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If x, y, z are integers, is xyz a multiple of 6?
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21 Sep 2018, 01:23
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[ Math Revolution GMAT math practice question] If \(x, y, z\) are integers, is \(xyz\) a multiple of \(6\)? 1) \(x+y+z\) is a multiple of \(6\) 2) \(x, y\), and \(z\) are consecutive integers
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Re: If x, y, z are integers, is xyz a multiple of 6?
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21 Sep 2018, 01:39
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(x, y, z\) are integers, is \(xyz\) a multiple of \(6\)? 1) \(x+y+z\) is a multiple of \(6\) 2) \(x, y\), and \(z\) are consecutive integers Question: is xyz multiple of 6 ?
Statement 1: x + y + z is multiple of 6. Not always.
Suppose,
x = 1 y=2 z=3
1 + 2+ 3 = 6. and 3*2*1 = 6.
case 2:
x = 10 y=1 z = 1
x + y+ z = 10 + 1 + 1 = 12 and 10*1*1= 10, which is not a multiple of 6.
Thus statement 1 is NOT sufficient.
Statement 2: x , y , z are consecutive integers.
the product of 3 consecutive integers is a multiple of 6. all time. Sufficient.
1*2*3 6
2*3*4 = 24
3*4*5 = 60
So , this statement is SUFFICIENT .
The best answer is B.



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Re: If x, y, z are integers, is xyz a multiple of 6?
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21 Sep 2018, 23:36
I thought consecutive integers include negatives too. In that case 1, 0 and 1 are also a set of three consecutive integers. And 1*0*1 is not a multiple of 6. Product of any three "positive consecutive integers" are. In absence of the term "positive", B should not be the OA.
OA should be C. Please correct if my understanding is wrong.



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Re: If x, y, z are integers, is xyz a multiple of 6?
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22 Sep 2018, 03:14
[quote="urvashis09"]I thought consecutive integers include negatives too. In that case 1, 0 and 1 are also a set of three consecutive integers. And 1*0*1 is not a multiple of 6. Product of any three "positive consecutive integers" are. In absence of the term "positive", B should not be the OA.
OA should be C.
0 is a multiple of every other integer. I your example 1*0*1 is 0. And 0 is a multiple of 6. So OA is B. Hope its clear.



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Re: If x, y, z are integers, is xyz a multiple of 6?
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22 Sep 2018, 03:38
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(x, y, z\) are integers, is \(xyz\) a multiple of \(6\)? 1) \(x+y+z\) is a multiple of \(6\) 2) \(x, y\), and \(z\) are consecutive integers Statement I: Consider, x,y,z,= 1,1,8. Not a multiple of 6. Statement II: As its a consecutive number, it will be divisible by 6.
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Re: If x, y, z are integers, is xyz a multiple of 6?
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22 Sep 2018, 05:50
rahul16singh28 wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(x, y, z\) are integers, is \(xyz\) a multiple of \(6\)? 1) \(x+y+z\) is a multiple of \(6\) 2) \(x, y\), and \(z\) are consecutive integers Statement I: Consider, x,y,z,= 1,1,8. Not a multiple of 6. Statement II: As its a consecutive number, it will be divisible by 6. Oh, correct! I missed that part. Thank you for pointing the flaw.



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Re: If x, y, z are integers, is xyz a multiple of 6?
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24 Sep 2018, 05:21
=> Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution. Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Since x and y are two consecutive integers, one of them, x or y is an even integer, and xy is a multiple of 2. Since x, y and z are three consecutive integers, one of them, x, y or z is a multiple of 3, and xyz is a multiple of 3. Therefore, xyz is a multiple of 6. Both conditions (together) are sufficient. Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B. Condition 1) If x = 1, y = 2 and z = 3, then xyz = 6 is not a multiple of 6. If x = 2, y = 2 and z = 2, then xyz = 8 is a multiple of 8. Since we do not have a unique answer, condition 1) is not sufficient. Condition 2) Using the same argument as the one given for both conditions together, condition 2) is sufficient. Therefore, B is the answer. Answer: B In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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