Number of integer solutions to the equation x = b*(sumofdigits(x) ^ a)+c

Suppose three integers a, b and c are given, and there is an equation x = b* (sumofdigits(x)^a) c. Here, sumofdigits(x) is the sum of all digits in x. We will explore various methods in C in order to find all possible integral solutions that satisfy the equation.

Input and output scenarios

Given below are the values ​​of a, b and c. Different integral solutions satisfying the equation x = b* (sumofdigits(x)^a) c are given as output.

Input: a = 2, b = 2, c = -3
Output: 125, 447, 575

In the above case, a has a value of 2, b has a value of 2, c has a value of -3, and the possible values ​​of x are 125, 447, and 575.

Consider the number 125. The sum of its digits is 8. If you substitute this value into the equation b*(sum(x)^a) c, the answer is 125, which is the same as x equal. Therefore, it is a possible solution to Eq.

NOTE- The integral solution of this equation is in the range 1 to 10⁹.

Use recursion

We can use recursive search to find the integral solution of a given equation.

We need to create a function called sumOfDigits() that will calculate the sum of the digits for any given number N.

• Iterate over N numbers using the modulo operator and division operator.

• The modulo operator is used to extract the last digit of N.

• After each iteration, add the numbers stored in the variable sum one by one.

We create an integralSolutions() function to calculate the integral solutions.

• It calculates the sum of the digits of x using the sumOfDigits function.

• Next, using a for loop we raise the sum to the power a.

• We evaluate the right side of the equation by multiplying b by power and adding c.

• If the value of x is equal to the value on the right, it is considered an integer solution.

Next, we have the recursive function to search for integral solutions within a specified range.

Example

#include <iostream>
using namespace std;

int sumOfDigits(int N) {
   int sum = 0;
   while (N != 0) {
      sum += N % 10; // addition of the last digit of N
      N /= 10;
   }
   return sum;
}
void integralSolutions(int x, int a, int b, int c) {
   int sum = sumOfDigits(x);
   int power = 1;
   for (int j = 0; j < a; j++) {
      power *= sum;
   }
   int rightHandSide = b * power + c;
   if (x == rightHandSide) {
      std::cout << "Integral solution: " << x << std::endl;
   }
}
void recursion(int start, int end, int a, int b, int c) {
   if (start > end) {
      return;
   }
   integralSolutions(start, a, b, c);
   recursion(start + 1, end, a, b, c);
}
int main() {
   int a = 1, b = 3, c = 5;
   recursion(1, 100000, a, b, c);
   return 0;
}

Output

Integral solution: 11
Integral solution: 38

Segmentation Fault This error occurs when the end value of the specified range in a recursive search exceeds 100000. So you can't have x-values ​​beyond that.

Use simple iteration

If you want an integer solution for x greater than 100000, then we don't use recursion. Here we will use a simple iteration of x from 1 to 109 and compare it to the value on the right side of the equation.

Example

#include <iostream>
using namespace std;

int sumOfDigits(int N) {
   int sum = 0;
   while (N != 0) {
      sum += N % 10;
      N /= 10;
   }
   return sum;
}

bool integralSolution(int x, int a, int b, int c) {
   int sum = sumOfDigits(x);
   int power = 1;
   for (int i = 0; i < a; i++) {
      power *= sum;
   }
   int rightHandSide = b * power + c;
   return x == rightHandSide;
}

int main() {
   int a = 3, b = 5, c = 8;
   // x ranges from 1 to 109
   for (int x = 1; x <= 1000000000; x++) {
      if (integralSolution(x, a, b, c)) {
         std::cout << "Integral solution: " << x << std::endl;
      }
   }
   return 0;
}

Output

Integral solution: 53248
Integral solution: 148963

in conclusion

We explored methods of finding integral solutions to the equation x = b* (sumofdigits(x)^a) c, including using recursion or simple iteration. Recursive methods allow you to flexibly specify the range of solutions. However, it increases time complexity and may show segmentation fault for larger range of values, resulting in stack overflow.

Iterative methods are efficient in terms of time complexity and memory usage. However, it offers limited flexibility and more complex code. Therefore, both methods have their own advantages and disadvantages. Depending on your needs, you can choose any of the methods.

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