If P = k^3 – k, where k and P are positive integers, and k is given by

k^3-k = k(k^2-1) = (k-1)k(k+1)

k = 10^n*x^n + 625. K is odd.
So the remainder of k when divided to 4 is either 1 or 3

* remainder = 1. K-1 = 4a
* remainder = 3. K+1 = 4a

IMO A.

So basically, k^3-k can be written as : k(k^2-1) = (k-1)(k)(k+1) - which is nothing but product of 3 consecutive numbers.
Now, an odd number will always leave a remainder of either 1 or 3 when divided by 4.
Eg : 15, 25,27..

Here, the product will always be divisible by 4 hence remainder is zero.
IMO A

If you can find the loophole you can be in and out in 1 minute.

P = (k)^3 - k

Since k = positive integer, P is the Product of 3 consecutive Integers, regardless of the value of X and N (positive integers) in:

k = (10x)^n + (5)^4


P = (k)^3 - k

P = (k) ( k^2 - 1 )

P = (k - 1) (k) (k + 1)


If K = odd integer, then the 3 Factors will be the product of Two Consecutive Even Integers, which will be divisible by 4


k = (10x)^n + (5)^4


Since the variables x and n must be positive integers ———> the base of 10x raised to any positive integer power will be EVEN

and (5)^4 = ODD

Therefore:

k = Even + Odd = Odd Integer

Therefore, P will be the Product of 3 consecutive integers, the middle of which is ODD

P = (even integer) (even integer + 1) ( even integer + 2


P will be a multiple of 4 and the remainder will be 0

A

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Quick one.
\(P = k^3 – k\) says that P is product of three consecutive integers. and \(k = (10x)^{n} + 5^{4}\) that the unit digit of k is 5. Thus P Must be a multiple of 8 since it's last two digits is a MULTIPLE OF 8.

4 is a factor of 8 thus A is the answer.


LET'S EXPLAIN IT ANOTHER WAY
The product of three consecitive positive integers must be a multiple of at least 6 and in one situation of 8. This is because any three consecutive postive integer must contain a multiple of 3(because you're counting up to three consecutiveness) and an even number which means a multiple of 2; \((3*2 = 6)\)
The only situation where consecutive integers (k-1), k, and (k+1) must be a multiple of 8 is when k has a unit digit of 5, as in 5 or 15 or 25 etc.. thus multiplying whatever that ends in 5 with the preceding consecutive number and next consecutive must contain up to three different 2s which is 8 when multiplied.
\(k = (10x)^{n} + 5^{4}\) plain tells you the unit digit of k is 5. So P MUST be divisible by 8.

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