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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

If k is a positive integer and n=(k-1)k(k+1), is n a multiple of 8?

1) k is an odd number
2) k = 1
Every other even integer is a MULTIPLE OF 4:
2, 4, 6, 8, 10, 12...
Thus, the product of two consecutive even integers = product of a multiple of 4 and an even non-multiple of 4 = MULTIPLE OF 8:
2*4 = 8
4*6 = 2*2*2*3 = 8*3
10*12 = 2*5*2*2*3 = 8*15

Statement 1:
Since k is odd, k-1 and k+1 constitute two consecutive even integers, with the result that their product is a multiple of 8.
Thus, n is a multiple of 8, and the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Since k is odd, we can apply the same reasoning applied to Statement 1.
Thus, the answer to the question stem is YES.

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

If k is a positive integer and n=(k-1)k(k+1), is n a multiple of 8?

1) k is an odd number
2) k = 1
$$k \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\left( * \right)$$
$${{\left( {k - 1} \right)k\left( {k + 1} \right)} \over {{2^3}}}\,\,\mathop = \limits^? \,\,{\rm{int}}$$

$$\left( 1 \right)\,\,k\,\, = 2M + 1\,\,\,\left( {M\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} } \right)$$
$$\left( {k - 1} \right)k\left( {k + 1} \right) = \left( {2M} \right)\left( {2M + 1} \right)\left( {2M + 2} \right) = 2M\left( {2M + 1} \right)2\left( {M + 1} \right) = {2^2}\left( {2M + 1} \right)\underbrace {M\left( {M + 1} \right)}_{{\rm{even}}\,\,{\rm{ = }}\,\,{\rm{2}}J,\,\,J\,\,{\mathop{\rm int}} }$$
$$?\,\,\,:\,\,\,{{{2^3} \cdot \left( {2M + 1} \right) \cdot J} \over {{2^3}}} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,\,k = 1\,\,\,\, \Rightarrow \,\,\,\,k - 1 = 0\,\,\,\, \Rightarrow \,\,\,\,{{\left( {k - 1} \right)k\left( {k + 1} \right)} \over {{2^3}}} = 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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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 2 variables (n and k) and 1 equation, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since k is an odd number, k - 1 and k + 1 are consecutive even integers.
Any product of consecutive even integers is a multiple of 8.
Thus, condition 1) is sufficient.

Condition 2)
Since k = 1, we have n = (k-1)k(k+1) = 0*1*2 = 0. 0 is a multiple of any number, so n = 0 is a multiple of 8.
Thus, condition 2) is sufficient.

Since this question is a CMT4(B) question. Condition 2) is easy to understand and condition 1) is hard. When one condition is easy to understand, and the other is hard, D is most likely to be the answer.

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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