It works for 32-bit unsigned integers and double precision floats.

For n < 19, "(double)n * (1.0 / 19.0)" evaluates to a double between 0.0 and 1.0, then it is truncated to 0 when it is implicitly converted to unsigned int.

Since there are only 2^32 values for 32-bit integers, it is possible to test all values in under a minute:

    #include <stdio.h>
    #include <stdint.h>

    int main() {
        uint32_t n = 0;
        do {
            uint32_t a = n / 19;
            uint32_t b = (double)n * (1.0 / 19.0);
            if (a != b) {
                printf("Not equal for n = %u\n", n);
            }
            ++n;
        } while (n != 0);
    }