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m05 #22

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Re: m05 #22 [#permalink] New post 02 Aug 2012, 06:07
Bunuel wrote:
thevenus wrote:
Bunuel, please throw some light here,
(c) is right as per many of us but why not (E)?


Is x divisible by 15?

(1) When x is divided by 10, the result is an integer --> \frac{x}{10}=integer --> x=10*integer. Now, if x=0 (in case integer=0), then the answer is YES but if x=10 (in case integer=1), then the answer is NO. Not sufficient.

From this statement though we can deduce that x is an integer (since x=10*integer=integer).

(2) x^2 is a multiple of 30 --> if x=0, then the answer is YES but if x=\sqrt{30}, then the answer is NO. Not sufficient.

(1)+(2) Since from (1) x=integer then x^2=integer, and in order x^2 to be divisible by 30=2*3*5, x must be divisible by 30 (x must be a multiple of 2, 3 and 5, else how can this primes appear in x^2?), hence x is divisible by 15 too. Sufficient.

Notice that x can be positive, negative or even zero, but in any case it'll be divisible by 30.

Answer: C.

Next, every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only). So, we edited this question and in the new GMAT Club tests this question reads:

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 --> since x is an integer, then x^2 is a perfect square. 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.

Hope it's clear.


Thanks a lot, Kudos for you +1 :-D
Wasn't the previous one was tougher? why did you changed / modified ? Now the GMAT can't put such an option (of choosing irrational no. at least if not negative numbers?)
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Re: m05 #22 [#permalink] New post 02 Aug 2012, 06:10
Expert's post
thevenus wrote:
Thanks a lot, Kudos for you +1 :-D
Wasn't the previous one was tougher? why did you changed / modified ? Now the GMAT can't put such an option (of choosing irrational no. at least if not negative numbers?)


Again, every GMAT divisibility question will tell you in advance that any unknowns represent positive integers, so the stem should have mentioned that x is a positive integer.
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Re: m05 #22 [#permalink] New post 02 Aug 2012, 10:25
Bunuel wrote:
Next, every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only). So, we edited this question and in the new GMAT Club tests this question reads:

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 --> since x is an integer, then x^2 is a perfect square. 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.

Hope it's clear.


Just thought I'd take a crack at the expanding on option B. I was left with my head scratching for a while when I read your explanation of choice B, but I worked it out (I hope so!)

x^2 is a multiple of 12.

Now 12 = 3*2*2

Therefore x^2 = 3*2*2*m

m must at least be a 3. Why? Because the prime numbers need to repeat at least once since we are multiplying the number by itself when we square it.

e.g.
6=3*2
6^2=3*2*3*2

Therefore, in order for x to multiply by itself, we need m to equal at least a 3. In which case x^2=36 and x=6. In which case \frac{6}{15} does not work!

However, if m=15, then the statement is sufficient. That's why B by itself is insufficient.

The correct answer of the modified question is therefore C.
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Re: m05 #22 [#permalink] New post 02 Aug 2012, 16:47
I went with (B) and realized my mistake. I should have plugged in more numbers so as to eliminate B.
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Re: m05 #22 [#permalink] New post 02 Aug 2013, 04:37
Well, its C.

I always end up messing with the premise is X an integer. :(

If X is an integer, the answer is B.
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Re: m05 #22 [#permalink] New post 03 Aug 2013, 03:45
it went for me in this way :

For x to be divisible my 15 - should have atleast one 3 and atleast one 5.

Statement 1 - stated x/10 is an integer, which means x is properly divisible by ten, whose factors are 2 and 5. Not sufficient alone. - gave me a 5

Statement 2 - stated x/30 is an integer hence by same logic factor of 30 are 2*3*5. But this has a twist that as 15 is smaller than 30 so 30 cannot divide 15 with 0 remainder.

Hence (1)+(2) - guides me to following conclusions :
1. We have ample number of factors to qualify.
2. It is smaller 30 and greater than 10.


Correct me, if I am wrong. :?:
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Re: m05 #22 [#permalink] New post 11 Aug 2013, 01:10
Expert's post
BELOW IS REVISED VERSION OF THIS QUESTION:

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] New post 18 Jul 2014, 07:56
Question: Is X divisible by 15?

S1: When X is divided by 10, the result is integer.
X = 20 => No to the question.
X = 30 => Yes to the question.
So, S1 is not sufficient. Eliminate A and D.

S2: X^2 is a multiple of 30.
X^2 = 60 => X = Square root of 60 => No to the question.
X^2 = 900 => X = 30 => Yes to the question.
So, S2 is not sufficient. Eliminate B.

Both S1 and S2:
If X is divisible by 10 then X^2 must at least be divisible by 100.
Additionally X^2 is multiple of 30.

Thus applying both S1 and S2, X^2 must be divisible by 300.
X^2 = 300 => X = Square root of 300 => No to the question.
X^2 = 900 => X = 30 => Yes to the question.
Therefore, Both S1 and S2 are also not sufficient.

Eliminate C. Correct answer is E.
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Re: m05 #22 [#permalink] New post 18 Jul 2014, 08:16
Let’s look at a different scenario: Let’s say the question includes a constraint that X is an integer and suppose that the two given statements remain unchanged.


Modified Question: Given that X is an integer, is X divisible by 15?


S1: When X is divided by 10, the result is integer.
X = 20 => No to the modified question.
X = 30 => Yes to the modified question.
So, S1 is not sufficient. Eliminate A and D.

S2: X^2 is a multiple of 30.
Per modified question if X is an integer, X^2 must be a perfect square that is a multiple of 30.
In such case the data such as X^2 = 30, 60, 90, 120, 180, and others which are not perfect square become invalid.

So, we are left with data such as X^2 = 900, 3600, 8100 which are the perfect square and are multiple of 30.
X^2 = 900 => X = 30 => Yes to the modified question.
X^2 = 3600 => X = 60 => Yes to the modified question.
X^2 = 8100 => X = 90 => Yes to the modified question.

So, S2 is sufficient. Correct answer is B for the above stated modified question.
Re: m05 #22   [#permalink] 18 Jul 2014, 08:16
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