$$P\ =\ m^3\ -\ m$$
$$m\ =\ \left(10x\right)^n\ +\ 5^4$$
x and n are said to be positive integers.
Let x and n = 1
$$m\ =\ \left(10x\right)^1\ +\ 5^4$$
$$=\ 10^1\ +\ 5^4$$
10 + 625
m = 635
$$p\ =\ \left(635\right)^3\ -\ 635$$
p = 256, 047, 240
What is the remainder when P is divided by 4
$$p\ =\ \frac{\left(256,\ 047,\ 240\right)}{4}$$
P = 64, 011, 810 remainder 0
Option B is CORRECT.

BTGmoderatorRO wrote:If P = m^3 - m and $$m=\left(10x\right)^n+5^4$$, where x and n are positive integers, what is the remainder when P is divided by 4 ?

A) 3
B) 0
C) 1
D) undetermined
E) 2

We cannot explore particular cases, because (D) is at stake.
(We can find many cases in which the answer is (say) zero, but... is there another possibility?)
\[x,n\,\,\, \geqslant 1\,\,\,{\text{ints}}\,\,\,\,\,\left( * \right)\]
\[m = {\left( {10x} \right)^n} + {5^{4\,}}\]
\[?\,\,\,:\,\,\,{\text{remainder}}\,\,\,{\text{of}}\,\,P\,\,{\text{by}}\,\,4\]
\[\left. \begin{gathered}
{\left( {10x} \right)^n}\,\,\mathop = \limits^{\left( * \right)} \,\,\,even \hfill \\
{5^4} = {\text{odd}}\,\,\,\left( {{\text{no}}\,\,{\text{factor}}\,\,2} \right) \hfill \\
\end{gathered} \right\}\,\,\, \Rightarrow \,\,\,\,m = {\text{odd}}\,\,\,\left( {**} \right)\]
\[P = m\left( {{m^2} - 1} \right)\mathop = \limits^{\left( {**} \right)} \underbrace {\left( {m - 1} \right)}_{{\text{even}}}m\underbrace {\left( {m + 1} \right)}_{{\text{even}}}\,\,\, \Rightarrow \,\,\,P\,\,{\text{multiple}}\,\,{\text{of}}\,\,4\,\, \Rightarrow \,\,\,? = 0\,\,\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.