π \pi madness 1

Computer Science Level 4

In the first 1 , 000 , 000 1,000,000 digits of π \pi , there are 2 2 digits that occur the same number of times. Let the number of times it occurs be A A , and the digits be B B and C C . Find A × B × C A\times B\times C

Feel free to search for any derivations of π \pi

Details and Assumptions \textbf{Details and Assumptions}

∙ \bullet The first 1 , 000 , 000 1,000,000 digits of π \pi includes the 3 3 in 3.1415... 3.1415... , so if the number is 3.1415 3.1415 there are 5 5 digits in that number.

∙ \bullet Supposing the 2 2 digits that occur the same number of times is 1 1 and 2 2 and the number of times it occurs is 100 100 times, the answer would be 1 × 2 × 100 = 200 1\times 2\times 100=\boxed{200}

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The answer is 1202760.

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Julian Poon by Julian Poon , 20, Singapore

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4 solutions

Anatoliy Razin
Nov 20, 2014

Amazing fact and good subject for algorithm based solution

Easy thing if using tools: DigitCount[IntegerPart[Pi*10^1000000]] - takes a second

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Shouldn't this be mentioned that digits include "3" (integral part) also.

Pranjal Jain - 6 years, 5 months ago
Gabriel Alves
Feb 7, 2015

My c++ solution: 

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#include <iostream>
#include <fstream>
#include <cstring>

using namespace std;

int main()
{

    string pi;

    fstream fin ("name_of_your_txt_file_with_1000000_digits_of_pi");

    fin>>pi;

    int rep[10]; //Number of repetitions of each digit
    int i;

    for(i=0;i<=9;i++)  rep[i]=0;

    for(i=0;i<=1000000;i++){
        cout<<pi[i]<<"\n";
        switch(pi[i]){
            case '0':rep[0]++;break;
            case '1':rep[1]++;break;
            case '2':rep[2]++;break;
            case '3':rep[3]++;break;
            case '4':rep[4]++;break;
            case '5':rep[5]++;break;
            case '6':rep[6]++;break;
            case '7':rep[7]++;break;
            case '8':rep[8]++;break;
            case '9':rep[9]++;break;
        }
    }

    for(i=0;i<=9;i++) cout<<i<<" repeats "<<rep[i]<<" times\n";

    return 0;
}

I downloaded a txt file with 1000000 digitis of Pi and used the fin command to read it.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Ilya Andreev
Dec 31, 2014

Here's my Pylution, and a link to 1 million digits of Pi 

countList = []
for i in range(10):
f = open('C:\Users\[your username]\Documents\Python\pi.txt', 'r')
totalCount = 0
    for line in f:
        count = 0
        for char in line:
            if char == str(i):
                count += 1
        totalCount += count
    countList.append(totalCount)

for i in range(9):
    for k in range(i+1,10):
        if countList[i] == countList[k]:
            print (i,k,countList[i],i*k*countList[i])
1 Helpful 0 Interesting 0 Brilliant 0 Confused

Pylution. Happy new year!

Julian Poon - 6 years, 5 months ago

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Hey, you too! ;]

Ilya Andreev - 6 years, 5 months ago
Aryan Gaikwad
Feb 27, 2015

Easier way - go to http://www.exploratorium.edu/pi/pi_archive/Pi10-6.html and then press Ctrl + F. A box will open. Type a digit and number of times it is repeated will be displayed on it. Find out for each digit.

Only problem was that I realized it after writing my algorithm.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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