The Quest for the Boolean of Truth

Step 1) Download file

Step 2) Open notepad of said file

Step 3) Copy a couple of first few 0's from the said notepad into Microsoft word

Step 4) Delete pages 2,3,4, and so on.

Step 5) Ctrl+F and replace "0" with "z"

Step 6) Note the total number of "0"s that can fit into one page: which is 3600.

Step 7) Calculate the number of zeros in one row: 80

Step 8) Copy the entire file into a new Microsoft Word spreadsheet

Step 9) Wait for awhile because it's a large file

Step 10) Ctrl+F: Search for number 1

Step 11) Wait again

Step 12) Record page number: 5389

Step 13): See page 5389

Step 14) Calculation time:

So there's 5388 pages full of zeros + the last page which has $80 \times 22 - 26$ zeros.

The number of zeros before the number 1 is $5388 \times 3600 + 80\times 22 -26 = 19398534$ .

But since we start with 0 instead of 1, we need to subtract one. Answer is $19398533$ .

Python one liner -

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open('sea.txt', 'r').read().index('1')

Java solution -

EDIT

Seems like Java also provides an index method -

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new IO("sea").read().indexOf('1');

Old solution -

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String n = (new IO("sea")).read(); //IO is a class written by me to read files.
for(int i = 0;; i++)
    if(n.charAt(i) == '1')
        System.out.println(i);

I wrote solutions in both languages because I considered the first index as $1$ instead of $0$ .

Python 3.3:

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position = 0
with open('sea.txt', 'r') as sea:
    while sea.read(1) != '1':
        position += 1
print("BOOLEAN OF TRUTH found at", position)

x=0

a=open('C:\Users\User\Downloads\sea.txt','r')

for b in a:

for c in b:

    if c=='1':

        print c,x

    x=x+1