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M05-22

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:25
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62% (00:53) correct 38% (01:07) wrong based on 195 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
[Reveal] Spoiler: OA

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16 Sep 2014, 00:25
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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.

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27 Sep 2014, 13:20
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: 44290

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27 Sep 2014, 13:45
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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.
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27 Sep 2014, 13:56
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.
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Joined: 17 May 2015
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17 Sep 2015, 08:24
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
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Posts: 44290

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18 Sep 2015, 09:17
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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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28 Jun 2016, 20:17
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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: 44290

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29 Jun 2016, 00:29
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.
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Joined: 01 Nov 2016
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06 May 2017, 21:46
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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Joined: 11 Jan 2015
Posts: 28

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07 May 2017, 01:16
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: 44290

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07 May 2017, 02:03
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.
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13 Oct 2017, 04:57
how sqrt(180) gives a integer??? we cant take 180 has a sol..
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Posts: 44290

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13 Oct 2017, 06:37
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.
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Re: M05-22   [#permalink] 13 Oct 2017, 06:37
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