Creating Primes!

The sum is equal to $S = 45 + 9T + 99H + \cdots$ where $T$ is the sum of digits in the tenths position, $H$ the sum of digits in the hundredths position, and so on.

Clearly, we want as few digits as possible in the tenths and higher position. Let's first assume that it is possible to avoid any three digits numbers or higher. Since it is inevitable that $4, 6, 8$ are in the tenths position, we have $T \geq 18$ and $S \geq 45 + 162 = 207$ .

A solution with $S = 207$ is easily constructed. In fact, there are three: $(2, 3, 5, 41, 67, 89);\ (2, 3, 5, 47, 61, 89); \ (2, 5, 7, 43, 61, 89).$ (A simple consideration in forming these solutions is, that 2, 5 may not be the unit digits in a two-digit number, and that 1, 9 may not stand alone. Since 49, 69 are not prime, we must include 89. The rest is straightforward.)

Thus we have not only found a solution but also proven it to be minimal.

@Satyajit Mohanty , you're back! This solution is for you.

I began this problem with two thoughts: because we want the sum to be as small as possible, we will try out best to keep large digits confined to the units place.

Secondly, the digits of 4, 6, and 8 must go in the tens place or higher because any number with a 4, 6, or 8 as the units digit cannot be prime. With that said, I created the primes 89, 61, and 47. The remaining digits (2, 3, and 5) are all prime so they can stand by themselves, giving us the answer of 89 + 61 + 47 + 2 + 3 + 5 = 207.

Welcome back, Satyajit.

Although you can use an algebraic method, I found another method that works. First I assumed that there would be no three digit numbers. I then found that the only two prime numbers you can make with an 8, are 89 and 83. 3 is prime on its own, so i choose 89 because that also gets rid of the 9, one of the larger numbers. From there, I broke the remaining numbers up into 2 categories, even(2,4,6) and odd(1,3,5,7). I kept the two separate because it is a small number, and the only even prime. I then looked at the 4, and found the smallest prime with a 4 in it is 41. Moving on to the 6, the lowest was 67. However, I could have used 47 and 61 and got to the same answer. That left me with 3 and 5 which are already prime. I added up my numbers, (89+67+41+5+3+2) and got 207.

I constructed five primes 5,29,41,83 and 67. My observation was that almost everytime the sum was coming out to be 225. So, I gave up on the idea that the primes should have 1,2,2,2,2 digits and instead tried for more single digit ones. I had already zeroed in on 5 ,to be alone because it cant fit at the last of any number and if it was the tens digit it would force us to make 1,2,2,2,2 digit primes. Now, since the odd- even should remain in pairs for two digit primes, I must remove two single digit numbers; one of which is 2. I chose 7 as the other and the primes were 2,5,7,43,61,89.

I constructed five primes 5,29,41,83 and 67.
My observation was that almost everytime the sum was coming out to be 225.
So, I gave up on the idea that the primes should have 1,2,2,2,2 digits and instead tried for more single digit ones. I had already zeroed in on 5 ,to be alone because it cant fit at the last of any number and if it was the tens digit it would force us to make 1,2,2,2,2 digit primes.
Now, since the odd- even should remain in pairs for two digit primes, I must remove two single digit numbers; one of which is 2. I chose 7 as the other and the primes were 2,5,7,43,61,89.