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

 It is currently 14 Nov 2018, 01:39

### 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

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# M05-22

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

15 Sep 2014, 23:25
2
4
00:00

Difficulty:

45% (medium)

Question Stats:

62% (00:54) correct 38% (01:06) wrong based on 203 sessions

### HideShow timer Statistics

If $$x$$ is a positive integer, is $$x$$ divisible by 15?

(1) $$x$$ is a multiple of 10

(2) $$x^2$$ is a multiple of 12

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

15 Sep 2014, 23:25
Official Solution:

If x is a positive integer, is x divisible by 15?

(1) $$x$$ is a multiple of 10. If $$x=10$$ then the answer is NO but if $$x=30$$ then the answer is YES. Not sufficient.

(2) $$x^2$$ is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

_________________
Current Student
Joined: 06 Mar 2014
Posts: 245
Location: India
GMAT Date: 04-30-2015

### Show Tags

27 Sep 2014, 12:20
Bunuel wrote:
Official Solution:

(1) $$x$$ is a multiple of 10. If $$x=10$$ then the answer is NO but if $$x=30$$ then the answer is YES. Not sufficient.

(2) $$x^2$$ is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming:
1) x multiple of 55
2) x^3 multiple of 110

Would be really grateful.
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

27 Sep 2014, 12:45
2
2
earnit wrote:
Bunuel wrote:
Official Solution:

(1) $$x$$ is a multiple of 10. If $$x=10$$ then the answer is NO but if $$x=30$$ then the answer is YES. Not sufficient.

(2) $$x^2$$ is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming:
1) x multiple of 55
2) x^3 multiple of 110

Would be really grateful.

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that $$x^2$$, where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

Similar questions to practice:
if-x-is-an-integer-is-x-3-divisible-by-165973.html
how-many-different-prime-numbers-are-factors-of-the-positive-126744.html
if-n-2-n-yields-an-integer-greater-than-0-is-n-divisible-by-126648.html
if-k-is-a-positive-integer-is-k-the-square-of-an-integer-55987.html
if-k-is-a-positive-integer-how-many-different-prime-numbers-95585.html
if-n-is-the-integer-whether-30-is-a-factor-of-n-126572.html
if-x-is-an-integer-is-x-2-1-x-5-an-even-number-104275.html

Hope it helps.
_________________
Current Student
Joined: 06 Mar 2014
Posts: 245
Location: India
GMAT Date: 04-30-2015

### Show Tags

27 Sep 2014, 12:56
Bunuel wrote:
earnit wrote:
Bunuel wrote:
Official Solution:

(1) $$x$$ is a multiple of 10. If $$x=10$$ then the answer is NO but if $$x=30$$ then the answer is YES. Not sufficient.

(2) $$x^2$$ is a multiple of 12. The least perfect square which is a multiple of 12 is 36. Hence, the least value of $$x$$ is 6 and in this case the answer is NO, but if for example $$x=12*15$$ then the answer is YES. Not sufficient.

Notice that from this statement we can deduce that $$x$$ must be a multiple of 3 (else how can this prime appear in $$x^2$$?).

(1)+(2) $$x$$ is a multiple of both 10 and 3, hence it's a multiple of 30, so $$x$$ must be divisible by 15. Sufficient.

I couldn't follow the highlighted portion actually.

I seem to find DS Questions with given statements such as below one little difficult/time consuming:
1) x multiple of 55
2) x^3 multiple of 110

Would be really grateful.

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that $$x^2$$, where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

Similar questions to practice:
if-x-is-an-integer-is-x-3-divisible-by-165973.html
how-many-different-prime-numbers-are-factors-of-the-positive-126744.html
if-n-2-n-yields-an-integer-greater-than-0-is-n-divisible-by-126648.html
if-k-is-a-positive-integer-is-k-the-square-of-an-integer-55987.html
if-k-is-a-positive-integer-how-many-different-prime-numbers-95585.html
if-n-is-the-integer-whether-30-is-a-factor-of-n-126572.html
if-x-is-an-integer-is-x-2-1-x-5-an-even-number-104275.html

Hope it helps.

THANKS a ton.
Really appreciate it.
Intern
Joined: 17 May 2015
Posts: 30

### Show Tags

17 Sep 2015, 07:24
Need great help , I don't understand why (2) can give both yes and no answer , many thanks
My understanding:
(2)X2 can be 36 or 144 which is x=6 or 12
Both are not divisible by 15
Many thsnks
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

18 Sep 2015, 08:17
1
apple08 wrote:
Need great help , I don't understand why (2) can give both yes and no answer , many thanks
My understanding:
(2)X2 can be 36 or 144 which is x=6 or 12
Both are not divisible by 15
Many thsnks

What if x = 12*15 = 180?
_________________
Intern
Joined: 11 Jan 2015
Posts: 34

### Show Tags

28 Jun 2016, 19:17
1
Plz let me know if my approach is correct:

In order to know if x is divisible by 15, x has to contain at least the prime factors of $$15$$ -> $$(3*5)$$.

From (1): We only know that x is similar to $$x*2*5$$ ->$$x$$ could be three. Not sufficient.

From (2): We only know that $$x^2$$ is similar to $$x*x*2*2*3$$ -> One of the $$x$$ could be $$5$$. Not sufficient.

From (1) + (2): We know that $$x$$ contains at least one 3 and one 5 what makes it divisible by $$15$$ -> $$(3*5)$$. Sufficient
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

28 Jun 2016, 23:29
Plz let me know if my approach is correct:

In order to know if x is divisible by 15, x has to contain at least the prime factors of $$15$$ -> $$(3*5)$$.

From (1): We only know that x is similar to $$x*2*5$$ ->$$x$$ could be three. Not sufficient.

From (2): We only know that $$x^2$$ is similar to $$x*x*2*2*3$$ -> One of the $$x$$ could be $$5$$. Not sufficient.

From (1) + (2): We know that $$x$$ contains at least one 3 and one 5 what makes it divisible by $$15$$ -> $$(3*5)$$. Sufficient

______________
Yes, that's correct.
_________________
Manager
Joined: 01 Nov 2016
Posts: 66
Concentration: Technology, Operations

### Show Tags

06 May 2017, 20:46
Bunuel wrote:
Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that $$x^2$$, where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

I don't think I understand this. Are you saying that if $$x^2$$ is divisible by 12 then x is divisible by 12?

So does that mean that if $$x^z$$ is divisible by y then x is divisible by y (as long as z is positive)?
Intern
Joined: 11 Jan 2015
Posts: 34

### Show Tags

07 May 2017, 00:16
joondez wrote:
Bunuel wrote:
Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that $$x^2$$, where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

So does that mean that if $$x^z$$ is divisible by y then x is divisible by y (as long as z is positive)?

...as long as y is prime your statement should be correct.
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

07 May 2017, 01:03
joondez wrote:
Bunuel wrote:
Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

We are given that $$x^2$$, where x is an integer, is a multiple of 12 = 2^2*3.This means that x must be a multiple of both 2 and 3. If they weren't how can x^2 have these primes?

So does that mean that if $$x^z$$ is divisible by y then x is divisible by y (as long as z is positive)?

...as long as y is prime your statement should be correct.

You should read more carefully. As correctly noted by paddy41, the property above is talking about primes.
_________________
Intern
Joined: 24 Jun 2017
Posts: 6

### Show Tags

13 Oct 2017, 03:57
how sqrt(180) gives a integer??? we cant take 180 has a sol..
Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

13 Oct 2017, 05:37
how sqrt(180) gives a integer??? we cant take 180 has a sol..

Please re-read the solution. (2) says that x^2 is a multiple of 12, not $$\sqrt{x}$$. Next, the solution then considers 180 as one of possible values for x.
_________________
Manager
Joined: 26 Feb 2018
Posts: 59
WE: Sales (Internet and New Media)

### Show Tags

02 Jun 2018, 10:45
Bunuel wrote:
If $$x$$ is a positive integer, is $$x$$ divisible by 15?

(1) $$x$$ is a multiple of 10

(2) $$x^2$$ is a multiple of 12

Hi Bunuel ,

I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n)

statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible .

Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 .

Can you please guide me ?
_________________

" Can't stop learning and failing"

Math Expert
Joined: 02 Sep 2009
Posts: 50578

### Show Tags

02 Jun 2018, 11:13
1
loserunderachiever wrote:
Bunuel wrote:
If $$x$$ is a positive integer, is $$x$$ divisible by 15?

(1) $$x$$ is a multiple of 10

(2) $$x^2$$ is a multiple of 12

Hi Bunuel ,

I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n)

statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible .

Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 .

Can you please guide me ?

Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, ....

(2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be:

36 (for x = 6);
144 (for x = 12);
324 (for x = 18);
576 (for x = 24);
900 (for x = 30);
1296 (for x = 36);
1764 (for x = 42);
2304 (for x = 48);
2916 (for x = 54);
3600 (for x = 60);
...

As you can see for some values, x is NOT a multiple of 15 but for some values x IS a multiple of 15 (for example, for x = 30 or x = 60)
_________________
Manager
Joined: 26 Feb 2018
Posts: 59
WE: Sales (Internet and New Media)

### Show Tags

02 Jun 2018, 11:20
Bunuel wrote:
loserunderachiever wrote:
Bunuel wrote:
If $$x$$ is a positive integer, is $$x$$ divisible by 15?

(1) $$x$$ is a multiple of 10

(2) $$x^2$$ is a multiple of 12

Hi Bunuel ,

I got a wrong answer , chose B ,as this a Yes / No question , to find whether X is divisible by 15 or not (y/n)

statement 2 : says , X^2 is a multiple of 12 , so in this case the only possibility is 6^2 = 36 which is a multiple of 12 , hence it is NO , this means X should also has to a perfect square , then only number left out is 12 itself , (12*12) i.e 144, in this case also 15 is not divisible .

Why are we considering x = 12 *15 as a possibility , where it is mentioned X^2 is a multiple of 12 .

Can you please guide me ?

Positive multiples of 12 are 12, 2*12 = 24, 3*12 = 36, 4*12 = 48, ....

(2) says that $$x^2$$ is a multiple of 12. Since x^2 is a perfect square, then we are looking for multiples of 12 which are also perfect squares. So, x^2 could be:

36 (for x = 6);
144 (for x = 12);
324 (for x = 18);
576 (for x = 24);
900 (for x = 30);
1296 (for x = 36);
1764 (for x = 42);
2304 (for x = 48);
2916 (for x = 54);
3600 (for x = 60);
...

As you can see for some values, x is NOT a multiple of 15 but for some values x IS a multiple of 15 (for example, for x = 30 or x = 60)

Oh , this clears up . I stopped my multiple list till 12*12 = 144 . Your explanation absolutely makes sense , I realised my mistake now . 30 & 60 , makes sense.

Faced this one in my GMAT club quant test. Now it's clear to me.

Thanks a lot Bunuel.
_________________

" Can't stop learning and failing"

Re: M05-22 &nbs [#permalink] 02 Jun 2018, 11:20
Display posts from previous: Sort by

# M05-22

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.