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.
Swetha said:
1 decade ago
Why can't we can take consecutive even integers as x, x+2, x+4....where x=2.
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.
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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