Aptitude - Numbers - Discussion

Thank you so much @Mahi.

Take 67 as common 67(1+1)= 134.

Now, divide 134 by 68 you get 66 as remainder, is this easy?

I didn't understand the concept. Please explain it for me.

Please give proper description. Can't understand the problem.

22 + 2 divided by 3.
= 4 + 2,
= 6.
so when 6 divided by 3 then remainder = 0.

So by doing calculations like this we can get:

22 + 2 divided by 3 then remainder = 0.
33 + 3 divided by 4 then remainder = 2.
44 + 4 divided by 5 then remainder = 0.
55 + 5 divided by 6 then remainder = 4.

So by observing above examples we can say;

xx + x is divided by (x+1) then the remainder is (x-1) where x is odd number.

So, now we can say when 6767 + 67 is divided by 68 then remainder is 66.

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.

(67^67+67) = 67 (1^67+1) = 67(1+1) = 134.

134%68 = 66. So answer = 66.

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.

We take the odd number 67 and divide it by even 68, the remainder is 1 less than the odd number.

So, the remainder when (67^67 + 67) is divided by 68 will be 67 -1 = 66.

Thanks @Lokesh.