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.
Satyendra said:
10 years ago
If we taking two consecutive number as 2 and 4.
Then, we get 12 which is divisible by 3 and 6 also then.
Why we choose 4 only.
16- 4 = 12.
12/3 = 0.
Please clear this doubt.
Then, we get 12 which is divisible by 3 and 6 also then.
Why we choose 4 only.
16- 4 = 12.
12/3 = 0.
Please clear this doubt.
(1)
Sobhit Kumar Sah said:
9 years ago
@Satyendra.
According to your opinion, there are two answers. Since we take next consecutive even integers 4 & 6.
According to your opinion, there are two answers. Since we take next consecutive even integers 4 & 6.
(1)
Kavitha said:
9 years ago
Agree @Prabhu.
a^2 - b^2 = (a + b)(a - b).
a^2 - b^2 = (a + b)(a - b).
Saymon said:
8 years ago
Is it applicable?
x & x+2
so, (x+x+2)^2 =x^2+2.x(x+2)+(x+2)^2 =x^2+2x^2+4x+x^2+2x2+4 = 4x^2+8x+4 = 4(x^2+2x+1).
x & x+2
so, (x+x+2)^2 =x^2+2.x(x+2)+(x+2)^2 =x^2+2x^2+4x+x^2+2x2+4 = 4x^2+8x+4 = 4(x^2+2x+1).
Sreemoy said:
5 years ago
-n +2>0 and 2n >4 the number of integer n satisfying.
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