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$a,b$ and $c$ are all digits. How many different combinations of $\left ( a,b,c \right )$ are there that makes $a^\left ({b^{c}} \right )$ a prime number?

Obviously $a$ has to be $2,3,5$ or $7$ and $b^{c}$ must equal $1$. When $b=1$, $c$ can be anything.

So $\left (4\cdot 1\cdot 10 \right ) = 40$ is my answer but the answer key says it is 72.

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    Please, if the answer is sitting in front of my eyes, feel free to *lead* me rather than giving me the answer right up. I tried on my own and failed. – 2017-01-02
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    Well, $a$ prime, $b=1$, $c$ could be anything. – 2017-01-02
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    $b$ cannot be zero, as $2^{(0^3)}=2^0=1,$ which is not prime. That still doesn't match your answer key, but I would answer $40$ – 2017-01-02
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    @RossMillikan is right, except that if both $b$ **and** $c$ are $0$ it is allowable. $0^0=1$. – 2017-01-02
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    @Wildcard: some define $0^0=1,$ some say it is undefined. We have questions on that on this site. – 2017-01-02
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    You guy are correct. Sorry for my mistake. Got upset so much that my mind is actually inaccurate. Edited the question to correct that. – 2017-01-02
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    Still, the closest we get is $40$. Should I assume the answer key is wrong? – 2017-01-02
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    @ZacharySelk It asks the total number of combinations that results in a prime number. – 2017-01-02
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    Working the answer key's answer in reverse, let's allow that $a$ must be one of $2, 3, 5, 7$. This means that the possibilities for $b$ and $c$ **according to your answer key** must be accounted for in $72/4=18$ possibilities. Not only is this incorrect, I can't even *guess* at what errors they are making in counting to wind up with this answer. (I'm usually pretty good at making such guesses.) – 2017-01-02
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    @Wildcard $18$ is indeed correct. – 2017-01-02
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    @ZacharySelk, haha, right you are. I almost had it. – 2017-01-02

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So we need $a$ to be prime, so $a=2,3,5,7$. We also need $b^c=1$. So either $b=1$ or $c=0$. If $b=1$ then $c$ can be $0,...,9$ so we have $10$ options. If $c=0$ then $b$ can be $1,...,9$ (as $0^0$ is undefined). However we double counted $1^0$, so we have $10+8=18$ ways for $b^c=1$.

So then pick an $a$ to get $4\cdot 18=72$ ways.

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    Wildcard almost had it lol – 2017-01-02