Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 30)
30.
What will be remainder when (6767 + 67) is divided by 68 ?
Answer: Option
Explanation:
(xn + 1) will be divisible by (x + 1) only when n is odd.
(6767 + 1) will be divisible by (67 + 1)
(6767 + 1) + 66, when divided by 68 will give 66 as remainder.
Discussion:
70 comments Page 1 of 7.
Clinton said:
3 months ago
@Ummar.
Please elaborate your answer.
Please elaborate your answer.
Sumit said:
9 months ago
(67^67+67)/68,
n(1^n +1)/68,
67(1^67+1)/68,
67(1+1)/68 = give reminder 66.
n(1^n +1)/68,
67(1^67+1)/68,
67(1+1)/68 = give reminder 66.
(9)
Ziyad said:
10 months ago
Thanks for your explanation @Ummar.
(1)
Gnanam said:
2 years ago
1 IS NOT DIVISIBLE BY 6 & 8.
(6+3) =9 -> 9 is not divisible by 6 & 8.
(6+7) -> 13 is not divisible by 6 & 8.
(6+6) -> 12 is divisible by 6 & 8 ===> ANSWER.
(6+3) =9 -> 9 is not divisible by 6 & 8.
(6+7) -> 13 is not divisible by 6 & 8.
(6+6) -> 12 is divisible by 6 & 8 ===> ANSWER.
(17)
Ummar said:
3 years ago
@All.
Here is the solution;
67 * 2 - 68 = remainder = 66.
Valid for any type of formation like the given one.
n*2-(n+1) = remainder.
Here is the solution;
67 * 2 - 68 = remainder = 66.
Valid for any type of formation like the given one.
n*2-(n+1) = remainder.
(34)
Saikumar Padi said:
4 years ago
Here, we can use n^n+n = (n+1) (n^n+1)/(n+1) + (n-1).
As division n=p*q+r.
As division n=p*q+r.
(7)
Prakash said:
4 years ago
Thank you so much @Mahi.
Prakash said:
4 years ago
Thank you so much @Mahi.
(3)
Srikanth said:
4 years ago
Thanks @Lokesh.
(1)
Pushpendra Shukla said:
4 years ago
(x^n - a^n) is divisible by (x - a) for all values of n.
(x^n - a*n) is divisible by (x + a) for even values of n.
(x^n + a^n) is divisible by (x + a) for odd values of n.
(x^n - a*n) is divisible by (x + a) for even values of n.
(x^n + a^n) is divisible by (x + a) for odd values of n.
(5)
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