I have the answer for the following problem, but I don't know how to mathematically or intuitively arrive at that answer.
For the quantity $2^a$$3^b$$5^c$, how many three-digit numbers of $abc$ would result in a prime number?
A three-digit number for $abc$ would be 100, but not 001, for example.
The way I understand it, since $abc$ must be a three-digit number, then 1-9 is possible for a and c, and 0-9 is possible for b. Out of those 810 combinations, what combinations of $abc$ result in $2^a$$3^b$$5^c$ as a prime number.
The book says that since $2^a$$3^b$$5^c$ is prime, it has only two factors, and so $2^a$$3^b$$5^c$ cannot be prime unless $abc$ has two zeros and one 1.
I think I understand why the book says the above, but without confirming, it seems to me that 100, 010, and 001 are just three possibilities, but not the only possibilities for $abc$ that make $2^a$$3^b$$5^c$ a prime.
Obviously I don't understand something basic that bypasses the need for trying all 810 combinations for $abc$ . Is there a better way to solve the above question?
If $a$ is greater than $1$ the number will be divisible by $4$ and not a prime. $b$ and $c$ are similar. If $a$ and $b$ are both $1$ the number will be divisible by $6$ and not a prime. The other two pairs are similar. If $a=b=c=0, 2^a3^b5^c=1,$ which is not prime. We must therefore have one of $a,b,c$ equal to $1$ and the other two equal to zero, as your book says. These possibilities give the primes $2,3,5$