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If x and y are integers, is x/3 an integer?
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Updated on: 02 Oct 2013, 23:28
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If x and y are integers, is x/3 an integer? (1) x + 14 = 2y (2) (2y+1)/3 is an integer.
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Originally posted by avohden on 02 Oct 2013, 15:54.
Last edited by Bunuel on 02 Oct 2013, 23:28, edited 1 time in total.
Edited the question.




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Re: If x and y are integers, is x/3 an integer?
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02 Oct 2013, 16:02
Official ExplanationCORRECT ANSWER IS C. The question is really asking whether or not x is divisible by 3. The stem doesn't offer much more information, so move on to the statements. Statement (1): insufficient. Solving for x, we have that x = 2y – 14. We thus know that x is divisible by 2 (since y is an integer), but we don't know whether x is also divisible by 3. Statement (1) is insufficient. Eliminate (A) and (D). Statement (2): insufficient. This equation does not contain x, so it can't tell us whether x is divisible by 3. Eliminate (B). Statements (1) and (2): sufficient. Both statements contain 2y, so use substitution to solve. Substituting x + 14 for 2y in the second expression, we have that is an integer. This simplifies to , and since is an integer, must be an integer. The answer to the question must be "yes," and choice (C) is correct.




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Re: If x and y are integers, is x/3 an integer?
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02 Oct 2013, 20:24
Thanks avohden! Good explanation.
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Re: If x and y are integers, is x/3 an integer?
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26 Nov 2013, 12:59
Given: X & Y are Integers Find: Is X/3=Integer ?
1) Not sufficient.Eliminate A&D 2) Not Sufficient. Eliminate B. 1+2) x=2y14  1 x=2y+115  Add +1 & 1 in RHS Therefore, 15 is divisible by 3 and 2y+1 is divisible by 3. Hence ,C



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Re: If x and y are integers, is x/3 an integer?
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26 Nov 2013, 21:33
avohden wrote: If x and y are integers, is x/3 an integer?
(1) x + 14 = 2y (2) (2y+1)/3 is an integer. From St 1, we have x= 2y14 and x/3 = (2y14)/3 Now if y= 7 then we have x/3 as 0/3 =0, which is an integer But if y=8 then we have x/3= 2/3 not an Integer so A and D ruled out From St 2, we know nothing about X and hence B also ruled out Combining the 2 equations we get x= 2y14 or x = (2y+1) 15 x/3 = (2y+1)/3  15/3> Now (2y+1)/3 is an integer and also 15/3 is an integer and therefore x/3 is an integer Hence E ruled. Ans C
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Re: If x and y are integers, is x/3 an integer?
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06 Jul 2017, 16:23
avohden wrote: If x and y are integers, is x/3 an integer?
(1) x + 14 = 2y (2) (2y+1)/3 is an integer. We are given that x and y are integers and need to determine whether x/3 is an integer. Statement One Alone: x + 14 = 2y The information in statement one is not sufficient to answer the question. For example, if x = 4 (and y = 9), x is not divisible by 3; however if x = 6 (and y = 10), then x is divisible by 3. Statement one alone is not sufficient to answer the question. Statement Two Alone: (2y+1)/3 is an integer. We see that y can be values such as 1, 4, 7, etc. Thus, we see that y = 3k  2, where k is an integer. However, since we do not know anything about x, statement two alone is not sufficient to answer the question. Statements One and Two Together: We can substitute 3k  2 for y in statement one and we have: x + 14 = 2(3k  2) x + 14 = 6k  4 x = 6k  18 x = 6(k  3) Since x is a multiple of 6, x/3 will always be an integer. Answer: C
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Re: If x and y are integers, is x/3 an integer?
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07 Jul 2017, 22:09
avohden wrote: If x and y are integers, is x/3 an integer?
(1) x + 14 = 2y (2) (2y+1)/3 is an integer. 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. There are 2 variables and 0 equation. By Variable Approach method, the answer is C most likely. The question asks if \(x\) is a multiple of \(3\). Condition 1) : \(x + 14 = 2y\) is equivalent to \(x = 2y  14\). If \(y = 0\), then \(x = 14\) is not a multiple of \(3\). If \(y = 7\), then \(x = 0\) is a multiple of \(3\). Thus the solutions are not unique. Hence this condition is not sufficient. Condition 2) : We don't have any information about \(x\). Thus this is not sufficient. Condition 1) & 2) The second condition that \(\frac{2y + 1}{3}\) is an integer is equivalent to the statement that \(2y + 1\) is a multiple of \(3\). If \(2y + 1\) is a multiple of \(3\), then \(2y\) has a remainder \(2\) when it is divided by \(3\), that is \(2y = 3k + 2\) for some an integer \(k\). Then \(x = 2y  14 = 3k + 2  14 = 3k  12 = 3(k4)\). Thus \(x\) is a multiple of \(3\). Therefore, C is the answer as expected. Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If x and y are integers, is x/3 an integer?
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04 Jan 2018, 01:30
this question is a common pattern in gmat. The question asks about the value of x / 3, not the value of x; hence, C is the answer even though the st 2 gives no value about x.



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Re: If x and y are integers, is x/3 an integer?
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19 Oct 2018, 17:02
The question is really asking whether or not x is divisible by 3. The stem doesn't offer much more information, so move on to the statements. Statement (1): insufficient. Solving for x, we have that x = 2y  14. We thus know that x is divisible by 2 (since y is an integer), but we don't know whether x is also divisible by 3. Statement (1) is insufficient. Eliminate (A) and (D). Statement (2): insufficient. This equation does not contain x, so it can't tell us whether x is divisible by 3. Eliminate (B). Statements (1) and (2): sufficient. Both statements contain 2y, so use substitution to solve. Substituting x + 14 for 2y in the second expression, we have that is an integer. This simplifies to (x + 15)/3 , and since (x/3) + 5 is an integer, (x/3 ) must be an integer. The answer to the question must be "yes," and choice (C) is correct. Posted from my mobile device
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