Aptitude - Numbers - Discussion

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

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.

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.

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.

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

+67/68 remainder is 67.

So answer 67-1=66.

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

134%68 = 66. So answer = 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.

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

What would be the solution when n is even?