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# Is the positive integer y a multiple of 12?

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Is the positive integer y a multiple of 12?  [#permalink]

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04 Jan 2014, 18:20
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Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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28 Nov 2016, 22:34
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Nightfury14 wrote:
GMATBeast wrote:
Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

Statement (1) -
Factors of 48 = 2*2*2*2*3
=> $$y^3$$= $$2^3$$*2*3*k ----- ($$y^3$$ is a multiple of 48 and k is the quotient when $$\frac{y^3}{48}$$)
Hence = $$2^2*3^2$$ to complete the statement.
Question - why can't k be = $$2^2*3^2*x^2$$ (where x is any integer)

Statement (2) -
Factors of 30 = 2*3*5
=> $$y^2$$ = 2*3*5*p ----- ($$y^2$$ is a multiple of 30 and p is the quotient when $$\frac{y^2}{30}$$)
Hence can be p = 2*3*5*$$x^2$$ to make the statement sufficient.

I understand that i am adding an extra prime factor of x in Statement - (1) & (2) - I am missing some simple logic.
Bunuel - kindly review and clarify.

Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48 --> $$y^3 = 2^4*3*k$$ --> the least value of k for which $$2^4*3*k$$ is a perfect cube is $$2^2*3^2$$, therefore the least value of y is $$\sqrt[3]{2^4*3*2^2*3^2}=12$$. Thus y must be divisible by 12. Sufficient.

(2) y^2 is a multiple of 30 --> $$y^2=2*3*5*m$$ --> the least value of m for which $$2*3*5*m$$ is a perfect square is $$2*3*5$$, therefore the least value of y is $$\sqrt{2*3*5*2*3*5}=30$$. Thus y may or may not be a multiple of 12. Not sufficient.

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Re: Is the positive integer y a multiple of 12?  [#permalink]

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16 Oct 2015, 18:35
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Quote:
Is the positive integer y a multiple of 12?

(1) $$y^3$$ is a multiple of 48.

(2) $$y^2$$ is a multiple of 30.

(Statement 1) $$y^3 = 48*q$$, where q is an integer greater than or equal to 0. --> $$y^3 = 2^4*3*q.$$ Given that y is an integer, a value of q should be chosen such that all the exponents of the right hand side are multiples of 3 and it should be the smallest value that accomplishes this, so that if you were to take the cubic root of $$y^3$$, you will be left with an integer. So $$q = 2^2*3^2*n^3$$, where n is an integer greater than or equal to 0.

Now you have $$y^3 = 2^6*3^3*n^3$$. To solve for y, take the cubic root of $$y^3$$ --> y = $$2^2*3*n$$ Therefore y is a multiple of 12. Statement 1 is sufficient

(Statement 2) $$y^2 = 30*q$$, where q is an integer greater than or equal to 0. --> $$y^2 = 2*3*5*q.$$ Given that y is an integer, a value of q should be chosen such that all the exponents of the right hand side are multiples of 2 and it should be the smallest value that accomplishes this, so that if you were to take the square root of $$y^2$$, you will be left with an integer. So $$q = 2*3*5*n^2$$, where n is an integer greater than or equal to 0.

Now you have $$y^2 = 2^2*3^2*5^2*n^2$$. To solve for y, take the square root of $$y^2$$ --> y = $$2*3*5*n$$ Therefore y is a multiple of 30. A multiple of 12 requires at least two 2's and one 3 at all times, but a multiple of 30 won't always have at least two 2's.

Don't analyze statement 1 and statement 2 the way I did, unless you are given that something is an integer. For example if y wasn't an integer, then statement 1 would have been insufficient because y could take the value $$y = 48^\frac{1}{3}$$
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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05 Jan 2014, 03:09
Ans A

Y^3=Y(Y²)=12(3K) K is a positive factor

Therefore Y is divided by 12.

Hope it helps
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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05 Jan 2014, 04:46
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GMATBeast wrote:
Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

Statement I is sufficient:

y^3 = (2^4)(3)(k)

Since y is an integer minimum value of k will be 2^2 x 3^2 hence minimum value of y will be cube root of it which is 2^3(3) = 24 hence it is definitely a multiple of 12

Statement II is insufficient
y^2 = 5 x 3 x 2k

Since y is an integer the minimum value of k is 5 x 3 x 2 hence minimum value for y is 30 which is not a multiple of 12

It can also hold a value which is a multiple of 12 as well since k can further increase to be any square of an integer.

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Re: Is the positive integer y a multiple of 12?  [#permalink]

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11 Jul 2015, 06:45
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GMATBeast wrote:
Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

Per statement 1, $$y^3 = 48p$$, p>0 integer ----> as y is an integer, $$p = 2^2*3^3*q^3$$. Thus y=12q ----> This statement is sufficient.

Per statement 2, $$y^2 = 30p$$, p>0 integer ----> as y is an integer, $$p = 2^2*3^2*5^2*q^3$$. Thus y=30q ----> Thus y = 30, 60, 90, $$120$$... Thus this statement is not sufficient.

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Is the positive integer y a multiple of 12?  [#permalink]

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16 Oct 2015, 14:40
Is the positive integer y a multiple of 12?

(1) y3 is a multiple of 48.

(2) y2 is a multiple of 30.

---------------------------
Its from a Kaplan CAT I just took, below is kaplans explanation.
I just cant understand how the get "So we know that y3 has at least four 2’s and at least one three among its prime factors" Can someone also point to a good source so I can get around this topic? I understand the basic primes and multiples properties but the exponent really toughed up the problem for me.
thanks.

Analyze the Question Stem:

If y were a multiple of 12, would have to equal an integer. That means that all the factors in 12 could be cancelled out by factors in y.

In other words asking whether y is a multiple of 12 is the same ask asking whether y has at least 22 and 31 as prime factors.

Since this is a Yes/No question, we need to know whether y definitely does or definitely does not have those factors.

Evaluate the Statements:

Statement (1): For y3 to be a multiple of 48, it must be that yields an integer. Let’s find the prime factors of 48.

48 = 8 × 6

48 = 23× 2 × 3

48 = 24 x 3

So we know that y3 has at least four 2’s and at least one three among its prime factors.

So our answer is definitely "yes," and Statement (1) is Sufficient. Eliminate choices (B), (C), and (E).

Statement (2): For y2 to be a multiple of 30, it must be that yields an integer. Again, let’s go through prime factorization.

30 = 6 × 5

30 = 2 × 3 × 5

So, y2 has at least one factor of 2, at least one factor of 3, and at least one factor of 5.

Just as y3 had to have all of its factors in groups of threes, y2 must have its factors in groups of twos. So we can deduce that y2 has at least two factors of 2, at least two factors of 3, and at least two factors of 5.

y will have the same factors as y2 has, but half as many of each. So, y must have at least one factor of 2, at least one factor of 3, and at least one factor of 5.

So, y definitely has one factor of 3. But it might have only one factor of 2, or it might have two factors of 2.

Statement (2) is Insufficient. Eliminate choice (D). Answer Choice (A) is correct.

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Re: Is the positive integer y a multiple of 12?  [#permalink]

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17 Oct 2015, 21:22
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

If we modify the original condition and the question, we ultimately want to know whether y=12int=(2^2)3int. It is important here that y is a positive integer. There is only 1 variable (y), so in order to match the number of variables and that of equations, we need one equation, but since 2 are given from the 2 conditions, there is high chance (D) will be our answer.
From condition 1, y^3=48n(n is an integer)= (2^3)(2)(3)n, and y becomes cube root[(2^3)(2)(3)n]=2cube root(6n). As y is told to be a positive integer, the cube root should be eliminated. Then, from y=2cube root(6n)=2cube root(6*6^2m^3)where m is some integer, the answer to the question becomes 'yes' from y=2*6*m=12m, so this condition is sufficient.
But from condition 2, we can see that the answer becomes 'no' when y=30^2, whereas it is 'yes' for y=(30^2)(2^2). This condition is insufficient, and the answer therefore becomes (A).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Status: My heart can feel, my brain can grasp, I'm indomitable. Affiliations: Educator Joined: 16 Oct 2012 Posts: 39 Location: Bangladesh WE: Social Work (Education) Re: Is the positive integer y a multiple of 12? [#permalink] ### Show Tags 17 Oct 2015, 23:08 1)..y^3=48k=2.2.2.2.3…=2^3.2^3.3^… y=12k …suff 2)y^2=30k=2.3.5…=4.9.25k not necessarily 12k.. inssuff Ans: A _________________ please press "+1 Kudos" if useful Retired Moderator Joined: 18 Sep 2014 Posts: 1117 Location: India Is the positive integer y a multiple of 12? [#permalink] ### Show Tags 20 Nov 2015, 22:18 2 Is the positive integer y a multiple of 12? (1) $$y^3$$ is a multiple of 48 (2) $$y^2$$ is a multiple of 30 we want to know whether $$y=12x$$ where x is an integer. From condition 1, $$y^3$$=48a (a is an integer) Therefore $$y= \sqrt[3]{48a}$$ =$$y= \sqrt[3]{8*6*a}$$=$$\sqrt[3]{(2^3)*6*a}$$=$$2*\sqrt[3]{6*a}$$ As y is told to be a positive integer, the cube root should be eliminated. Therefore a needs to have a value of $$6*(b^2)$$ where b is an integer. hence $$y=2*6*m=12m$$ thereby provides us the required answer. So B/E/C are ruled out. we have to evaluate statement 2 for D. From condition 2, $$y^2=30p$$ (p is an integer) so $$y=\sqrt{30p}=\sqrt{2*3*5*p}$$ As y is told to be a positive integer, the square root should be eliminated. Therefore p needs to have a value of $$2*3*5*(q^2)$$ where b is an integer. hence $$y=2*3*5*q=30q$$ This condition is insufficient, and the answer therefore becomes (A) ruling out D. I took 115s for the same in exam and still got this question wrong.(First qn) Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6645 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is the positive integer y a multiple of 12? [#permalink] ### Show Tags 22 Nov 2015, 01:32 1 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is the positive integer y a multiple of 12? (1) y^3 is a multiple of 48 (2) y^2 is a multiple of 30 There is one variable (y) and 2 equations are given by the 2 conditions, so there is high chance (D) will be our answer. For condition 1, y^3=48t(t=positive integer), y=cube root (48t)=cube root (2^3*6t)=12s, where s is a positive integer. This is a 'yes' and the condition is sufficient. For condition 2, it is a 'yes' for y=60, but 'no' for y=30, so the answer becomes (A). This is a recent type of question. For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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22 Sep 2016, 11:47
To be honest I really do not understand it completely. Can someone answer me the following question concerning statement 1:

If y^3 is a multiple of 48*t (t some integer). Then the prime factorization gives: (2^4*3) * t. Sooo how is that exactly sufficient? I mean it is for y^3 and not for y right? Isn't it possible that if you take the cube root of y^3 the outcome is different then with y^3?
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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28 Nov 2016, 09:43
GMATBeast wrote:
Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

Statement (1) -
Factors of 48 = 2*2*2*2*3
=> $$y^3$$= $$2^3$$*2*3*k ----- ($$y^3$$ is a multiple of 48 and k is the quotient when $$\frac{y^3}{48}$$)
Hence = $$2^2*3^2$$ to complete the statement.
Question - why can't k be = $$2^2*3^2*x^2$$ (where x is any integer)

Statement (2) -
Factors of 30 = 2*3*5
=> $$y^2$$ = 2*3*5*p ----- ($$y^2$$ is a multiple of 30 and p is the quotient when $$\frac{y^2}{30}$$)
Hence can be p = 2*3*5*$$x^2$$ to make the statement sufficient.

I understand that i am adding an extra prime factor of x in Statement - (1) & (2) - I am missing some simple logic.
Bunuel - kindly review and clarify.
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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23 Feb 2017, 05:15
1
Prompt analysis
Y is a positive integer

Superset
The answer will be either yes or no

Translation
In order to find the answer, we need:
1# exact value of y
2# factors of y
3# some other information to predict the value of factors of y

Statement analysis

St 1: 48 = 2^4 x3 . so y^3 = 2^4 x 3. y^3 can be minimum 2^6 x 3^3. There fore y can be minimum 12. ANSWER

St 2: 30 = 2 x 3 x 5. y^2 can be minimum ( 2 x 3 x 5 )^2. Hance y can be minimum 30. Not a factor. INSUFFICIENT.

Option A
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Re: Is the positive integer y a multiple of 12?  [#permalink]

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13 Dec 2017, 15:48
GMATBeast wrote:
Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

We need to determine whether y/12 is an integer.

Statement One Alone:

y^3 is a multiple of 48.

Since y^3/48 = integer, we can say that the product of 48 and some integer n is equal to a perfect cube. In other words, 48n = y^3.

We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So let’s break down 48 into primes to help determine what extra prime factors we need to make 48n a perfect cube.

48 = 16 x 3 = 2^4 x 3^1

In order to make 48n a perfect cube, we need two more 2s and two more 3s. Thus, the smallest perfect cube that is a multiple of 48 is 2^6 x 3^3.

To determine the least possible value of y, we can take the cube root of 2^6 x 3^3 and we have:

2^2 x 3 = 12

Thus, the minimum value of y is 12, so y/12 is an integer. Statement one alone is sufficient to answer the question.

Statement Two Alone:

y^2 is a multiple of 30.

Since y^2/30 = integer, we can say that the product of 30 and some integer m is equal to a perfect square. In other words, 30m = y^2.

We must remember that all perfect squares break down to unique prime factors, each of which has an exponent that is a multiple of 2. So let’s break down 30 into primes to help determine what extra prime factors we need to make 30m a perfect square.

30 = 5 x 3 x 2

In order to make 30m a perfect square, we need one more 2, one more 3, and one more 5. Thus, the smallest perfect square that is a multiple of 30 is 2^2 x 3^2 x 5^2.

To determine the least possible value of y, we can take the square root of 2^2 x 3^2 x 5^2 and we have:

2 x 3 x 5 = 30

Thus, the minimum value of y is 30; however if y is 30, then y/12 is not an integer; and if y is 60, then y/12 is an integer. Statement two alone is not sufficient to answer the question.

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