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

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Re: If x and y are integers and 2 < x < y, does y = 16 ?  [#permalink]

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New post 18 Aug 2017, 17:10
1
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'.

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If x and y are integers and 2 < x < y, does y = 16 ?  [#permalink]

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New post 26 Nov 2018, 18:51
Archit143 wrote:
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.

\(2 < x < y\,\,{\text{ints}}\)

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


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


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


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


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

Regards,
Fabio.
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Re: If x and y are integers and 2 < x < y, does y = 16 ?  [#permalink]

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New post 26 Nov 2018, 18:58
grassmonkey wrote:
When I see GCD, I think, "the most they have in common." When I see LCM, I think, "the least you need to make [satisfy] all of them." Helps me, maybe it can help someone else.

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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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)
Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount!

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Re: If x and y are integers and 2 < x < y, does y = 16 ? &nbs [#permalink] 26 Nov 2018, 18:58

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