Find if digital root of n is prime or not. DigitalRoot of a number is the repetitive sum of its digits until we get a single digit number.?

The digital root of a positive number n is the almost the same as the remainder from division by 9. The only difference is that a positive multiple of 9 has a digital root of 9, but a remainder of 0.

Since neither 0 nor 9 are prime numbers, that doesn't make any difference for your final output. For all integers n>0, the digital root of n is prime if and only if (n%9) is prime.

But, if you want to display the correct value of the digital root, it does matter. An easy way to get the root without looping in C-like languages:

int root = (n + 8)%9 + 1;

If you've learned the switch statement, you can use that with cases 2, 3, 5, 7 coded to handle a prime digital root, and the other (default:) cases handling a non-prime root.

A faster method, with maybe even smaller code+data program size, is to use an array as a lookup table:

const char sieve[] = "nnPPnPnPnn";

...

bool is_prime = (n>0) && (sieve[n%9] == 'P');

Add <stdbool.h> to your includes at the top to get the standard bool type, plus the constants true and false, defined. (For old, pre-C99 compilers use int instead of bool.)