Using "test" it for consecutive integer question

[jaiyeolab]

I know how to solve the below using math. I was however curious to know whether it is possible to solve this question or something similar by testing values? I am training myself to try to solve most questions that include variables, by testing values.

Is positive integer n - 1 a multiple of 3?

(1) n^2 - n is a multiple of 3

(2) n^3 + 2n^2+ n is a multiple of 3

[Brent@GMATPrepNow]

Is positive integer n-1 a multiple of 3?

(1) n³ - n is a multiple of 3
(2) n³ + 2n² + n is a multiple of 3

Target question: Is positive integer n-1 a multiple of 3?

Statement 1: n³ - n is a multiple of 3
Factor: n³ - n = n(n² - 1) = n(n-1)(n+1) = (n-1)(n)(n+1)
Notice that n-1, n and n+1 are three CONSECUTIVE INTEGERS.
IMPORTANT: Statement 1 is simply telling us that the product of 3 consecutive integers is divisible by 3. This is not new information. The product of ANY 3 consecutive integers will always be divisible by 3. In fact, there's a rule that says, "The product of n consecutive integers is divisible by n, n-1, n-2, . . . 2, 1"
Since statement 1 is just some rule that already exists in mathematics, we already knew this information BEFORE we even examined statement 1. So, there's no way that statement 1 could possibly add any information to help us answer the target question.
As such, statement 1 is NOT SUFFICIENT

Statement 2: n³ + 2n²+ n is a multiple of 3
Factor: n³ + 2n² + n = n(n² + 2n + 1) = n(n+1)(n+1)
This means that EITHER n is a multiple of 3 OR n+1 is a multiple of 3.
Let's examine each possible case:
case a: If n is a multiple of 3, then we can find other multiples of 3 by adding or subtracting multiples of 3 to n. So, for example, n+3 and n+6 will be also be multiples of 3. Likewise, n-3 and n-6 will be also be multiples of 3. Since n-1 is just 1 less than n, n-1 cannot be a multiple of 3 .
case b: If n+1 is a multiple of 3, then n-1 cannot be a multiple of 3 , Since n-1 is just 2 less than n+1.
Since both possible cases yielded the same answer to the target question, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent

[GMATGuruNY]

Is positive integer n-1 a multiple of 3?

(1) n³ - n is a multiple of 3
(2) n³ + 2n² + n is a multiple of 3

Statement 1:
n³ - n = n(n² - 1) = (n)(n+1)(n-1).

Case 1: n=2, with the result that (n)(n+1)(n-1) = 2*3*1.
In this case, n-1 = 2-1 = 1, which is NOT a multiple of 3.

Case 2: n=3, with the result that (n)(n+1)(n-1) = 3*4*2.
In this case, n-1 = 3-1 = 2, which is NOT a multiple of 3.

Case 3: n=4, with the result that (n)(n+1)(n-1) = 4*5*3.
In this case, n-1 = 4-1 = 3, which IS a multiple of 3.

Since n-1 is a multiple of 3 in Case 3 but not a multiple of 3 in Cases 1 and 2, INSUFFICIENT.

Statement 2:
n³ + 2n² + n = n(n² + 2n + 1) = (n)(n+1)(n+1).

Case 1: n=2, with the result that (n)(n+1)(n+1) = 2*3*3.
In this case, n-1 = 2-1 = 1, which is NOT a multiple of 3.

Case 2: n=3, with the result that (n)(n+1)(n+1) = 3*4*4.
In this case, n-1 = 3-1 = 2, which is NOT a multiple of 3.

n=4 implies that (n)(n+1)(n+1) = 4*5*5.
Since n=4 does not satisfy the constraint that (n)(n+1)(n+1) is a multiple of 3, n=4 is not a valid option in Statement 2.

Case 3: n=5, with the result that (n)(n+1)(n+1) = 5*6*6.
In this case, n-1 = 5-1 = 4, which is NOT a multiple of 3.

Case 4: n=6, with the result that (n)(n+1)(n+1) = 6*7*7.
In this case, n-1 = 6-1 = 5, which is NOT a multiple of 3.

In every case, n-1 is NOT a multiple of 3.
SUFFICIENT.

The correct answer is B.

[Brent@GMATPrepNow]

I know how to solve the below using math. I was however curious to know whether it is possible to solve this question or something similar by testing values? I am training myself to try to solve most questions that include variables, by testing values.

Is positive integer n - 1 a multiple of 3?

(1) n^2 - n is a multiple of 3

(2) n^3 + 2n^2+ n is a multiple of 3

Testing values is tricky for this question, plus you have the potential to choose bad numbers and draw an incorrect conclusion. See my article for more on this: https://www.gmatprepnow.com/articles/dat ... lug-values

Cheers,
Brent

[jaiyeolab]

You guys are awesome. Thank you so much for the quick responses. I am going to review your article, Brent; the question I have right now is how to determine when testing values would take me longer. I will check your article to see if that question is answered and then post back here if I still have questions. Thanks to you both for the responses!

[ceilidh.erickson]

Here's my general rule: if you can see and understand the underlying concept, that will likely be more effective that testing numbers. You should use number testing if you can't identify any particular rule or concept at play.

In this case, for example, recognizing that the question is testing products of consecutive integers would be much faster and easier than testing numbers. You can quickly ID everything you know about the product of 3 consecutives: it will always be even and it will always be a multiple of 3, etc.

Testing numbers is most effective on topics like digit placement (if the digits in a certain 2-digit number were reversed...) in which there is no universal rule to apply. On topics such as divisibility, odds/evens, or positives/negatives, it's best to learn the underlying concepts well, then commit the rules to memory.

[Max@Math Revolution]

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 equations ensures a solution.



=> For the remaining questions it is best to substitute the values directly. In the original condition, there is only 1 variable and we only need 1 equation. Thus it is likely that the answer is D.

1) n(n-1)=3*int ==> n=3 no, n=4 yes, NOT sufficient
2) n(n+1)^2=3*int ==> n=2,3,5,6,8,9.......... all the cases are "NO", sufficient

Therefore the best is B


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