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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.

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    Hint (for the divisibility): For any integer $j$, $2^j$ is either one more or one less than a multiple of $3$. Convince yourself that this means that all three of $2^m,2^n,2^p$ have to be the same $\pmod 3$. – 2017-01-09
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    Also: not following your count. You didn't say anything about $m,n,p$ being distinct....so why wouldn't the unconstrained count just be $25\times 50\times 100$? – 2017-01-09
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    I am a bit confused whether to consider (1,1,1), (2,2,3) etc. as ordered triples; that's why I thought it would be 25*49*98. – 2017-01-09
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    "ordered" just means that we keep track of which one is $m$, etc. So, yes...both $(1,1,1)$ and $(2,2,3)$ are ordered triples (though the latter doesn't pass the divisibility test). – 2017-01-09

1 Answers 1

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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 $k=5$ .