We Can Indeed Figure Something Out

• Five cannot be written as the sum of three distinct positive integers.

• The only way to write six as a sum of three distinct positive integers is $\{1, 2, 3 \}$ $\implies$ product = 6.

• The only way to write seven as a sum of three distinct positive integers is $\{1, 2, 4 \}$ $\implies$ product = 8.

Since the product cannot be 6, the product is 8. Therefore the numbers are $\{1, 2, 4\}$ and the median is the central value, that is $\boxed{2}$ $_\square$

For it to be a product of one of those three integers, you must look at its factors. Decomposing each into its primes:
8 = 4x2x1 = 2x2x2x1,
9 = 3x3x1, and
10 = 5x2x1.

Looking at the prime factorization of these numbers, only one of them has a sum of factors between 5 and 7 and has them all be distinct. The factors of 8. So the three numbers are 1, 2, and 4. The median is 2.

I went with the process of elimination. (1,2,3) has a sum of 6. However its product of 6 does not meet the criteria. The only possible combination is (1,2,4) because of its sum of 7 and product of 8, which both meet the criteria.

Let the integers be $a,b,c$ . then we have $abc$ as 8 or 9 or 10. since minimum value of $abc=8$ , then at least one of them is $1$ . Suppose $a=1$ . Since maximum of $a+b+c$ is $7$ and $bc$ can be 8 or 10 we have $(b,c)=\{(2,4),(2,5)\}$ but if we have $(2,5)$ is considered, $a+b+c=1+2+5=8>7$ thus $(2,4)$ is only possible. Hence $(a,b,c)=(1,2,4)$ which have $3$ terms. Thus median term $=\dfrac{3+1}{2}=2$ and the second term in $1,2,4$ is $\boxed{2}$ which is the median.