Do I have to check for all prime numbers?

$19 \rightarrow 1 + 9 = 10 \rightarrow 1 + 0 = 1$

$2 \rightarrow 2$

$3 \rightarrow 3$

$13 \rightarrow 1 + 3 = 4$

$5 \rightarrow 5$

$6 \rightarrow none$

$7 \rightarrow 7$

$17 \rightarrow 1 + 7 = 8$

$9 \rightarrow none$

If you recall, all multiples of 3 have a final digital sum, or a digital root , of 3, 6, or 9. The vice versa is also true, namely that all numbers that have a digital root of 3, 6, or 9 (excluding the case of 3) are divisible by 3. Therefore, no number can have a digital root of 6 or 9. However, 3 has a digital root of 3, and is prime. Above are listed examples showing that every digital root can be formed with primes, excluding 6 and 9.

Wrote and checked with Python3 program, 6 and 9 are the exceptions.

Answer=2

Of course, tested only for prime numbers between 1 and 10001.

For instance, 2 and 3 are within first 10.

Remaining show a steady pattern with exception of 6 and 9. To be sure, I was only verifying my guess of 6 and 9. Therefore, answer is 2.

It is a small program with two functions, primegenerator and digitsum and one driver for testing in a specified range.

Program below Python3 Program - Print digit sum of Prime Numbers

def digsum(n):

if(n==0):

    return 0

if(n%9==0):

    return 9

else:

    return (n%9)

print("Enter 'x' for exit.");

print("Enter starting and ending number: ");

start = input();

if start == 'x':

exit();

else:

end = input();

lower_number = int(start);

upper_number = int(end);

print("\nPrime Numbers between the given range:")

for num in range(lower_number, upper_number+1):

        if num>1:

            for i in range(2, num):

                if(num%i)==0:

                    break;

            else:

                print(num,digsum(num));

sample output between 1801 and 2001

Enter 'x' for exit.

Enter starting and ending number:

1801

2001

Prime Numbers between the given range:

1801 1

1811 2

1823 5

1831 4

1847 2

1861 7

1867 4

1871 8

1873 1

1877 5

1879 7

1889 8

1901 2

1907 8

1913 5

1931 5

1933 7

1949 5

1951 7

1973 2

1979 8

1987 7

1993 4

1997 8

1999 1

There is a very close relationship between the modulo 9 equivalents of numbers and their digit sums

In mathematical terminology there is an isomorphism between digit sum arithmetic and modular 9 arithmetic

Use the following Wolfram Alpha script to print digit sum of first 200 prime numbers

Table[Prime[n],{n,200}] mod 9

{2, 3, 5, 7, 2, 4, 8, 1, 5, 2, 4, 1, 5, 7, 2, 8, 5, 7, 4, 8, 1, 7, 2, 8, 7, 2, 4, 8, 1, 5, 1, 5, 2, 4, 5, 7, 4, 1, 5, 2, 8, 1, 2, 4, 8, 1, 4, 7, 2, 4, 8, 5, 7, 8, 5, 2, 8, 1, 7, 2, 4, 5, 1, 5, 7, 2, 7, 4, 5, 7, 2, 8, 7, 4, 1, 5, 2, 1, 5, 4, 5, 7, 8, 1, 7, 2, 8, 7, 2, 4, 8, 2, 1, 5, 4, 8, 5, 8, 1, 1, 7, 8, 5, 2, 4, 1, 2, 8, 5, 7, 4, 1, 5, 7, 1, 2, 4, 8, 5, 2, 4, 7, 2, 8, 7, 8, 7, 8, 7, 4, 1, 5, 4, 1, 5, 4, 8, 4, 5, 8, 1, 2, 4, 8, 1, 2, 7, 2, 4, 8, 4, 8, 1, 5, 7, 2, 1, 2, 1, 5, 2, 8, 4, 8, 5, 2, 1, 7, 1, 5, 2, 4, 5, 7, 4, 5, 7, 8, 1, 7, 7, 2, 4, 8, 5, 2, 1, 7, 4, 8, 1, 2, 1, 2, 8, 5, 4, 7, 2, 8}

• If the digital sum of a number is 3 , 6, or 9 then the original number is divisible by 3.
• 3 is the only prime number divisible by 3
• Therefore the digital sum of any prime number can never be a 6 or 9, so the solution is 2

I have a question. What about 0? It can not result from any combination, but it was not counted. So it should have been 0,6,9 all three. And the answer should be 3. Maybe I’m wrong but I would like to know :) thanks

Let's start with the first possible output, 1. Then, 19 gives digit sum 1; 2, 3, 5, 7 are themselves prime, so at least one number gives them as digit sum, i.e. the digit themselves. Now, 13 has digit sum 4; No.s with digit sum 6 and 9 can never be prime, as they are digit sum divisible by 3, so, their no.s must be divisible by 3. And, 17 has digit sum 8. Hence, 6 and 9 are two impossible outputs of the program.