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n is a positive integer. Is n(n+1)(n+2)/4 an even integer?

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n is a positive integer. Is n(n+1)(n+2)/4 an even integer?

by Max@Math Revolution » Thu Dec 20, 2018 11:27 pm

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[Math Revolution GMAT math practice question]

n is a positive integer. Is n(n+1)(n+2)/4 an even integer?

1) n is an even integer
2) 1238 ≤ n ≤ 1240
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Source: — Data Sufficiency |
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by deloitte247 » Sun Dec 23, 2018 8:48 am

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$$check\ if\ \ \frac{n\left(n+1\right)\left(n+2\right)}{4}$$
n (n+1) (n+2) is the product of three consecutive integers.
The product of three consecutive integers is always divisible by 2 and 3 because the product of k consecutive integers is always divisible by k!

Statement 1
n is an even integer.
if n is an even integer, then n (n+1) (n+2) will be product of three consecutive integers that will be divisible by the multiple of 2
if n = 2

$$\frac{2\left(2+1\right)\left(2+2\right)}{4}$$
$$\frac{2\cdot3\cdot4}{4}=\frac{24}{4}=6$$
Hence, n (n+1) (n+2) will always be even if n is an even integer ;
Statement 1 is INSUFFICIENT.

Statement 2
$$1238\le n\le1240$$
$$hence,\ n\le1239$$
$$1239\ is\ a\ multiple\ of\ \ 3,\ hence\ n\left(n+1\right)\left(n+2\right)\ is\ divisible\ by\ 4$$
$$\frac{1239\left(1239+1\right)\left(1239+2\right)}{4}$$
$$\frac{1239\left(1240\right)\left(1241\right)}{4}$$
$$\frac{1906622760}{4}$$
$$476655690\ which\ is\ an\ even\ integer\ $$
statement 2 is INSUFFICIENT.

$$answer\ is\ Option\ D$$
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by Max@Math Revolution » Sun Dec 23, 2018 6:06 pm

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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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Asking for n(n+1)(n+2)/4 to be an even integer is equivalent to asking for n(n+1)(n+2) to be a multiple of 8. If n is an even integer, n and n+2 are consecutive even integers and a product of two consecutive even integers is a multiple of 8. Thus, condition 1) is sufficient.

Condition 2)
If n = 1238, n(n+1)(n+2)=1238*1239*1240 is a multiple of 8 since 1240 is a multiple of 8.
If n = 1239, n(n+1)(n+2)=1239*1240*1241 is a multiple of 8 since 1240 is a multiple of 8.
If n = 1240, n(n+1)(n+2)=1240*1241*1242 is a multiple of 8 since 1240 is a multiple of 8.
Thus, condition 2) is sufficient.

Therefore, the answer is D.
Answer: D

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