Aptitude - Numbers - Discussion

Any other method please.

Simplify please.

Cannot understand the concept please xplain.

Not understand.

When x pow even then. what's the procedure?

Find out unity digit of 67^67....i e unit digit is 3
n 3+67=70
and when divide by 68 remainder 2...ie 68-2=66

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.

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.