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# M05-22

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Math Expert
Joined: 02 Sep 2009
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M05-22  [#permalink]

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15 Sep 2014, 23:25
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45% (medium)

Question Stats:

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

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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

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Joined: 02 Sep 2009
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M05-22  [#permalink]

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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.

Answer: C
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Re: M05-22  [#permalink]

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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.

Answer: C

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

Please suggest if there are any familiar types your aware of.

Would be really grateful.
Math Expert
Joined: 02 Sep 2009
Posts: 50578
Re: M05-22  [#permalink]

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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.

Answer: C

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

Please suggest if there are any familiar types your aware of.

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.
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Joined: 06 Mar 2014
Posts: 245
Location: India
GMAT Date: 04-30-2015
Re: M05-22  [#permalink]

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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.

Answer: C

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

Please suggest if there are any familiar types your aware of.

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
M05-22  [#permalink]

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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
Re: M05-22  [#permalink]

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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?
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Joined: 11 Jan 2015
Posts: 34
Re: M05-22  [#permalink]

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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
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Re: M05-22  [#permalink]

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28 Jun 2016, 23:29
paddy41 wrote:
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.
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M05-22  [#permalink]

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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)?
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Posts: 34
Re: M05-22  [#permalink]

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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
Re: M05-22  [#permalink]

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07 May 2017, 01:03
paddy41 wrote:
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.
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Re: M05-22  [#permalink]

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13 Oct 2017, 03:57
how sqrt(180) gives a integer??? we cant take 180 has a sol..
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Posts: 50578
Re: M05-22  [#permalink]

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13 Oct 2017, 05:37
Bhadri1199 wrote:
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.
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Re: M05-22  [#permalink]

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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 ?
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Re: M05-22  [#permalink]

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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)
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Re: M05-22  [#permalink]

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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.
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Re: M05-22 &nbs [#permalink] 02 Jun 2018, 11:20
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# M05-22

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