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 2 of 7.
Nikesh Singh said:
1 decade ago
Nothing complexity with the solution.
From the formula : (x^n+1) is always divisible by (x+1) only when n is odd.
In the above problem n is 67 i.e, n is odd.
So, we can write (67^67+67) as ((67 ^ 67)+1) + 66.
Why because making the given expression as convenient to us.
Thank you.
From the formula : (x^n+1) is always divisible by (x+1) only when n is odd.
In the above problem n is 67 i.e, n is odd.
So, we can write (67^67+67) as ((67 ^ 67)+1) + 66.
Why because making the given expression as convenient to us.
Thank you.
Ravi said:
1 decade ago
I support deepak but in additional I would like you to know of this 67^67 units digit is to b found out. By cyclicity of numbers concept 67/4 remainder 3. So units digit is units digit of 7^3=**3 so adding 3+67=70.
When divide by 68 remainder 2. Ie 68-2=66.
When divide by 68 remainder 2. Ie 68-2=66.
Kamal hachan said:
5 years ago
(67^67+67)%68 where % is mod operator. it gives remainder.
= 67^67%68+67%68
=(68-1)^67%68 +(68-1)%68
=(-1)^67+(-1)
= -1-1
=-2 (Since remainder can't be negative )
=68-2
= 66
believe me this is best trick for remainder
If any doubt please email me.
= 67^67%68+67%68
=(68-1)^67%68 +(68-1)%68
=(-1)^67+(-1)
= -1-1
=-2 (Since remainder can't be negative )
=68-2
= 66
believe me this is best trick for remainder
If any doubt please email me.
(8)
Anil said:
1 decade ago
Just use remainder theorem see how,
Write down the given expression in the form of polynomial of x :67^67+67 = x^67+x, now by remainder theorem (x-(-1)) i.e. 68,
So x^67+x /(x-(-1)) will give you remainder as -2, add this to 68 to get answer 66.
Write down the given expression in the form of polynomial of x :67^67+67 = x^67+x, now by remainder theorem (x-(-1)) i.e. 68,
So x^67+x /(x-(-1)) will give you remainder as -2, add this to 68 to get answer 66.
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)
Vipin raj said:
7 years ago
X^n + 1 is divisible by x+1 always whatever the value of n.
X^n -1 is divisible by x+1 and x +1 and x -1.
That same the given value 67^67. +1, +66 extra port of this no will be remainder.
X^n -1 is divisible by x+1 and x +1 and x -1.
That same the given value 67^67. +1, +66 extra port of this no will be remainder.
(1)
Anusha said:
1 decade ago
(67^67+67)/68.
--> (67*67*67*...67)/68+(67)/68.
--> 67-68 = -1.
--> (-1*-1*-1*....*-1)/68+(67)/68.
--> (-1/68)+67/68.
--> 68-1= 67.
--> (67+67)/68 = 2.
68-2 = 66.
--> (67*67*67*...67)/68+(67)/68.
--> 67-68 = -1.
--> (-1*-1*-1*....*-1)/68+(67)/68.
--> (-1/68)+67/68.
--> 68-1= 67.
--> (67+67)/68 = 2.
68-2 = 66.
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)
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.
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