If x and y are integers and 2

Question

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.

mikemcgarry · Accepted Answer

Dear accincognito , I'm happy to help. :-)

This is a hard problem! See this post for some insight: https://magoosh.com/gmat/2012/gmat-math-factors/ Statement #1 : The GCF of x and y is 2 This leave open a wide array of possibilities. All we know is that x and y are two even numbers, both bigger than 2, with no common factors other than two: they could be x = 4, y = 6 x = 6, y = 8 x = 6, y = 10 x = 6, y = 16 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient . Statement #2 : The LCM of x and y is 48 Without any other information, we could have x = 3, y = 16 x = 4, y = 48 So, it's possible for y to equal 16 or equal something else. This statement, alone and by itself, does not give us sufficient information, so it is insufficient . Combined : this is where it gets interesting. The GCF of x and y is 2 The LCM of x and y is 48 This is a tricky combination. First, let's list all the factors of 48 --- in order to have a LCM of 48 with another number, each number must be a factor of 48. factors of 48 = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} Those are the possible candidates for x & y. We can eliminate 1 & 2, because x > 2, and we can eliminate 3, because that cannot have a GCF of 2 with anything else. Possibilities for x & y = {4, 6, 8, 12, 16, 24, 48} If y = 48, then every other number in the set is factor of 48, so the GCF would be the smaller number --- e.g. the GCF of 6 and 48 is 6. Therefore, we can't use 48. If y = 24, then the first four numbers are factors of 24, so they don't work, and the GCF of 16 & 24 is 8. Therefore, we can't use 24. Possibilities for x & y = {4, 6, 8, 12, 16} Suppose y = 16 x = 4, y = 16 ===> GCF = 4, doesn't work x = 6, y = 16 ===> GCF = 2 --- this is one possible pair!! x = 8, y = 16 ===> GCF = 8, doesn't work x = 12, y = 16 ===> GCF = 4, doesn't work Suppose y = 8 x = 4, y = 8 ===> GCF = 4, doesn't work x = 6, y = 8 ===> GCF = 2, but LCM = 24, doesn't work Suppose y = 6 x = 4, y = 6 ===> GCF = 2, but LCM = 12, doesn't work So, after all that, the only pair that satisfies both statements is x = 6, y = 16, so it turns out, y does in fact equal 16. We are able to give a definitive answer to the prompt question, so the combined statements are sufficient . Answer = (C) Does all this make sense? Mike :-)

KarishmaB · Answer

Given: 2 < X < Y
Question: Does Y = 16?

(1) The GCF of X and Y is 2.
X = 2a; Y = 2b (a and b are co prime integers)
Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 6, Y = 8 etc

(2) The LCM of X and Y is 48.
 48 = 2^4 * 3 
One of X and Y must have 2^4 = 16 as a factor and one must have 3 as a factor. Again, Y may or may not be 16 e.g. X = 6, Y = 16 OR X = 1, Y = 48 etc

Using both together, X = 2a, Y = 2b.
Since a and b need to be co-prime and both X and Y need to be greater than 2, one of a and b must be 3 and the other must be 8 (since X and Y already have a 2 to make 16).
X = 6, Y = 16 (since X is less than Y).
Y must be 16.

Answer (C)

suk1234 · Answer

If GCF is 2 and LCM is 48, than the Pair of X and Y can be:
 X=2,Y=48 and X=6,Y=16 and X=16, Y=6 and X=48, Y=2

I think the answer is  E

Bunuel · Answer

Please read the stem carefully. It's given that 2<x<y.

HKHR · Answer

I am not getting the right answer for some reason. Can you please tell me the set of numbers for which both the statements are satisfied?

thank you.

Bunuel · Answer

x = 6 and y = 16 --> the greatest common factor of 6 and 16 is 2, and the least common multiple of 6 and 16 is 48.

jlgdr · Answer

Again @Mike, No need for all the fuzz, let's see x,y are positive integers and 2

MathRevolution · Answer

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.

If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.

In the original condition, there are 2 variables(x,y) and 1 equation(2<x<y), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1equation, which is likely to make D the answer.
For 1), (x,y)=(4,6) -> no, (x,y)=(6,16) -> yes, which is not sufficient.
For 2), (x,y)=(3,48) -> no, (x,y)=(6,16) -> yes, which is not sufficient.
When 1) & 2), (x,y)=(6,16) -> yes, which is sufficient. Therefore, the answer is C.

 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.

grassmonkey · Answer

When I see GCD, I think, "the most they have in common." When I see LCM, I think, "the least you need to make all of them." Helps me, maybe it can help someone else.

Statement 1: Insufficient, lots of number greater than 2 have a GCD of 2: 8 and 6 for instance or 16 and 14.

Statement 2: LCM of x and y is 48, which is (2^5)*3. This is also going to be insufficient. You want to think of different combinations of those prime factors. For instance, we can have x = 3 and y = 32. Can we have anything else (remember the only stipulation is that x<y)? x = 6 and y = 16. These give us two different answers.

Together:

There are only two examples from the second statement that are possible: x = 6, y = 16 and x = 3, y = 32. Because 3 and 32 don't have a common factor of 2, we can exclude that case and we are left with x = 6, y = 16 and the answer is 'yes'.

Sufficient --> C

fskilnik · Answer

2 < x < y\,\,{	ext{ints}}

y\,\,\mathop = \limits^? \,\,16 = {2^4}

\left( 1 ight)\,\,GCF\left( {x,y} ight) = 2\,\,\,\left\{ \begin{gathered}
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {{2^2},2 \cdot 3} ight)\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{No}}} ightangle \hfill \
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {2 \cdot {{3,2}^4}} ight)\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{Yes}}} ightangle \hfill \ 
\end{gathered} ight.

\left( 2 ight)\,\,LCM\left( {x,y} ight) = {2^4} \cdot 3\,\,\,\left\{ \begin{gathered}
 \,\left( {\operatorname{Re} } ight){	ext{Take}}\,\,\left( {x,y} ight) = \left( {2 \cdot {{3,2}^4}} ight)\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{Yes}}} ightangle \hfill \
 \,{	ext{Take}}\,\,\left( {x,y} ight) = \left( {{2^2}{{,2}^4} \cdot 3} ight)\,\,\, \Rightarrow \,\,\,\left\langle {{	ext{No}}} ightangle \hfill \ 
\end{gathered} ight.

\left( {1 + 2} ight)\,\,\,\left\{ \begin{gathered}
 \,\left( 1 ight) \cap \left( 2 ight)\,\,\,\, \Rightarrow \,\,\,\,xy = GCF\left( {x,y} ight) \cdot LCM\left( {x,y} ight) = 3 \cdot {2^5}\,\,\,\left( * ight) \hfill \
 \,\left( 1 ight)\,\, \Rightarrow \left\{ \begin{gathered}
 x = 2 \cdot M \hfill \
 y = 2 \cdot N \hfill \ 
\end{gathered} ight.\,\,\,\,\,GCF\left( {M,N} ight) = 1\,\,,\,\,\left( * ight)\,\, \Rightarrow \,\,3 \cdot {2^3} = MN\,\,\,\,\left( {M,N \geqslant 2} ight) \hfill \
 \left( {M,N} ight) = \left( {{2^3},3} ight)\,\,\, \Rightarrow \,\,\,\left( {x,y} ight) = \left( {{2^4},2 \cdot 3} ight)\,\,\, \Rightarrow \,\,\,x > y\,\,{	ext{impossible}} \hfill \
 	herefore \left( {M,N} ight) = \left( {{{3,2}^3}} ight)\,\,\, \Rightarrow \,\,\,\left( {x,y} ight) = \left( {2 \cdot {{3,2}^4}} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{	ext{Yes}}} ightangle \hfill \ 
\end{gathered} ight.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

fskilnik · Answer

Excellent,  grassmonkey . This is essentially what is going on. (I prefer the "satisfy" expression - I mentioned it in red- , but the meaning is the same.)

You got a kudos of mine for this observation!

Congrats and success in your studies,
Fabio.

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If x and y are integers and 2 < x < y, does y = 16 ?

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If x and y are integers and 2 < x < y, does y = 16 ?

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