To solve this problem, we can follow these steps: 1. First, we need to create a function named `isUnitary` that takes an integer as input. This function will check if the number is a Unitary Perfect Number. 2. In the `isUnitary` function, we initialize a variable `sum` to 0. This variable will hold the sum of the proper divisors of the input number. 3. We then run a loop from 1 to the input number. For each iteration, we check if the current number is a divisor of the input number. If it is, we add it to `sum`. 4. After the loop, we check if `sum` is equal to the input number. If it is, the function returns true, indicating that the number is a Unitary Perfect Number. Otherwise, it returns false. 5. In the main function, we take an integer as input from the user. We then call the `isUnitary` function with this number. 6. Based on the return value of the `isUnitary` function, we print the appropriate message. Here is a C++ code snippet that implements the above steps: ```cpp #include using namespace std; bool isUnitary(int *n) { int sum = 0; for(int i = 1; i n; if(isUnitary(&n)) { cout

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To solve this problem, we can follow these steps: 1. First, we need to create a function named `isUnitary` that takes an integer as input. This function will check if the number is a Unitary Perfect Number. 2. In the `isUnitary` function, we initialize a variable `sum` to 0. This variable will hold the sum of the proper divisors of the input number. 3. We then run a loop from 1 to the input number. For each iteration, we check if the current number is a divisor of the input number. If it is, we add it to `sum`. 4. After the loop, we check if `sum` is equal to the input number. If it is, the function returns true, indicating that the number is a Unitary Perfect Number. Otherwise, it returns false. 5. In the main function, we take an integer as input from the user. We then call the `isUnitary` function with this number. 6. Based on the return value of the `isUnitary` function, we print the appropriate message. Here is a C++ code snippet that implements the above steps: ```cpp #include using namespace std; bool isUnitary(int *n) { int sum = 0; for(int i = 1; i < *n; i++) { if(*n % i == 0) { sum += i; } } return sum == *n; } int main() { int n; cin >> n; if(isUnitary(&n)) { cout << n << " is a unitary perfect number." << endl; } else { cout << n << " is not a unitary perfect number." << endl; } return 0; } ``` This program works by taking an integer as input, checking if it's a Unitary Perfect Number using the `isUnitary` function, and printing the appropriate message.

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To solve this problem, we can follow these steps: 1. First, we need to create a function named `isUnitary` that takes an integer as input. This function will check if the number is a Unitary Perfect Number. 2. In the `isUnitary` function, we initialize a variable `sum` to 0. This variable will hold the sum of the proper divisors of the input number. 3. We then run a loop from 1 to the input number. For each iteration, we check if the current number is a divisor of the input number. If it is, we add it to `sum`. 4. After the loop, we check if `sum` is equal to the input number. If it is, the function returns true, indicating that the number is a Unitary Perfect Number. Otherwise, it returns false. 5. In the main function, we take an integer as input from the user. We then call the `isUnitary` function with this number. 6. Based on the return value of the `isUnitary` function, we print the appropriate message. Here is a C++ code snippet that implements the above steps: ```cpp #include using namespace std; bool isUnitary(int *n) { int sum = 0; for(int i = 1; i < *n; i++) { if(*n % i == 0) { sum += i; } } return sum == *n; } int main() { int n; cin >> n; if(isUnitary(&n)) { cout << n << " is a unitary perfect number." << endl; } else { cout << n << " is not a unitary perfect number." << endl; } return 0; } ``` This program works by taking an integer as input, checking if it's a Unitary Perfect Number using the `isUnitary` function, and printing the appropriate message.

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