Aptitude - Numbers - Discussion

a^b is a raise to the power b.

To understand this question we must take a simpler example.
for ex- (3x3)/4

3 = (4-1),
Therefore the question becomes
(4-1)(4-1)/4.
On multiplying (4^2-4-4+1)/4
Note that all the terms in the expansion are completely divisible by 4 except 1.hence Remainder will be 1.

Now take expression for example- 3x3x3 /4.
again (4-1)(4-1)(4-1)/4,
(4^2-4-4+1)(4-1) / 4,
= 4x(4^2-4-4+1)-1x(4^2-4-4+1),
= 4x(4^2-4-4+1) + (-4^2)+4+4-1.

Here the first term is divisible by 4 and in the second term after multiplication by -1 only +1 remains which is not divisible.

Carrying out this process only 1 will remain as the remainder and its sign will depend on the power of the numerator. when the power of 3 was 2, the remainder was +1 and when the power was 3, the remainder was -1.

This means when the power of numerator is odd, -1 will remain and when the power will be even, +1 will remain.

Using this in our example:- (67^67)/68.

-1 will remain because the power of the numerator is odd.
and we also have a +67 in the numerator. therefore, total remaining in the numerator is -1+67 = 66. Which is the remainder and our answer.

Thanks for your explanation @Ummar.

@Kanak.

It was very helpful.

Thanks for the clear explanation, @Padmaja.

67 is not odd it is prime.

Just take the negative remainder so the new equation will be (-1) power 67 +67== (-1+67) ==66.

(X pwr n + 1) , if the n value is an even number then what should I do? Can anyone help me?

Trick : the number 67 to the power is given 67 so the power 67 is divisible by 2 remainders comes 1 then 67 of the power is take 1 and then solve i.e.
(67^1+67) divided by 68 = 66.

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

a^n/(a+1) = a when n is odd.
= 66(ans),
= 1 when n is even.

Simple.

For odd = (x^n+1)+(n-1).
For even= (x^n+1)-(n+1).

That's all.