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  1. Prepare
  2. Mathematics
  3. Number Theory
  4. Identify Smith Numbers
  5. Discussions

Identify Smith Numbers

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  • 3699classical
     + 0 comments

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     + 0 comments

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  • 2k23_csai2310233
     + 0 comments
    public static int solve(int nn) {
    // Write your code here
                  int ds=0;
        int temp=nn;
        int n=nn;
        List<Integer> a=new ArrayList<>();
        while(temp>0){
            ds+=temp%10;
            temp/=10;
        }
        
         while (n % 2 == 0) {
           a.add(2);
            n /= 2;
        }
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            while (n % i == 0) {
                a.add(i);
                n /= i;
            }
        }
 
        if (n > 2)
           a.add(n);
        
        
        int fsum=0;
        
        for(int i:a)
        {
            while(i>0){
                fsum+=i%10;
                i=i/10;
            }
        }
        //System.out.println(a);
        if(fsum==ds)
            return 1;
        return 0;
    }

  • 2k23_cscys231212
     + 0 comments
    #include <vector>
#include <cmath>
using namespace std;

// Function to calculate the sum of digits of a number
int sumOfDigits(int n) {
    int sum = 0;
    while (n > 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

// Function to find the prime factors of a number
vector<int> primeFactors(int n) {
    vector<int> factors;
    for (int i = 2; i <= sqrt(n); ++i) {
        while (n % i == 0) {
            factors.push_back(i);
            n /= i;
        }
    }
    if (n > 1) {
        factors.push_back(n);
    }
    return factors;
}

// Function to check if a number is a Smith number
bool isSmithNumber(int n) {
    if (n < 2) return false; // Numbers less than 2 are not Smith numbers

    vector<int> factors = primeFactors(n);

    if (factors.size() == 1) {
        return false; // Prime numbers are not Smith numbers
    }

    int digitSum = sumOfDigits(n);
    int factorsDigitSum = 0;

    for (int factor : factors) {
        factorsDigitSum += sumOfDigits(factor);
    }

    return digitSum == factorsDigitSum;
}

int main() {
    int n;
    cin >> n;

    if (isSmithNumber(n)) {
        cout << 1 << endl;
    } else {
        cout << 0 << endl;
    }

    return 0;
}

  • zee296770
     + 0 comments

    import math

    def check_condition(n): # Function to compute the sum of digits of a number def sum_of_digits(num): return sum(int(digit) for digit in str(num))

    # Compute the sum of digits of (n-1)
sum_digits_n_minus_1 = sum_of_digits(n - 1)

# Initialize sum of prime factors
prime_factors_sum = 0

# Check for factors of 2
while n % 2 == 0:
    prime_factors_sum += 2
    n //= 2

# Check for odd factors from 3 upwards
for i in range(3, int(math.sqrt(n)) + 1, 2):
    while n % i == 0:
        prime_factors_sum += i
        n //= i

# If n is still greater than 2, it's prime and should be added
if n > 2:
    prime_factors_sum += n

# Calculate the sum of digits of the sum of prime factors
sum_digits_prime_factors_sum = sum_of_digits(prime_factors_sum)

# Compare the two sums modulo 9 + 1
if sum_digits_n_minus_1 == sum_digits_prime_factors_sum:

    Variable Naming: Renamed variables for clarity ( [sum](https://r2park.net/) _of_digits_of_numbers to sum_digits_n_minus_1, etc.).
    return 1
else:
    return 0