If n is an integer, is n(n+2) divisible by 8?

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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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by Brent@GMATPrepNow » Wed Sep 19, 2018 5:57 am

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Max@Math Revolution 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.
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 8
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: n is a multiple of 4
This 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 8
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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Brent
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by fskilnik@GMATH » Wed Sep 19, 2018 5:50 pm

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Max@Math Revolution 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.
$$n\,\,{\mathop{\rm int}} $$
$${{n\left( {n + 2} \right)} \over 8}\,\,\mathop = \limits^? \,\,{\mathop{\rm int}} $$

$$\left( 1 \right)\,\,n = 2M,\,\,M\,\,{\mathop{\rm int}} $$
$${{n\left( {n + 2} \right)} \over 8} = {{2M \cdot 2 \cdot \left( {M + 1} \right)} \over 8} = {{M\left( {M + 1} \right)} \over 2}\mathop = \limits^{\left( * \right)} \,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

(*) 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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by Max@Math Revolution » Fri Sep 21, 2018 12:15 am

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=>
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.