Olympiad number theory question

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 ???

Answer 1

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 $A$...

Answer 2

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

$$2,3,5,41,67,89$$ $$2,3,5,47,61,89$$ $$2,5,7,43,61,89$$

Answer 3

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.

Answer 4

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 $666$.