Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 28)
28.
If n is a natural number, then (6n2 + 6n) is always divisible by:
Answer: Option
Explanation:
(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.
Discussion:
27 comments Page 3 of 3.
Hussain said:
1 decade ago
Who told you that 540 is not divisible by 12?
540/12 = 45.
540/12 = 45.
Priya said:
1 decade ago
But, If we put n=9 in 6n2+6n.
i.e, 6(9)2+6(9)
= 6(81)+54
= 486+54
= 540
But, here 540 is not divisible by 12.
So, how?
i.e, 6(9)2+6(9)
= 6(81)+54
= 486+54
= 540
But, here 540 is not divisible by 12.
So, how?
K.pavan said:
1 decade ago
ELIMINATION METHOD...
PUT n=1. we get 12.
12 is divisible by both 12, 6.
PUT n=1. we get 12.
12 is divisible by both 12, 6.
Sandeep said:
1 decade ago
Product of even, odd is always even.
Akhil said:
1 decade ago
How is n(n+1) proved an even numer?
Neo said:
1 decade ago
If a number is divisible by 12 or 18 then it is always divisible by 6.
So option C and D are eliminated...
We can see the no. 6n(n+1) is divisible by 6 so we have to check for 12
Now a number would be divisible by 12 if it is multiplication of 6 and a even number. (e.g. (6*even number)
Now n(n+1) is even number.
So answer is B
So option C and D are eliminated...
We can see the no. 6n(n+1) is divisible by 6 so we have to check for 12
Now a number would be divisible by 12 if it is multiplication of 6 and a even number. (e.g. (6*even number)
Now n(n+1) is even number.
So answer is B
Karma said:
1 decade ago
I don't understand this problem please help.
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