If P = k^3 – k, where k and P are positive integers, and k is given by the expression k=(10x)n+54k=(10x)n+54, where x and n are positive integers, then what is the remainder when P is divided by 4 ?	Choose the correct choice from the following answers.
A) 0
B) 1
C) 2
D) 3
E) cannot be determined
p=k^3-k   =k(k^2-1)   =(k-1)k(k+1)   as can be seen, p is a product of three consecutive integers. now k can be both even and odd.   now look into the given expression of k   k=(10x)n+54k=(10x)n+54   5454 will always be odd. and (10x)n(10x)n will always be even   thus k= even + odd =odd   0lso, k =1 is not possible ( because p is a positive integer)   so lets try k=3   p=(3-1)(3)(3+1)   =2*3*4   0NSWER:0
<result>A</result>
If P = k^3 – k, where k and P are positive integers, and k is given by the expression k=(10x)n+54k=(10x)n+54, where x and n are positive integers, then what is the remainder when P is divided by 4 ?