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Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 81)
Numbers
81.
The difference of the squares of two consecutive even integers is divisible by which of the following integers ?
3
4
6
7
Answer: Option
Explanation:

Let the two consecutive even integers be 2n and (2n + 2). Then,

(2n + 2)2 = (2n + 2 + 2n)(2n + 2 - 2n)

     = 2(4n + 2)

     = 4(2n + 1), which is divisible by 4.

Discussion:
15 comments Page 1 of 2.
Marthandan said:   1 decade ago
Please explain (2n+2) ^2 = (2n+2+2n) (2n+2-2n).
Balaji said:   1 decade ago
Please explain (2n+2) ^2 = (2n+2+2n) (2n+2-2n).
Prabhu said:   1 decade ago
The difference between two consecutive even no

(2n+2)^2-(2n)^2= (2n+2+2n) (2n+2-2n) (that is a^2-b^2=(a+b)(a-b))

=(4n+2)2

=4(2n+1)
Naveen said:   1 decade ago
Here missing it should be (2n+2)^2-(2n)^2
Priya said:   1 decade ago
(2n+2)^2 - (2n)^2 = 4n^2+8n+4 - 4n^2

= (8n+4)

= 4(2n+1)

Which is divisible by 4.
Shamili said:   1 decade ago
(2n+2)^2!=(2n+2+2n)(2n+2-2n)

It can be write as
(2n+2)^2-(2n)^2=(2n+2+2n)(2n+2-2n)
=(8n+4)
=4(2n+1)

Which is divisible by 4
(1)
Anusha said:   1 decade ago
Eg: Two consecutive even numbers are 6, 8 and 2, 4.

36-64 = 28.
28 is divisible by 4.

4-16 = 12.
12 is divisible by 4.
Swetha said:   1 decade ago
Why can't we can take consecutive even integers as x, x+2, x+4....where x=2.
Rumpa said:   1 decade ago
Let two consecutive even integer as x and x+2.

= (x+2)^2-x^2.
= x^2+4x+4-x^2.
= 4x+4.
= 4(x+4).

Divisible by 4.
(4)
Payal said:   10 years ago
But again if we take for odd integers and where x=1 then again the answer is 8 which does not come.
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