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If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 02:01
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[ Math Revolution GMAT math practice question] If \(n\) is an integer, is \(n(n+2)\) divisible by \(8\)? 1) \(n\) is an even number. 2) \(n\) is a multiple of \(4\).
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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 03:06
From statement 1: n is even. The least even number is zero. n=0 n(n+2) = 0. Which is divisible by 8. 1 is sufficient. From statement 2: n is a multiple of 4. The least value of n is 4. n=4 n(n+2) = 8. Which is divisible by 8. 2 is also sufficient. D is the answer.
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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 06:58
MathRevolution wrote: [ Math Revolution GMAT math practice question] If \(n\) is an integer, is \(n(n+2)\) divisible by \(8\)? 1) \(n\) is an even number. 2) \(n\) is a multiple of \(4\). Target question: Is n(n+2) divisible by 8? Statement 1: n is an even number. Let's examine some CONSECUTIVE even numbers: 6, 8, 10, 12, 14, 16, 18, 20, 22, etc Notice that the numbers increase by 2 with each subsequent value. So, if n is EVEN, then we know that n+2 is also even. Also, notice that for any pair of CONSECUTIVE even integers, one value is divisible by 2 and the other is divisible by 4. That is, for any pair of CONSECUTIVE even integers, one value can be written as (2)(some integer) and the other value can be written as (4)(some integer)In other words, we can write: n(n+2) = (2)(some integer)(4)(some integer)= ( 8)(some integer) This means, n(n+2) IS divisible by 8Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: n is a multiple of 4This tells is that n is even. So, n and n+2 are CONSECUTIVE even numbers This means we can apply the exact same logic we applied with statement 1 and conclude that n(n+2) IS divisible by 8Since we can answer the target question with certainty, statement 2 is SUFFICIENT Answer: D Cheers, Brent
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If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 07:12
Zero is also a multiple of 4. No? Afc0892Afc0892 wrote: From statement 1:
n is even. The least even number is zero. n=0 n(n+2) = 0. Which is divisible by 8. 1 is sufficient.
From statement 2:
n is a multiple of 4. The least value of n is 4. n=4 n(n+2) = 8. Which is divisible by 8. 2 is also sufficient.
D is the answer.



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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 09:45
I did not understand. What if n=2, then n(n+2) becomes 0, which is not divisible by 8. Am I missing something here? Thanks.
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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 09:48
bharu130 Zero is divisible b 8 since zero is a multiple of all numbers. Hope this helps. Cheers! bharu130 wrote: I did not understand. What if n=2, then n(n+2) becomes 0, which is not divisible by 8. Am I missing something here? Thanks.
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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 18:20
funsogu wrote: Zero is also a multiple of 4. No? Afc0892Afc0892 wrote: From statement 1:
n is even. The least even number is zero. n=0 n(n+2) = 0. Which is divisible by 8. 1 is sufficient.
From statement 2:
n is a multiple of 4. The least value of n is 4. n=4 n(n+2) = 8. Which is divisible by 8. 2 is also sufficient.
D is the answer. Yes, zero is a multiple of every number. Posted from my mobile device
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Re: If n is an integer, is n(n+2) divisible by 8?
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19 Sep 2018, 18:49
MathRevolution wrote: If \(n\) is an integer, is \(n(n+2)\) divisible by \(8\)?
1) \(n\) is an even number. 2) \(n\) is a multiple of \(4\).
\(n\,\,{\mathop{\rm int}}\) \({{n\left( {n + 2} \right)} \over 8}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}}\) \(\left( 1 \right)\,\,n = 2M,\,\,M\,\,{\mathop{\rm int}}\) \({{n\left( {n + 2} \right)} \over 8} = {{2M \cdot 2 \cdot \left( {M + 1} \right)} \over 8} = {{M\left( {M + 1} \right)} \over 2}\mathop = \limits^{\left( * \right)} \,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\) (*) The product of two consecutive integers is even (odd*even or even*odd), hence the integer M(M+1) has (at least) one factor 2 in its decomposition... \(\left( 2 \right)\,\,n = 4J\,\,,\,\,J\,\,{\mathop{\rm int}}\) \({{n\left( {n + 2} \right)} \over 8} = {{4J \cdot 2 \cdot \left( {2J + 1} \right)} \over 8} = J \cdot \left( {2J + 1} \right) = {\mathop{\rm int}} \cdot {\mathop{\rm int}} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\) This solution follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: If n is an integer, is n(n+2) divisible by 8?
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21 Sep 2018, 01:22
=> 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 \(1\) variable (n) and \(0\) equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first. Condition 1) If \(n\) is an even integer, then \(n\) and \(n + 2\) are two consecutive even integers. Products of two consecutive even integers are multiples of \(8\) since one of them must be a multiple of \(4\), and the other a multiple of \(2\). Condition 1) is sufficient. Condition 2) Since \(n\) is a multiple of \(4, n + 2\) is an even integer. Thus, \(n(n+2)\) is a multiple of \(8\). Condition 2) is sufficient. Therefore, D is the answer. Answer: D If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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