Olympiad number theory question Question -
Some sets of prime numbers, such as $\{7,83,421,659\},$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes can have?
Solution: The answer is $207$. Note that digits $4,6$ and $8$ cannot appear in the units digit. Hence the sum is at least $40+60+80+1+2+3+5+7+9=207.$
how they found that this is the least sum ???
 A: They basically constructed the set with the minimum sum, which is
$$
A = \{1,2,3,5,7,9\}
$$
but it is missing $4,6,8$ since the numbers ending on them cannot be prime, so they have to add it into tens instead of units digits. Since 9 is not prime you add one of them to 9 to make sure you get a prime (and the only one that works is an $8$ since $49=7 \cdot 7$ and $69 = 3 \cdot 13$), ending up with, for example,
$$
\{41,2,3,5,67,89\}
$$
Whichever way $4,6,8$ are added,they will contribute to the total sum $40+60+80$, and hence, the total sum is $40+60+80$ and the sum of elements in $A$...
A: They found a lower bound and then looked for an example.
In such an example, 


*

*the $2$ and $5$ would be single digit numbers, 

*the $1$ cannot be a single digit number; nor can it be in $81$ so must be in $41$ or $61$

*the $9$ cannot be a single digit number; nor can it be in $49$ or $69$ so must be in $89$

*$41$ and $43$ and $47$ are prime

*$61$ and $67$ are prime


So the possibilities are 
$$2,3,5,41,67,89$$
$$2,3,5,47,61,89$$
$$2,5,7,43,61,89$$
A: The remaining task is to find a representation of $207$. 
This is easy, for example,
$$
207=2+3+5+41+67+89.
$$
More interesting would be the question what the minimum of four such
primes is, such as suggested already in the text, with $
7,83,421,659$ (which have sum $1170$). Then the minimum is bigger.
A: A combination with all ten digits and what could be a minimal sum is $2,5,83,109,467$ with a devilish (or not) sum of $666$.
