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If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 02:04
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If y is an integer, is y^3 divisible by 9? (1) y is divisible by 4. (2) y is divisible by 6. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifyisanintegerisy3divisibleby127796.html
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 02:45
Yes this can be done by plugging numbers. (1) Take multiples of 4 and 9 and the answer is yes, but say for numbers that are only multiples of 4, the answer is No (2) is sufficient because y is divisible by 6 means that y has 3 as factor, hence y^3 will have 3^3 as factor => y^3 is divisible by 9. Answer  B
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 03:09
siddhans wrote: Can someone please explain in detail? Can this be done using random/smart numbers?
If y is an integer, is y^3 divisible by 9?
(1) y is divisible by 4. (2) y is divisible by 6. I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem. Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading In this question the best strategy is to understand the logic; to be divisible by 9 .. u need atleast two 3's st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's. hence sufficient.



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 03:18
sudhir18n wrote: siddhans wrote: Can someone please explain in detail? Can this be done using random/smart numbers?
If y is an integer, is y^3 divisible by 9?
(1) y is divisible by 4. (2) y is divisible by 6. I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem. Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading In this question the best strategy is to understand the logic; to be divisible by 9 .. u need atleast two 3's st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's. hence sufficient. The question says divisible by 6... can you write divisible by 6 to be a multiple of 6? Also, in 1 you have written: st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's. You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ... In st2 you have written : St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's. How can y^3 have atleast three 2's???



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 03:42
siddhans wrote: sudhir18n wrote: siddhans wrote: Can someone please explain in detail? Can this be done using random/smart numbers?
If y is an integer, is y^3 divisible by 9?
(1) y is divisible by 4. (2) y is divisible by 6. I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem. Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading In this question the best strategy is to understand the logic; to be divisible by 9 .. u need atleast two 3's st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's. hence sufficient. The question says divisible by 6... can you write divisible by 6 to be a multiple of 6? Also, in 1 you have written: st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's. You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ... In st2 you have written : St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's. How can y^3 have atleast three 2's??? What I meant in st:1 regarding 4, was some multiples of 4 will give you 3. not all ... st2: since 6= 2*3 ... any multiple of 6 will have atleast one 2 and atleast one 3..test numbers here 6,12,24,36,42..( all will have atleast one 2 and one 3) thus if you do a cube of these numbers .. you are bound to have three 2's and three 3's test numbers again: 12*12*12 = 2^2*3 * 2^2*3 * 2^2*3 = 2^6*3^3 ... same is the case with any multple. any number which is divisible by 6 .. is a multiple of 6.



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 08:18
siddhans wrote: Can someone please explain in detail? Can this be done using random/smart numbers?
If y is an integer, is y^3 divisible by 9?
(1) y is divisible by 4. (2) y is divisible by 6. Taking 1, y = 4x y^3 = (4^3)(x^3) = 64(x^3) can't say whether divisible by 9 or not Taking 2, y = 6x y^3 = (6^3)(x^3) = (8*27)(x^3) so divisible by 9 Hence B.



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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08 Jul 2011, 20:43
1. Not sufficient
if y =4 then y^3 is not divisible by 9. if y = 12 then y^3 is divisible by 9.
2. Sufficient
3 is a factor of 6.
so 3 must be there in all the multiples of 6.
so y^3 is divisible by 9 .
Answer is B.



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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16 Jul 2011, 01:34
[quote="sudhir18n"
I clearly see your interest in smart numbers. Though my first instinct is also to pick smart numbers ( blame MGMAt for that:) ), but u also need to understand the undelying logic in each problem. Smart number help a lot , especially in VIC problem, but sometimes its difficult to pick a smart number and sometimes smart numbers are misleading
In this question the best strategy is to understand the logic; to be divisible by 9 .. u need atleast two 3's
st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.............henc insufficent
St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's.
hence sufficient.[/quote]
The question says divisible by 6... can you write divisible by 6 to be a multiple of 6?
Also, in 1 you have written:
st1: multiple of 4 .. can give u atleast one 3 ..hence Y^3 will have three 3's but its not always the case.Ex: 8 , this will not have any 3's.
You said it gives you atleast one 3 ....I dont think so its always the case...only in some cases it gives you 3...like lets says 4*3 = 12 ...
In st2 you have written :
St2: Y a multiple of 6.. any multiple of 6 will atleast have one 2 and atleast one 3. thus Y^3 at minimum will have atleast three 2's and three 3's.
How can y^3 have atleast three 2's???[/quote]
What I meant in st:1 regarding 4, was some multiples of 4 will give you 3. not all ...
st2: since 6= 2*3 ... any multiple of 6 will have atleast one 2 and atleast one 3..test numbers here
6,12,24,36,42..( all will have atleast one 2 and one 3) thus if you do a cube of these numbers .. you are bound to have three 2's and three 3's test numbers again: 12*12*12 = 2^2*3 * 2^2*3 * 2^2*3 = 2^6*3^3 ... same is the case with any multple.
any number which is divisible by 6 .. is a multiple of 6.[/quote]
Thanks!



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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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20 Aug 2013, 06:26
Can someone clarify if my approach is correct for this problem.. ( initial part) we need to know if\(Y^3 = 9 I\) We can further solve this as \(Y^3 = 3^2 * I\) Now given that Y is an integer there must be an additional 3 present for Y to be an integer so \(Y^3 = 3^2 * 3 * I\) when we further solve this we get \(\frac{Y}{3} = Integer\) Statement 1 says \(\frac{Y}{4} = Integer\) Y could be 4, Y could be 36 insufficient Statement 2 says \(\frac{Y}{6} = Integer\) So the values 6,12,18 etc all divisible by 3 sufficient! Answer is B
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Re: If y is an integer, is y^3 divisible by 9? [#permalink]
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20 Aug 2013, 06:31
fozzzy wrote: Can someone clarify if my approach is correct for this problem.. ( initial part)
we need to know if\(Y^3 = 9 I\)
We can further solve this as \(Y^3 = 3^2 * I\) Now given that Y is an integer there must be an additional 3 present for Y to be an integer
so \(Y^3 = 3^2 * 3 * I\) when we further solve this we get \(\frac{Y}{3} = Integer\)
Statement 1 says \(\frac{Y}{4} = Integer\) Y could be 4, Y could be 36 insufficient
Statement 2 says \(\frac{Y}{6} = Integer\) So the values 6,12,18 etc all divisible by 3 sufficient!
Answer is B Yes, the question basically asks whether y is divisible by 3. If y is an integer is y^3 divisible by 9?(1) y is divisible by 4. Clearly insufficient: y can be 4 or 9*4. (2) y is divisible by 6 > since y is divisible by 6 then it's divisible by 3 too: y=3k. So, y^3 will definitely have more than two 3's and thus will be divisible by 9: y^3=(3k)^3=9*(3k^3). Sufficient. Answer: B. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifyisanintegerisy3divisibleby127796.html
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