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.
Fhici said:
5 years ago
Assume n is a odd number then n-1 will be even and if odd number multiply with even it will be even. So that's for n(n-1) is a even number.
If you assume n is an even number then (n-1) will be odd number..Then multiplication of odd and even number will be also even number. So n(n-1) is always even.
If you assume n is an even number then (n-1) will be odd number..Then multiplication of odd and even number will be also even number. So n(n-1) is always even.
Onkar said:
5 years ago
When N is odd. Then it's not divisible by 12.
So, it's divisible by only 6.
So, it's divisible by only 6.
Habib said:
4 years ago
@Onkar @Narra Rakesh @Emi.
If n is odd then (n+1) is even and if n is even then (n+1) is odd.
Multiplication of an odd and an even number is always even.
So that the number is divisible by 6 and 12.
If n is odd then (n+1) is even and if n is even then (n+1) is odd.
Multiplication of an odd and an even number is always even.
So that the number is divisible by 6 and 12.
(2)
Hemanth said:
4 years ago
6n^2 + 6n = 6 n(n+1) ->n(n+1) is always even
= 6 2(2+1), Ex:2(2+1) = 6.
= 36.
->I think everyone knows divisibility of 6 and 12.
= 6 2(2+1), Ex:2(2+1) = 6.
= 36.
->I think everyone knows divisibility of 6 and 12.
(1)
Bapan mondal said:
3 years ago
Solve with prove
(6n2 + 6n).
1st
6n(n+1) put the value n=6.
66(6+1)= 66(7)=462/6=77 clearly divisible.
2nd
6n(n+1)= put the value n=12.
612(12+1)=612(13)=7956/12=633 clearly divisible.
Exception
6n(n+1)= put the value 18.
618(18+1)=618(19) = 11742/18 it's not divisible.
(6n2 + 6n).
1st
6n(n+1) put the value n=6.
66(6+1)= 66(7)=462/6=77 clearly divisible.
2nd
6n(n+1)= put the value n=12.
612(12+1)=612(13)=7956/12=633 clearly divisible.
Exception
6n(n+1)= put the value 18.
618(18+1)=618(19) = 11742/18 it's not divisible.
(5)
Shruti said:
3 years ago
If we assume that n = 1.
Then,
6n^2+6n,
6*1*1+6*1 = 12,
So from the given options, 6 and 12 is always divisible.
Then,
6n^2+6n,
6*1*1+6*1 = 12,
So from the given options, 6 and 12 is always divisible.
(3)
Ajay said:
1 year ago
@All.
Let us assume n ( it can be any number)
I'm taking n as 1.
6n² + 6n = 6 × 1² + 6×1 = 12
So, 12 is divisible by both 6 and 12.
Answer: Option B
Let us assume n ( it can be any number)
I'm taking n as 1.
6n² + 6n = 6 × 1² + 6×1 = 12
So, 12 is divisible by both 6 and 12.
Answer: Option B
(4)
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