2017, 06:07
8
1
MSarmah wrote:
Didnot get how the answer is C

Brunnel can you explain !?

This is a hard problem and the solutions above are not correct.

a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$.

Notice that since x and a are positive integers, the above inequality holds true for any values except when x = 1.

If x = 1, then $$x^{2a} + \frac{1}{x^{2a}} = 1 + 1 = 2$$
If x > 1, then $$x^{2a} + \frac{1}{x^{2a}} = (integer \ greater \ than \ 2) + (positive \ number \ less \ than \ 1) >2$$

So, this statement tells us that x ≠ 1. Not sufficient.

(2) $$x^{k^3-k} = x^a$$

If x = 1, then a and k can be any integers, because 1^anything = 1, no matter what k^3 - k and a are.
If x ≠ 1, then we can equate the powers: $$a = k^3 - k = k(k^2 - 1)= (k - 1)k(k + 1)$$. So, a is a product of three consecutive integers, thus one of them is a multiple of 3. Also, since given that k is odd, then (k - 1) and (k + 1) are consecutive even integer, which imply that one of them is a multiple of 4, so a is a multiple of 2*4 = 8. Thus, in this case a becomes a multiple of both 3 and 8, so a multiple of 3*8 = 24.

Not sufficient.

(1)+(2) From (1) x ≠ 1, thus from (2) a multiple of 3*8 = 24. Sufficient.

_________________
##### General Discussion
Retired Moderator
Joined: 22 Aug 2013
Posts: 1428
Location: India
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

11 Dec 2017, 04:20
Bunuel wrote:
a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$

(2) $$x^{k^3-k} = x^a$$

24 = 2^3 * 3
So for any integer to be divisible by 24, that number must have at least three 2's and one 3 in its prime factorisation.

(1) If x is any positive integer apart from 1, then x^2a + 1/x^2a will ALWAYS be greater than 2 (given that 'a' is also a positive integer). For x=1, x^2a + 1/x^2a will be 1^2a + 1/x^2a = 1+1=2. But for x=2, x^2a + 1/x^2a= 2^2a + 1/2^2a. No matter which value of 'a' (positive integer) we put here, it will come out to be > 2. And with increasing values of x and 'a', the value of x^2a + 1/x^2a will keep on increasing further, and always > 2. So we cant say whether a is divisible by 24 or not. Insufficient.

(2) From this statement we can conclude that a = k^3 - k = k (k^2 - 1) = (k-1)*k*(k+1). Now 'a' is a product of three consecutive integers thus it will be divisible by 3. Also given that k is odd, so both k+1 and k-1 will be even, and also, one of them will also be a multiple of 4. So this guarantees three 2's also. Eg, take k=3, then k-1 and k+1 are 2 & 4 respectively and their product will have three 2's. Take k=5, then k-1 and k+1 will be 4 and 6 respectively and their product will also have three 2's.. So for any odd integer value of k, the product (k-1)*k*(k+1) will have at least one 3 and three 2's , thus will be divisible by 24. Sufficient.

Manager
Joined: 25 Jul 2017
Posts: 55
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

11 Dec 2017, 04:44
1
From 1:

X^ (2a) = (X^a)^2 ==> positive number
1/ (X^2a) = (1/x^a)^2 ==> positive number
==> X^ (2a) + 1/ (X^2a) = (X^a)^2 + (1/x^a)^2 is always greater than 2, no matter which value of a
(1) is sufficient

From 2 ==> K^3 - K = a = (K-1)*K* (K+1).

K - 1; K and K+1 is 3 consecutive interger ==> (k-1)K(k+1) is multiple of 3

K is odd ==> K-1 and K+1 must be multiple of 8
==> A = (k-1)k(K+1) is multiple of 24

Intern
Joined: 19 Aug 2017
Posts: 15
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

20 Dec 2017, 04:03
Didnot get how the answer is C

Brunnel can you explain !?
Intern
Joined: 19 Aug 2017
Posts: 15
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

21 Dec 2017, 00:07
1
Bunuel wrote:
MSarmah wrote:
Didnot get how the answer is C

Brunnel can you explain !?

This is a hard problem and the solutions above are not correct.

a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$.

Notice that since x and a are positive integers, the above inequality holds true for any values except when x = 1.

If x = 1, then $$x^{2a} + \frac{1}{x^{2a}} = 1 + 1 = 2$$
If x > 1, then $$x^{2a} + \frac{1}{x^{2a}} = (integer \ greater \ than \ 2) + (positive \ number \ less \ than \ 1) >2$$

So, this statement tells us that x ≠ 1. Not sufficient.

(2) $$x^{k^3-k} = x^a$$

If x = 1, then a and k can be any integers, because 1^anything = 1, no matter what k^3 - k and a are.
If x ≠ 1, then we can equate the powers: $$a = k^3 - k = k(k^2 - 1)= (k - 1)k(k + 1)$$. So, a is a product of three consecutive integers, thus one of them is a multiple of 3. Also, since given that k is odd, then (k - 1) and (k + 1) are consecutive even integer, which imply that one of them is a multiple of 4, so a is a multiple of 2*4 = 8. Thus, in this case a becomes a multiple of both 3 and 8, so a multiple of 3*8 = 24.

Not sufficient.

(1)+(2) From (1) x ≠ 1, thus from (2) a multiple of 3*8 = 24. Sufficient.

Thanks Brunnel !!
Intern
Joined: 17 Sep 2017
Posts: 45
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

21 Dec 2017, 07:53
I thought 0 (zero) is divisible by every integer. Therefore I chose B.
Math Expert
Joined: 02 Sep 2009
Posts: 58453
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

21 Dec 2017, 07:57
lichting wrote:
I thought 0 (zero) is divisible by every integer. Therefore I chose B.

Yes, 0 is divisible by every integer, except 0 itself but what this has to do with the question? How did you get that a = 0?
_________________
Intern
Joined: 30 Sep 2016
Posts: 5
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

28 Jan 2018, 19:19
Hello Bunuel can you please explain me how did you get this (k−1)k(k+1)?. I don't know if I'm missing a step or rule on this one. Although it is straightforward how you got to a=k3−k=k(k2−1).

Thank you.
Math Expert
Joined: 02 Aug 2009
Posts: 7978
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

28 Jan 2018, 19:39
3
Bunuel wrote:
a, x, and k are positive integers. k is odd. Is a divisible by 24?

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$

(2) $$x^{k^3-k} = x^a$$

(1) $$x^{2a} + \frac{1}{x^{2a}}>2$$....
Let $$x^{2a}=y$$...
$$x^{4a} + 1>2x^{2a}......y^2-2y+1>0.....(y-1)^2>0$$....
Since y is POSITIVE integer, y>1, or $$x^{2a}>1$$
This means x>1..
Nothing much
Insufficient

(2) $$x^{k^3-k} = x^a$$...
If x >1...
$$x^{k^3-k} = x^a......a=k^3-k=k(k^2-1)=(k-1)k(k+1)$$...
Sum of three CONSECUTIVE integers where k is odd...
So one of the digits will be MULTIPLE of 3, one of them will be ATLEAST div by 2 and other even number by ATLEAST 4, so a is div by 2*3*4=24..yes
If x=1
K and a can be any values as 1^anything is 1
Insufficient

Combined
x>1, so a is div by 24 as shown from statement II
Suff..

Tavo, $$k^3-k$$, take out k common from both terms.
So k^3-k=k*k^2-k*1=k(k^2-1)...
Now a^2-b^2=(a-b)(a+b).... So k(k-1)(k+1)..
_________________
Intern
Joined: 30 Sep 2016
Posts: 5
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

29 Jan 2018, 13:48
Thank You Chetan2u, It was much easier than I thought. Great!!!
Non-Human User
Joined: 09 Sep 2013
Posts: 13259
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?  [#permalink]

### Show Tags

09 Jul 2019, 11:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: a, x, and k are positive integers. k is odd. Is a divisible by 24?   [#permalink] 09 Jul 2019, 11:05
Display posts from previous: Sort by