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 ???
Best Answer
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 $A$...