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Arrange given numbers to the equations, so if we turn the equations upside down, they are still right?

+1 vote
235 views

Numbers to be arranged

_ + _ = _ 
_ + _ = _ 
_ + _ = _ 
_ + _ = _ 

Fill the blanks with numbers above, so:
* The 4 additions are right.
* All numbers must be used.
* If we turn the additions upside down, they are still right.
* You can rotate any numbers too, example 1091 -> 1601.

posted Dec 5, 2016 by anonymous

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1 Answer

0 votes

158 + 1998 = 2156
851 + 8661 = 9512 (flipped)

1601 + 1051 = 2652
1091 + 1501 = 2592 (flipped)

1011 + 1811 = 2822
1101 + 1181 = 2282 (flipped)

1081 + 1901 = 2982
1801 + 1061 = 2862 (flipped)

answer Dec 6, 2016 by Tejas Naik
Used the following c code but the output has 8 solutions with 4 of them repeats:
#include<stdio.h>
#include<math.h>
int ud(int c);
int main()
{
    int i,j,x,y,c,check,check1,k;
   
    int a[]={851,1091,1101,1181,1501,1801,1901,1998,2652,2822,2982,9512,2156,2862,2282,2592,8661,1061,1081,1051,1811,1011,1601,158};
   
    for(i=0;i<24;i++)
    {
        for(j=(i+1);j<24;j++)
        {
            if(i!=j)
            {
           
            c=(a[i]+a[j]);
            x=a[i];
            y=a[j];
            c=ud(c);
            x=ud(x);
            y=ud(y);
           
            check=0;
            check1=0;
            for(k=0;k<24;k++)
            {
                if(a[k]==c)
                {
                    check=1;
                }
                if(a[k]==(a[i]+a[j]))
                {
                    check1=1;
                }
            }
       
                if(c==(x+y)&&(check==1)&&(check1==1))
                {
                printf("%d + %d = %d\n",x,y,c);
                printf("%d + %d = %d\n\n",a[i],a[j],(a[i]+a[j]));
               
                }
            }
        }
    }
}

int ud(int c)
{
    int i,udc,countc,returnc;
    udc=c;
    returnc=0;
    countc=0;
   
    while(udc>=1)
    {
        countc=countc+1;
        udc=udc/10;
    }
   
    udc=c;
    i=0;
    while(countc>0)
    {
        if(int(fmod((udc/pow(10,(countc-1))),10))!=6&&int(fmod((udc/pow(10,(countc-1))),10))!=9)
        {
        udc=c;
        returnc=(returnc+(int(fmod((udc/pow(10,(countc-1))),10))*pow(10,i)));
        i++;
        countc--;
        }
        if(int(fmod((udc/pow(10,(countc-1))),10))==6)
        {
        udc=c;
        returnc=(returnc+(9*pow(10,i)));
        i++;
        countc--;   
        }
        if(int(fmod((udc/pow(10,(countc-1))),10))==9)
        {
        udc=c;
        returnc=(returnc+(6*pow(10,i)));
        i++;
        countc--;   
        }
       
   
   
    }
   
    return returnc;
   

}



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