Aptitude - Numbers - Discussion

What would be the solution when n is even?

Simple method of this problem is 67+67 = 134-68 = 66.

Dividend = Divisor*Quot+Remainder.

Here, Dividend is '67^67 + 67' and divisor is 68.

i.e 67^67 + 67 = 68*Quot+Remainder.....(1).

Now, we know that (x^n + 1) will be divisible by (x + 1) only when n is odd. --> Theory.

We express Dividend as follows: (67^67+1)*1+66.....(2).

i.e. 67^67+67 = 68*Quot+Remainder.....(from 1).

Comparing the above.

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

134%68 = 66. So answer = 66.

67/68 remainder -1 then (-1) to the power 67 is -1.

+67/68 remainder is 67.

So answer 67-1=66.

Hi guys,

Here you can use concept of negative reminders.

68/67 here positive reminder as all we know 68, but negative reminder is -1(67-68).

For example 11/3 positive reminder 2 and negative reminder -1(11-12).

Here is our question.

Rem of (67^67+67)/68 = (-1)^67+(-1) = -2.

So the answer is 68-2 = 66.

Hi guys,

Here you can use concept of negative reminders.

68/67 here positive reminder as all we know 68, but negative reminder is -1(67-68).

For example 11/3 positive reminder 2 and negative reminder -1(11-12).

Here is our question.

Rem of (67^67+67)/68 = (-1)^67+(-1) = -2.

So the answer is 68-2 = 66.

It's quite simple. Find unit digit of power value add it with their next value. Divide with divisor. Then subtract remainder from original divisor.

For ex: (3^3+3)/4.

Unit digit of 3^3 is 7. Add with 3 as it is given 7+3=10.

Divide 10/4 remainder=2.

Now finally subtract 2 with 4 so 4-2=2.

Let's see this one: (67^67+67)/68.

Unit digit of 67^67 is 3 as (7^16*7^3). Add this with 67.67+3 = 70.

Divide 70/68 = remainder = 2.

Finally subtract 2 with 68 so 68-2 = 66 answer.

67^67 = (68^67-1^67).

So (68^67)+67-1^67.

= (68^67)+66.

= (68^67) is completely divisible by 68. So remainder is 66.

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