If n is a positive integer greater than 6, what is the remai

Q:If n is a positive integer greater than 6, what is the remainder when n is divided by 6?

(1) n2 – 1 is not divisible by 3.

(2) n2 – 1 is even.

Ans is C and here is how

St 1 tells us that n^2-1 is not divisible by 3 and we know n>6 so possible values of n =9,12,15,18......
Now if n =12,18,24... then n/6 gives remainder as 0 but if n=9,15,21 then remainder is 3

So from St1 we see that remainder will follow a pattern of 0,1,0,1....

Not sufficient

from St 2 we we have n^2-1=even or n^2= Even +1 =odd, so we see that n is odd.
Now if n=7 then n/6 gives remainder 1
but if n=9 then remainder is 3
if n=11 then remainder is 5

So St2 is not sufficient alone

Combining we see that n is odd and n>6 so possible value of n satisfying the above conditions are n=9,15,21,27.....so on
Remainder will be 3

Ans is C

N is a positive integer > 6, what is the remainder when N is divided by 6?

(1) N^2 – 1 is not divisible by 3.

--> N^2 - 1 = (N-1)*(N+1) , and this is not divisible by three.

Notice that (N-1)*N*(N+1) is the product of three consecutive positive integers. Such a product will always contain exactly one multiple of 3. If this multiple of three is not (N-1) or (N+1), then the multiple of 3 must be N.

If N = 3*3=9, the remainder when divided by 6 is 3.
If N = 3*4=12, the remainder when divided by 6 is 0.
--> Not sufficient.

(Note: The product of K consecutive positive integers is always divisible by K!)

(2) N^2 – 1 is even

--> N^2 = even + 1 = odd --> N is odd.

If N = 7, the remainder when divided by 6 is 1.
If N = 9, the remainder when divided by 6 is 3.
--> Not sufficient.


(1) and (2)
N is an odd multiple of 3 greater than 6.
--> N = 3*(2k-1) = 6k - 3 for k >1.

Thus, since N is three less than a multiple of six, the remainder is always 3. Sufficient.

Thanks for the explanation. I can definitely follow what you have outlined above but I wouldn't have been able to derive it on my own.

Two questions:
1) Is it better to plug in numbers for these type of problems or work through the theory?
2) How do I get better at these specific types of problems? Is it just practice and if so, can you suggest similar problems please?

Thanks!

1. Questions on remainders are good for plug-in, but not all of them. It depends on a question and your skills/preferences what approach to choose.

2. By studying theory and practicing:


Hope this helps.

Here is my take!

Statement 1: n^2 - 1 is not divisible by 3 --> (n-1)(n+1) is not divisible by 3 ---> n is divisible by 3 --> insuff

Statement 2: n^2 - 1 is odd --> (n-1)(n+1) is odd ---> n -1 is even ---> n is odd ---> insuff

Statement 1+2: n is divisible by 3 ---> n = 3a

n is even ---> n = 2b +1

So, n = 6c +3 ---> remainder when n is divided by 6 is 3

Hence, C!

Goal: We need to know the remainder when n is divided by 6.

Statement 1: (N^2-1)/3 =int. So (n+1)(n-1)/3 is not an integer. Thus, n must be a multiple of three because every three consecutive integers has at least one multiple of three within it. Not sufficient because all it tells us is that n is a multiple of 3, so its remainder when divided by 6 has multiple potential values.

Statement 2: N^2 - 1 is even. Hence (n+1)*(n-1) is even, so n must be odd. Not sufficient because some odd numbers have different remainders when divided by 6.

Combined: We know that N is an integer greater than 6 that is odd and a multiple of 3. Start testing cases: n=9 (r=3); n=15 (r=3); and so on. Sufficient.

The problem can be solved algebraically or using smart numbers (testing cases). Both methods are shown below.

(1) NOT SUFFICIENT:
Algebra
The expression n^\(2\) – 1 = (n – 1)(n + 1). Together with n, the three expressions represent three consecutive integers: n – 1, n, n + 1. Statement 2 indicates that the product (n – 1)(n + 1) is not divisible by the prime number 3; therefore, neither n – 1 nor n + 1 is a multiple of 3.

Since every third integer is a multiple of 3, it follows that n must be a multiple of 3. This fact alone is not sufficient, though, since different multiples of 3 can give different remainders upon division by 6. For example, 9/6 gives a remainder of 3, and 12/6 gives a remainder of 0.

Testing Cases
Find values of n > 6 that satisfy statement 1.
The value n = 7 does not satisfy the statement, since 72 – 1 = 48 is divisible by 3.
The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is divisible by 3.
The value n = 9 satisfies the statement, since 92 – 1 = 80 is not divisible by 3. The remainder of 9/6 is 3.
The value n = 10 does not satisfy the statement, since 102 – 1 = 99 is divisible by 3.
The value n = 11 does not satisfy the statement, since 112 – 1 = 120 is divisible by 3.
The value n = 12 satisfies the statement, since 122 – 1 = 143 is not divisible by 3. The remainder of 12/6 is 0.

At least two different remainders are possible, so the statement is insufficient.

(2) NOT SUFFICIENT:
Algebra
If n\(2\) – 1 is even, then \(n2\) must be odd, so n itself is odd. This fact alone is not sufficient, though, since different odd numbers can give different remainders upon division by 6. For example, 7/6 gives a remainder of 1, and 9/6 gives a remainder of 3.

Testing Cases
Find values of n > 6 that satisfy the statement.
The value n = 7 satisfies the statement, since 72 – 1 = 48 is even. The remainder of 7/6 is 1.
The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is not even.
The value n = 9 satisfies the statement, since 92 – 1 = 80 is even. The remainder of 9/6 is 3.
At least two different remainders are possible, so the statement is insufficient.

(1) AND (2) SUFFICIENT: Statement (1) indicates that n is a multiple of 3 and statement (2) indicates that n is odd. From this point, you can either use algebra or test numbers.
Algebra
According to the two statements, n must be 3 times some odd integer, so it can be written as n = 3(2k + 1), where k is an integer. Distributing gives n = 6k + 3, which is 3 more than a multiple of 6. Therefore, the remainder upon dividing n by six must be 3.

Testing Cases
Test numbers that are multiples of 3, odd, and greater than 6:
If n = 9, then 9 / 6 = 1 remainder 3.
If n = 15, then 15 / 6 = 2 remainder 3.
If n = 21, then 21 / 6 = 3 remainder 3

The problem can be solved algebraically or using smart numbers (testing cases). Both methods are shown below.

(1) NOT SUFFICIENT:
Algebra
The expression n2 – 1 = (n – 1)(n + 1). Together with n, the three expressions represent three consecutive integers: n – 1, n, n + 1. Statement 2 indicates that the product (n – 1)(n + 1) is not divisible by the prime number 3; therefore, neither n – 1 nor n + 1 is a multiple of 3.

Since every third integer is a multiple of 3, it follows that n must be a multiple of 3. This fact alone is not sufficient, though, since different multiples of 3 can give different remainders upon division by 6. For example, 9/6 gives a remainder of 3, and 12/6 gives a remainder of 0.

Testing Cases
Find values of n > 6 that satisfy statement 1.
The value n = 7 does not satisfy the statement, since 72 – 1 = 48 is divisible by 3.
The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is divisible by 3.
The value n = 9 satisfies the statement, since 92 – 1 = 80 is not divisible by 3. The remainder of 9/6 is 3.
The value n = 10 does not satisfy the statement, since 102 – 1 = 99 is divisible by 3.
The value n = 11 does not satisfy the statement, since 112 – 1 = 120 is divisible by 3.
The value n = 12 satisfies the statement, since 122 – 1 = 143 is not divisible by 3. The remainder of 12/6 is 0.

At least two different remainders are possible, so the statement is insufficient.

(2) NOT SUFFICIENT:
Algebra
If n2 – 1 is even, then n2 must be odd, so n itself is odd. This fact alone is not sufficient, though, since different odd numbers can give different remainders upon division by 6. For example, 7/6 gives a remainder of 1, and 9/6 gives a remainder of 3.

Testing Cases
Find values of n > 6 that satisfy the statement.
The value n = 7 satisfies the statement, since 72 – 1 = 48 is even. The remainder of 7/6 is 1.
The value n = 8 does not satisfy the statement, since 82 – 1 = 63 is not even.
The value n = 9 satisfies the statement, since 92 – 1 = 80 is even. The remainder of 9/6 is 3.
At least two different remainders are possible, so the statement is insufficient.

(1) AND (2) SUFFICIENT: Statement (1) indicates that n is a multiple of 3 and statement (2) indicates that n is odd. From this point, you can either use algebra or test numbers.
Algebra
According to the two statements, n must be 3 times some odd integer, so it can be written as n = 3(2k + 1), where k is an integer. Distributing gives n = 6k + 3, which is 3 more than a multiple of 6. Therefore, the remainder upon dividing n by six must be 3.

Testing Cases
Test numbers that are multiples of 3, odd, and greater than 6:
If n = 9, then 9 / 6 = 1 remainder 3.
If n = 15, then 15 / 6 = 2 remainder 3.
If n = 21, then 21 / 6 = 3 remainder 3.

Will this always return a remainder of 3? Yes! An odd multiple of 3 will always have a remainder when divided by 6 (because 6 is even). Further, an even multiple of 3 will always have a remainder of zero (because an even multiple of 3 is divisible by both 2 and 3). Therefore, the odd multiple of 3, which is always 3 more than an even multiple of 3, will always have a remainder of 3.

The correct answer is (C).

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