Numbers of another world

Computer Science Level 4

JBEMRA ‾ × PTPMIR ‾ = IBSMTBRERMAS ‾ \displaystyle \large \overline {\text{ JBEMRA} } \times \overline { \text{PTPMIR} } =\overline {\text{ IBSMTBRERMAS }}

Given above is a mathematical expression in base 10 10 . The expression is written in some alien language where each digit is replaced with a particular letter.

Find the number associated with EABMRPSTJIRB ‾ \overline { \text{EABMRPSTJIRB} }

Inspiration.


The answer is 926754018356.

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Raghav Vaidyanathan by Raghav Vaidyanathan , 23, USA

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3 solutions

Pranjal Jain
Mar 22, 2015 
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from itertools import permutations
for i in permutations('0123456789'):
    x=''.join(i)
    J=x[0]
    B=x[1]
    M=x[2]
    E=x[3]
    S=x[4]
    T=x[5]
    A=x[6]
    P=x[7]
    R=x[8]
    I=x[9]
    if int(J+B+E+M+R+A)*int(P+T+P+M+I+R)==int(I+B+S+M+T+B+R+E+R+M+A+S):
        print(int(E+A+B+M+R+P+S+T+J+I+R+B))

Returns 926754018356

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Aditya Raut 's algorithm.

Raghav Vaidyanathan - 6 years, 2 months ago

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Yeah right! I use just the same in all such problems.... in this problem, it took 18 seconds to answer....

Aditya Raut - 6 years, 2 months ago

@Pranjal Jain you don't have to import * from math or itertools, you'll have to import just "permutations" from itertools. The multiplication use of Asterisk (*) is inbuilt in python, so it'll do that even if you import just permutations and nothing else...

Aditya Raut - 6 years, 2 months ago

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I agree there is no use to import math, but I usually import math (always) in case I need some basic functions somewhere. Same with itertools, and laziness also, len('*')<len('permutations') Lol XD

Pranjal Jain - 6 years, 2 months ago

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Hey wait!!!! NO! You have to import that asterisk(*) as well as permutations (from itertools) if you go by what you had previously written (b4 edit)

So ` len('*') < len('permutations') is a sentence of no use, though it meant about typing number of characters :P ...

Aditya Raut - 6 years, 2 months ago

Anyways, I have edited it out.

Pranjal Jain - 6 years, 2 months ago

Here is the genetic algorithm code I used to solve this problem: 

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#include<fstream.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>

class indi       //Objects of this class are the individuals
    {
    char x[11];  //genetic code/DNA
    public:
    indi()
        {
        strcpy(x,"EBMJPRSTIA"); //initial DNA
        }
    void mutate() //causes random switching to genetic code
        {
            int a=(rand()%10),b=(rand()%10);
            char t=x[a];
            x[a]=x[b];
            x[b]=t;
        }
    void show() //prints DNA
        {
        cout<<x<<endl;
        }
    int retlet(char a)//returns respective digit that a letter represents
        {
        for(int i=0;i<10;i++)
            if(x[i]==a)
                return i;
        }

    };

double fitness(indi I,char c1[],int n1,char c2[],int n2,char c3[],int n3)
    {
    //used to calculate the fitness level of the DNA
    //more fitness implies better evolved
    double a=0,b=0,c=0;
    for(int i=n1-1;i>=0;i--)
        a+=pow(10,n1-i-1)*I.retlet(c1[i]);
    for(int i=n2-1;i>=0;i--)
        b+=pow(10,n2-i-1)*I.retlet(c2[i]);
    for(int i=n3-1;i>=0;i--)
        c+=pow(10,n3-i-1)*I.retlet(c3[i]);
    double f=abs((a*b)-c);
    return f;
    }

void nature(char S1[],char S2[],char S3[]) //implements natural selection
    {
    indi I[100]; //first 100 DNA
    int ss[3];
    ss[0]=strlen(S1);
    ss[1]=strlen(S2);
    ss[2]=strlen(S3);
    for(int gen=0;;gen++)
        {
        double b=fitness(I[99],S1,ss[0],S2,ss[1],S3,ss[2]);int a=99;
        for(int i=0;i<100;i++)
           {
           I[i].mutate();  //random mutation
           double k=fitness(I[i],S1,ss[0],S2,ss[1],S3,ss[2]);
           if(k<b)
               {
               b=k;a=i;      //fittest DNA chosen
               }
           }
        if(b==0) //if DNA is perfect, print DNA
            {
            I[a].show();
            cout<<endl<<gen;
            break;
            }
        indi II=I[a];
        for(int i=0;i<100;i++)
           I[i]=II;     //fittest DNA of current generation is cloned 100 times
        }
    }

int main()
    {
    clrscr();
    char a[]="PTPMIR",b[]="JBEMRA",c[]="IBSMTBRERMAS";
    nature(a,b,c);
    system("pause");
    return 0;
    }

1 Helpful 0 Interesting 0 Brilliant 0 Confused 
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from itertools import permutations

for j,b,e,m,r,a,p,t,s,i in permutations('0123456789',10):
  if int(j+b+e+m+r+a)*int(p+t+p+m+i+r)==int(i+b+s+m+t+b+r+e+r+m+a+s):
    print(e+a+b+m+r+p+s+t+j+i+r+b)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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