Numbers .. similar the GMTPrep Q

[Yaj]

Hi Guys,

Need some help with this one:

What is the remainder when (n-1)*(n+1) is divided by 24?
(1) When n is divided by 3, the remainder is 1.
(2) n is odd.

[spoiler]OA:C[/spoiler]

[vinay1983]

It can be rephrased to n^2-1/24=?

1. When n is divided by 3, the remainder is 1.

Numbers cab be 4, 7, 10, 13.
4= 4^2-1/24=16-1/24=15/24---some value
7=7^2-1/24=49-1/24=48/24=2

So 2 answers,not sufficient

2. N is odd

Numbers can be 3, 5, 7, 11 etc

3=3^2-1/24=9-1/24=8/24=1/3
5=5^2-1/24=25-1/24=24/24=1

2 answers not sufficient

Combine both statements

7, 13, 19

But these numbers when divided by 24 yield different remainders such as 1, 7 etc

I feel the question is asking "what", hence it should be specific value.

How can the answer be C?


Experts should reply to this.

[ganeshrkamath]

Hi Guys,

Need some help with this one:

What is the remainder when (n-1)*(n+1) is divided by 24?
(1) When n is divided by 3, the remainder is 1.
(2) n is odd.

[spoiler]OA:C[/spoiler]

Statement 1: n = 3a+1
(n-1)*(n+1) = (3a)(3a+2) = 9a^2 + 6a
Not sufficient.

Statement 2: n = 2b+1
(n-1)*(n+1) = (2b)(2b+2) = 4b^2 + 4
Not sufficient.

Combination: n = 7, 13, 19, ...
n = 6c + 1
(n-1)(n+1) = 6c(6c+2) = 36c^2 + 12c = 12c(3c + 1)
Now if c is even, 12c is divisible by 24.
If c is odd, (3c+1) becomes even. 12(3c+1) is divisible by 24.
So the remainder is always 0.

Choose C

Cheers

[vinay1983]

@ganeshrkamat,

Thanks for the reply, but i need to know, whether my method has some flaws in it. Where did i go wrong?
Can it be solved by that method?

[ganeshrkamath]

@ganeshrkamat,

Thanks for the reply, but i need to know, whether my method has some flaws in it. Where did i go wrong?
Can it be solved by that method?

First, you have calculated the quotient and not the remainder.

Second, they have not asked the remainder when n is divided by 24.
They have asked the remainder when (n-1)*(n+1) is divided by 24.

Correct these two things and you should arrive at the right solution.

Cheers