Sum of Pandigital Triplet of 3-Digit Primes
Theorem
The smallest integer which is the sum of a set of $3$ three-digit primes using all $9$ digits from $1$ to $9$ once each is $999$:
- $149 + 263 + 587 = 999$
Proof
All three-digit primes end in $1, 3, 7, 9$.
Suppose $1$ is used as the units digit of a prime.
Since the digit $1$ cannot be used again, the sum of the primes is at least:
- $221 + 333 + 447 = 1001$
so $1$ cannot be used as a units digit .
The units digits of the primes are $3, 7, 9$.
To minimise the sum, the hundreds digits must be $1, 2, 4$.
This leaves $5, 6, 8$ be the tens digits.
The primes satisfying these conditions are:
- $157, 163, 167$
- $257, 263, 269, 283$
- $457, 463, 467, 487$
Only $269$ contain $9$, so we must choose it.
Only $157$ does not contain $6$, so must choose it next.
But then all primes beginning with $4$ have some digit coinciding with the above primes.
Hence the next minimal sum of the primes (if they exist) is:
- $10^2 \paren {1 + 2 + 5} + 10 \paren {4 + 6 + 8} + \paren {3 + 7 + 9} = 999$
and we have shown that these primes do exist.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $999$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $999$