numbers for a certain requirements

[zandiago]

There are two sections to this program.
Section one:
Write a C++ program that determines which four digit numbers meet the following conditions:
Add the first two digits as a number to the last two digits as a number. The square of this sum equals the original number.
Example:
Original number is 4321. Determine if 43 + 21 which equals 63, when squared, is 3969 is equal to the original number (4321).
Section two:
The program continues by determining which 2, 3, and 4 digit numbers are equal to the sum of the cubes of their digits.
Example:
If the original number is 234, determine if 2 cubed (8) + 3 cubed (27) + 4 cubed (64) which is 99 is equal to the original 234.

So far:

 #include <cstring>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
#include <string>
#include <ctime>
 
using namespace std;
 
int main ()
{
    int secA;
 
    for(int j=1000; j<9999; j++)
    {
 
        secA =partA(j) +  partB(j);
 
    if ((temp * temp) == j)
 
        cout<<"The number is: "<<j<<endl;
    }
 
    return 0;
}

For now, I'd like for the variables partA & partB, to be the two first and last two digits repectively. That solves section one the problem. I'm gonna need a jump-start on section two of the question. Your assistance is appreciated.

[Ganon11]

You correctly set the variable secA to the sum of partA and partB, but then you check whether the square of temp is equal to j. Where is temp even declared or used?

[zandiago]

#include <cstring>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
#include <string>
#include <ctime>
 
using namespace std;
 
int main ()
{
    int secA;
 
    for(int j=1000; j<9999; j++)
    {
 
        secA =partA(j) +  partB(j);
 
    if ((secA * secA) == j)
 
        cout<<"The number is: "<<j<<endl;
    }
 
    return 0;
}

[lumpybanana247]

Well, this would be a simple way, but proably a lot longer.
load the number as a string[string1], sepreate the sring [at pos 1 or 2]
then youd have two strings[string2 and string3]
then you could convert the two strings each to ints (int1 and int2)
then int3=int1+int2
then int4=int3*int3
then convert int4 to a string[string4]
if(string4==string1){THEY ARE EQUAL!}

[Ganon11]

OK, this looks good so far. Now what ideas fdo you have for writing partA and partB? How can you manipulate a number to get the first 2 digits and the last 2 digits?

HINT: If I take any 4 digit number (say 3954), I can break it down like this:

(3 * 1000) + (9 * 100) + (5 * 10) + (4 * 1)

(3 * 10**3) + (9 * 10**2) + (5 * 10**1) + (4 * 10**0)

where 10**n is 10 raised to the nth power.

[zandiago]

i was thinking something like this for section 1:

#include <cstring>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
#include <string>
#include <ctime>
 
using namespace std;
 
int main ()
{
    int secA, partA, partB;
 
    for(int j=1000; j<9999; j++)
    {
        partA = j/100;
        partB = j%100;
 
        secA =partA(j) +  partB(j);
 
    if ((secA * secA) == j)
 
        cout<<"The number is: "<<j<<endl;
    }
 
    return 0;
}

[Ganon11]

That will work, but the parentheses you have after partA and partB makes the computer think you are using a function. Get rid of those parentheses and you'll be OK.

SIDE NOTE: With your for loop running with (int j = 1000; j < 9999; j++), you're getting values 1000 to 9998, but 9999 is also a 4 digit number. Maybe you should change that loop.

[zandiago]

#include <cstring>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
#include <string>
#include <ctime>
 
using namespace std;
 
int main ()
{
    int i, j, k, m, digits;
    int number = 0;
    int sum_cubes=0;
 
    for( int i = 1 ; i < 10 ; ++i )
    for( int j = 0 ; j < 10 ; ++j )
    for( int k = 0 ; k < 10 ; ++k )
    for( int m = 0 ; m < 10 ; ++m )
      {
          int number = i*1000 + j*100 + k*10 + m ;
          int digits = i*10 + j + k*10 + m ;
          if( digits * digits == number )
          cout<<"A # for Section 1 : "<<i<<j<<k<<m<<endl<<endl;
      }
 
   for( int i = 0 ; i < 10 ; ++i )
   for( int j = 0 ; j < 10 ; ++j )
   for( int k = 0 ; k < 10 ; ++k )
   for( int m = 0 ; m < 10 ; ++m )
      {
          int number = i*1000 + j*100 + k*10 + m ;
          if( number < 10 ) continue ;
          int sum_cubes = i*i*i + j*j*j + k*k*k + m*m*m ;
          if( sum_cubes == number )
          cout<<number<<sum_cubes<<endl;
      }
 
 
 
    return 0;
}

I made a revision to the code. A few things tho:
1. The first section of the question works, however section 2: The program continues by determining which 2, 3, and 4 digit numbers are equal to the sum of the cubes of their digits.
Example:
If the original number is 234, determine if 2 cubed (8) + 3 cubed (27) + 4 cubed (64) which is 99 is equal to the original 234.
My current output for section two, shows only 3 digit numbers.

[Ganon11]

#include <cstring>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <fstream>
#include <string>
#include <ctime>
 
using namespace std;
 
int main ()
{
    int i, j, k, m, digits;
    int number = 0;
    int sum_cubes=0;
 
    for( int i = 1 ; i < 10 ; ++i )
    for( int j = 0 ; j < 10 ; ++j )
    for( int k = 0 ; k < 10 ; ++k )
    for( int m = 0 ; m < 10 ; ++m )
      {
          int number = i*1000 + j*100 + k*10 + m ;
          int digits = i*10 + j + k*10 + m ;
          if( digits * digits == number )
          cout<<"A # for Section 1 : "<<i<<j<<k<<m<<endl<<endl;
      }
 
   for( int i = 0 ; i < 10 ; ++i )
   for( int j = 0 ; j < 10 ; ++j )
   for( int k = 0 ; k < 10 ; ++k )
   for( int m = 0 ; m < 10 ; ++m )
      {
          int number = i*1000 + j*100 + k*10 + m ;
          if( number < 10 ) continue ;
          int sum_cubes = i*i*i + j*j*j + k*k*k + m*m*m ;
          if( sum_cubes == number )
          cout<<number<<sum_cubes<<endl;
      }
 
 
 
    return 0;
}

I made a revision to the code. A few things tho:
1. The first section of the question works, however section 2: The program continues by determining which 2, 3, and 4 digit numbers are equal to the sum of the cubes of their digits.
Example:
If the original number is 234, determine if 2 cubed (8) + 3 cubed (27) + 4 cubed (64) which is 99 is equal to the original 234.
My current output for section two, shows only 3 digit numbers.

This code looks good. Have you considered the possibility that only 3 digit numbers satisfy this requirement?

[zandiago]

Well....true, however, one of the actual answers is : 351...however, 153 isn't listed..similarly for section 1.

[Ganon11]

(3 * 3 * 3) + (5 * 5 * 5) + (1 * 1 * 1) = 27 + 125 + 1 = 153, not 351. Maybe you're getting your number in improper order...Check which of i, j, k, and m you are using for each digit.

[zandiago]

(3 * 3 * 3) + (5 * 5 * 5) + (1 * 1 * 1) = 27 + 125 + 1 = 153, not 351. Maybe you're getting your number in improper order...Check which of i, j, k, and m you are using for each digit.

ok kool-thanks so much...problem solved...time to move on to my new post.(C++ shape calculation).