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 4 of 7.
Vicky said:
8 years ago
Just take the negative remainder so the new equation will be (-1) power 67 +67== (-1+67) ==66.
Siva ram said:
8 years ago
Take 67 as common 67(1+1)= 134.
Now, divide 134 by 68 you get 66 as remainder, is this easy?
Now, divide 134 by 68 you get 66 as remainder, is this easy?
(2)
Satheesh Kumar said:
8 years ago
(X pwr n + 1) , if the n value is an even number then what should I do? Can anyone help me?
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)
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)
Sandeep jaiswal said:
10 years ago
(67^67+67) = 67 (1^67+1) = 67(1+1) = 134.
134%68 = 66. So answer = 66.
134%68 = 66. So answer = 66.
(1)
Lokesh said:
7 years ago
Simple.
For odd = (x^n+1)+(n-1).
For even= (x^n+1)-(n+1).
That's all.
For odd = (x^n+1)+(n-1).
For even= (x^n+1)-(n+1).
That's all.
Meghana said:
9 years ago
Why do we subtract 2 the remainder from 68, can anyone explain this?
Random retard said:
6 years ago
Thanks to all the good souls who took time to explain the answer.
Pooja said:
1 decade ago
Please give proper description. Can't understand the problem.
(1)
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