Aptitude - Numbers - Discussion

Take two simpler numbers, like 2 and 2+1=3.

If we take (2^2 + 2)/3, that is seen to be 6/3, so the remainder is 0.
Is it 0? Try the next number.

If we take (3^3+3)/4, that is seen to be (27+3)/4 = 30/4, so the remainder is 2.

This one is 1/2, so it is not always 0.

If we take (4^4 + 4)/5, we get (256 + 4)/5 = 260/5 = 0.
Now from this one, make it is 0 for even's, 2 for odds.

If we take (5^5 + 5)/6, we get (3125+5)/6 = 3130/6 = 521, remainder 4.
Since we have an odd, we'll keep going with odds. So far we have the remainder is 1 less than the starting number.

If we take (7^7 + 7)/8, we get (823,543+7)/8 = 823,550/9 = 102943 remainder 6.

Again, we get a remainder of 6, which is 7-1.

To try the next odd, take (9^9-9)/10 = (387,420,489+9)/10 = 38,420,498/10, which has a remainder of 8, which is, again one less than what we started with.

Given this, we known that (67^67 + 67) mod 68 is 67-1 = 66.

(67^67+67)/68 = 67^67/68+67/68.

If we 67/68 = -1 remainder.

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

i.e.(-2)/68 = remainder can't negative then 68-2 = 66 answer.

67((i^67+1)) = 67(1+1) => 67*2 = 134 => 134-68 = 66.

Please give me answer in more better way?

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.

Check numbers from 3, 4, 5, ... , in the above given format. We will get remainder 1 less than the number, so for 67^67+67 = 66.

(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.

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.

If (67^67 + 1) + 66, when divided by 68 will give 66 as remainder.

Again if you use 63 instead of 67 (67^67 + 1) + 63, when divided by 68 will give 63 as remainder.

Same goes with (67^67 + 1) + 67, when divided by 68 will give 67 as remainder.

WHAT LOGIC IS THERE? AND THE QUESTION IS DIRECTLY MENTION THAT (67^67+1) +67 IS DIVIDE BY 68.

FIND THE REMAINDER THEN WHY ARE WE ADDING EXTRA NUMBERS ON IT AGAIN?

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.