If n is a positive integer, is n 1 divisible by 3?

Question

If n is a positive integer, is n – 1 divisible by 3?

(1) n^2 + n is not divisible by 6.
(2) 3n = k + 3, where k is a positive multiple of 3.

chetan2u · Accepted Answer

1) n^2 + n =n(n+1) 
Now, n and n+1 are consecutive integers, so one of them is surely even. 
Therefore,  if n(n+1) is divisible by 2 but not by 6, it means n(n+1)is not div by 3. 
n(n+1) not divisible by 3 means that the consecutive numbers n and n+1 are not div by 3.
As we have to have one multiple of 3 in every set of 3 consecutive integers, n-1 must be div by3.
Sufficient

2)3n = k + 3
3n-3=k
3(n-1)=k...
we are given k is a multiple of 3, which can be seen from the result but n-1 can take any value.
Insufficient

A

ManojReddy · Answer

If n is a positive integer, is n – 1 divisible by 3?

(1) n^2 + n is not divisible by 6.

n^2+n = n(n+1)

Lets assume n-1 divisible by 3, n = 4,7,10,13,16 .......

For all the values of n, n(n+1) is not divisible by 6.
So our assumption n-1 is divisible by 3 is  true 
 Sufficient.

(2) 3n = k + 3, where k is a positive multiple of 3.

Since K is a positive multiple of 3, the above equation can be rewritten as 
3n = 3K+3, where k =1,2,3,4,5,6,7.......
3n= 3(K+1)
n=K+1
n-1 = k

So Clearly n-1 is divisible by 3.

Sufficient

Ans:D

-Manoj Reddy

shriramvelamuri · Answer

I Think answer is A.

Option 1

N^2+ N not divisible by 6 => N can be 1, 4, 7 and 10.

From the above N-1= 0, 3, 6, 9 (All are divisible by 3) Sufficient.

Option 2

3N= K+3 => K can be 3, 6, 9 and 12 etc.

Then N can be equal to 2, 3, 4, 5 Respectively.

Of the above numbers, N-1 => 1 and 2 are not divisible by 3 but 4-1 is divisible by 3. Hence not sufficient.

Option A.

chetan2u · Answer

Hi,
you have gone wrong by cancelling out 3 from both sides..
 3n = k + 3
3n-3=k
3(n-1)=k...
we are given k is a multiple of 3, which can be seen from the result but n-1 can take any value..
hope it is clear

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: An accurate yes/no rephrase is the following: Is (n - 1)/2 = integer? Is n – 1 = 3 × integer? Is n = 3 × integer + 1? Is the positive integer n one greater than a multiple of 3? This narrows down our values of interest to a certain type of number, which follows a pattern: 1, 4, 7, 10, etc. A value question such as “What is n?” would be unnecessarily picky. We don't need to know the exact value, just whether n is a certain type of number. (1) SUFFICIENT: If n^2 + n = n(n + 1) is not divisible by 6, we can rule out certain values for n. https://gmatclub.com/forum/download/file.php?id=27034 The correct yes/no rephrasing directs us to look for a certain pattern. We see that pattern here: n can only be 1, 4, 7, 10, etc., all integers that are one greater than a multiple of 3. (2) INSUFFICIENT: If 3n = 3 × pos integer + 3, then n = pos integer + 1. Therefore, n is an integer such that n ≥ 2. This does not resolve whether n is definitely one greater than a multiple of 3. The correct answer is A. 2015-06-22_1714.png

mvictor · Answer

I answered A, but now don't seem so confident...

1. n(n+1) not divisible by 6. or not divisible by 2 nor 3.
we have consecutive numbers. we can have 1x2, 2x3, 3x4, 4x5,
or we can have 7x8, 10x11..
where n=7 or 10.
7-1=6 and is divisible by 3.
10-1=9, and is divisible by 3.

2. 3n=k+3, where k is multiple of 3.
if k=3, then n=2 -> n-1 not divisible by 3.
if k=9, then n=4 -> n-1 divisible by 3.
2 different answers, so can't be B.

p.s. thanks to the above comments..
I indeed remembered that
x(x+1)(x-1) is always divisible by 3.
if x(x+1) is not divisible by 3, then x-1 MUST be divisible by 3.

MathRevolution · Answer

Forget conventional ways of solving math questions. 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 independent equations ensures a solution.

If n is a positive integer, is n – 1 divisible by 3?

(1) n^2 + n is not divisible by 6.
(2) 3n = k + 3, where k is a positive multiple of 3.

In the original condition, there is 1 variable(n), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), since n(n+1) is not multiple of 6, it becomes an integer of n=3m+1 like n=1,4,7,10..... Thus, it becomes n-1=3m, which is yes and sufficient.
For 2), 3n=3m+3 -> n=m+1 -> n-1=m. Thus, m=2 -> no and m=3 -> yes, which is not sufficient.
Therefore, the answer is A.

 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

stonecold · Answer

Note => On out of every three consecutive integers is divisible by 3 
so statement is sufficient

Aamirso · Answer

Statement 1: If n(n+1) is not divisible by 6, neither n nor n+1 is a multiple of 3 (note that one of n and n+1 has to be even, so it means neither is a multiple of n). One of every 3 consecutive integers is a multiple of 3. Hence, if n and n+1 are not multiple of 3, n-1 is. Sufficient.

Statement 2: In other words, 3n = 3p + 3 where p is a positive integer. So, n=p+1. So n-1 may or may not be a multiple of 3, depending on the value of p. Insufficient.

A is my answer.

Prakruti_Patil · Answer

so simply explained -- thankyou so much!

If n is a positive integer, is n 1 divisible by 3?

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