Question

Найти количество натуральных чисел в интервале от 1 до 1000, у которых сумма цифр-простое число. Решить с помощью паскаля , вложенные циклы.

Answer 1

var

 i,j,count,sp,s:integer;

begin

 count:=0;

 for i:=2 to 1000 do

   begin

     s:=i; sp:=0;

     while s>0 do

       begin

         sp:=sp + s mod 10;

         s:=s div 10;

       end;

     if sp>=2 then

     begin

     s:=1; j:=2;

     while (sqr(j)<=sp) and (s=1) do

       if sp mod j = 0 then

         s:=0

           else inc(j);        

     if s=1 then  

       inc(count);

     end;

   end;

 writeln(count);

end.

Answer 2

Finding Natural Numbers with Prime Digit Sums

To find the number of natural numbers in the range from 1 to 1000, where the sum of the digits is a prime number, we can use nested loops in C++ to iterate through the numbers and calculate their digit sums.

Here's a sample code snippet in C++ to achieve this:

#include

// Function to check if a number is prime
bool isPrime(int n) {
    if (n <= 1) {
        return false;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

int main() {
    int count = 0;
    for (int num = 1; num <= 1000; num++) {
        int sum = 0;
        int temp = num;
        while (temp > 0) {
            sum += temp % 10;
            temp /= 10;
        }
        if (isPrime(sum)) {
            count++;
        }
    }
    std::cout << "The number of natural numbers with prime digit sums from 1 to 1000 is: " << count << std::endl;
    return 0;
}

In this code, we iterate through the numbers from 1 to 1000, calculate the sum of their digits, and then check if the sum is a prime number. If it is, we increment the count. Finally, we output the total count of such numbers.

This approach efficiently solves the problem using nested loops and a prime number checking function.

Let me know if you need further assistance with this or any other topic!