Number of ordered triples with given condition We are given that there are  $6250k$ ordered triples $(m,n,p)$ where $m,n,p$ are integers such that $$1≤m≤100,\;1≤n≤50,\;1≤p≤25\quad \& \quad  3\,|\,2^m+2^n+2^p$$  Find $k$.
With the given range of values of $m, n,p$, I guess the possible numbers of ordered triples without any conditions would be $25\times (50-1)\times (100-2)$, i.e., $25\times 49\times 98$. But after this I am unable to detect the relationship between m, n and p for the given divisibility condition.
Please help.

Given that $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ and $  3\,|\,2^m+2^n+2^p$  where $m,n,p$ are integers.
Now if $2^m+2^n+2^p$ is divisible by $3$, then each of $2^m$, $2^n$, $2^p$ is divided by $3$. 
Remainder of ${2^m}$ when it is divisible by $3$ is $(-1)^m$.
So remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}$. Now if ${(-1)^m+(-1)^n+(-1)^p}$ is divisible by $3$, the we are done.
Case I: If $m$ is odd and $n$ and $p$ are even, then  remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1+1+1}={1}$ which is not divisible by $3$.
Case II: If $m$ is even and $n$ and $p$ are odd, then  remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1-1-1}={-1}$ which is not divisible by $3$.
Case III: If $m$, $n$ and $p$ are even, then  remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1+1+1}={3}$ which is divisible by $3$.
Case IV: If $m$, $n$ and $p$ are odd, then  remainder of ${2^m+2^n+2^p}$  when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1-1-1}={-3}$ which is divisible by $3$.
So we are interested in the last two cases.
Now $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ 
In the given range, the number of possible choice that all $m, n, p$ are even is $50, 25, 12$ respectively and the number of possible choice that all $m, n, p$ are odd is $50, 25, 13$ respectively.
So the total number of ordered triples $(m,n,p)$ where $m,n,p$ are integers  that $ 3\,|\,2^m+2^n+2^p$ is $(50 \times 25 \times 12)+(50 \times 25 \times 13)=50 \times 25 \times 25= 31250 = 5\times 6250$.
So the possible value of $k=5$ .
