Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 107)
107.
The difference of the squares of two consecutive odd integers is divisible by which of the following integers ?
3
6
7
8
Answer: Option
Explanation:

Let the two consecutive odd integers be (2n + 1) and (2n + 3). Then,

(2n + 3)2 - (2n + 1)2 = (2n + 3 + 2n + 1) (2n + 3 - 2n - 1)

     = (4n + 4) x 2

     = 8(n + 1), which is divisible by 8.

Discussion:
7 comments Page 1 of 1.

Hema said:   2 years ago
What do we take 2n+1 and 2n+3? Anyone, please explain.

Saloni Bagwani said:   5 years ago
What if we take (n+2) and (n+4) as the two consecutive odd integers?
(1)

Reddivari radha gayathri said:   5 years ago
The first odd number is 1 and the second odd number is 3.

Square of those odd numbers are 9-1 (have to sub big num - small num to get ans).

So Ans=8.
(5)

Naam me kya rakha hai said:   8 years ago
(a^2-b^2) = (a+b) (a-b).
So,

(2n + 3)^2 - (2n + 1)^2 = (2n + 3 + 2n + 1) (2n + 3 - 2n +1).
The last 1 is (+1) Not ( - 1).

Please explain clearly.

Sasi said:   10 years ago
It is very simple. Take any two consecutive odd integers and square them. And then difference is divisible by 8.

Ex = 3, 5 numbers; 25-9 = 16; which is divisible by 8.
(2)

Asdf said:   1 decade ago
As they mentioned for two consecutive number x+1 and x+3 cannot be taken. 2n+1 and 2n+3 must be taken.

Dibakara nayak said:   1 decade ago
Take 3, 5 ( both are odd integer )
for this case it will fail.
Take number x+1 , x+3
we will get -4(x+2) , then it mean ans is 4 which will satify all cases

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