Made the first one as anonymous, sorry! This is a program is my solution to a Google c++ problem with the following bounds: This is a number analogy to a famous card trick. Ask the user to enter a three-digit number. Think of the number as ABC (where A, B, C are the three digits of the number). Now, find the remainders of the numbers formed by A…

gistfile1.txt

#include <iostream>
using namespace std;
 
int firstNumber(int x) // from left to right. Finds the number in the hundreds place.
{
    int p, q;
    p = x % 100;
    q = ((x - p) / 100);
    return q;
 
 
}
 
int secondNumber(int x) // finds the number in the tens place
{
    int p;
    p = ((x % 100) * 10) / 100;
    return p;
}
 
int thirdNumber(int x) // finds the number in the ones place
{
    int p;
    p = x % 10;
    return p;
}
 
int findRemainder(int hundreds, int tens, int ones) // finds the remainder of the different Number orders entered and checks      // for factors of 11
{
    int newNumber;
    hundreds = hundreds * 100;
    tens = tens * 10;
    newNumber = (hundreds + tens + ones) % 11;
 
    return newNumber;
 
}
 
bool isOdd(int x) // checks to see if the number is odd or even
{
    if((x % 2) == 0)
        return false;
    else
        return true;
}
 
int lessThan20(int x) // Checks that the Number entered is less than 20 or equal to 9 and adds or substracts 11. 
{
   if(x < 9)
        x = (x + 11) / 2;
    else if (x > 9)
        x = (x - 11) / 2;
    else if(x == 9)
        return x;
    return x;
}
int equation( int X, int Y, int Z) // a function to solve the equation after splitting the number in 3
{
    int U,P,S; // created ups because I don't want to get the variable values confused
    U = X + Y;
    P = Y + Z;
    S = Z + X;
    if( isOdd(U) == true)
    {
        U = lessThan20(U);
    }
    else
        U = U / 2;
    if( isOdd(P) == true)
    {
        P = lessThan20(P);
    }
    else
        P = P / 2;
    if( isOdd(S) == true)
    {
        S = lessThan20(S);
    }
    else
        S = S / 2;
 
 
    cout << "A: " << U << endl;
    cout << "B: " << P << endl;
    cout << "C: " << S << endl;
 
    return 3;
 
}
int main() {
  int ABC, A, B, C, X, Y, Z;
  cout << "Enter a 3 digit Number: ";
  cin >> ABC;
  if((ABC > 99) && (ABC < 1000)) // Error checking. Makes sure the user enters a 3 digit number
  {
 
    A = firstNumber(ABC);
    B = secondNumber(ABC);
    C = thirdNumber(ABC);
    X = findRemainder(A, B, C);
    Y = findRemainder(B, C, A);
    Z = findRemainder(C, A, B);
    equation(X, Y, Z);
    return 1;
  }
  else
  {
      cout << "The number must be 3 digits!" << endl;
      return 0;
  }
 
 
}