Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 81)
81.
The difference of the squares of two consecutive even integers is divisible by which of the following integers ?
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 2 of 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.
= (8n+4)
= 4(2n+1)
Which is divisible by 4.
Naveen said:
1 decade ago
Here missing it should be (2n+2)^2-(2n)^2
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)
(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)
Balaji said:
1 decade ago
Please explain (2n+2) ^2 = (2n+2+2n) (2n+2-2n).
Marthandan said:
1 decade ago
Please explain (2n+2) ^2 = (2n+2+2n) (2n+2-2n).
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