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 5 of 7.
Rudra said:
8 years ago
Trick : the number 67 to the power is given 67 so the power 67 is divisible by 2 remainders comes 1 then 67 of the power is take 1 and then solve i.e.
(67^1+67) divided by 68 = 66.
(67^1+67) divided by 68 = 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?
Vicky said:
8 years ago
Just take the negative remainder so the new equation will be (-1) power 67 +67== (-1+67) ==66.
Pranoti said:
7 years ago
@Kanak.
It was very helpful.
It was very helpful.
Falandu said:
7 years ago
67 is not odd it is prime.
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.
Deepak Parmar said:
7 years ago
How to find unit digit of any number? Please tell me.
Raji said:
7 years ago
(x^n+1) is divisible by (x+1), if not is odd.
So, ( 67^67+1) is divisible by 68 (i.e.)67 + 1.
(67^67+1)÷68 provides remainder 0.
Given: (67^67+67)÷68 ->(67^67 + 1 + 66)÷68.
->provides remainder 66.
So, ( 67^67+1) is divisible by 68 (i.e.)67 + 1.
(67^67+1)÷68 provides remainder 0.
Given: (67^67+67)÷68 ->(67^67 + 1 + 66)÷68.
->provides remainder 66.
Prateeksharma said:
7 years ago
As given (x^n +1) is divisible by (x+1) if and only if n is odd,
In the question we have (67^67 + 67).
So we can write it as( 67^67+ 1+66) = (67^67+1)+66. So that's why we get remainder 66.
In the question we have (67^67 + 67).
So we can write it as( 67^67+ 1+66) = (67^67+1)+66. So that's why we get remainder 66.
(1)
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