Project Euler : Finding The nth Digit Of The Fractional Part Of Irrational Number

The Problem 40 of Project Euler required just a pencil and a paper to get the answer. We are given an irrational decimal fraction by concatenating consecutive integers:

0.123456789101112131415 …

The $12^{th}$ digit of this irrational number is 1 and is denoted as $d_{12}$ . The problem requires us to determine the product:

$d_1$ x $d_{10}$ x $d_{100}$ x $d_{1000}$ x $d_{10000}$ x $d_{100000}$ x $d_{1000000}$

A simple enumeration of digits and their positions works out easily.

To begin with, there are only 9 single digit numbers, namely, 1 – 9. So the first 9 places are occupied by single digit numbers. Next, there are 90 double digit numbers from 10 – 99. So these occupy a total of 180 places. We have now accounted for 189 places in out irrational number.

Determine the digits $d_{100}$ and $d_{1000}$ are as simple as illustrated below:

For $d_{100}$ :

Excess from last known digit position = 100 – 9 = 91

Numbers crossed in 91 positions = 91 / 2 = 45

Since the remainder is 1, we want the 1st digit of the 46th number from the one that appeared on position 9.

Thus the number we seek is 9 + 46 = 54, and the digit is therefore 5.

For $d_{1000}$ :

Excess from last known digit position = 1000 – 189 =18 1

Numbers crossed in 811 positions = 811 / 3 = 270

Since the remainder is 1, we want the 1st digit of the 271st number from the one that appeared on position 189.

Thus the number we seek is 99 + 271 = 370, and the digit is therefore 3.

Repeating the same for all the digits we needed, the result is:

$d_{1}$ = 1

$d_{10}$ = 1

$d_{100}$ = 5

$d_{1000}$ = 3

$d_{10000}$ = 7

$d_{100000}$ = 2

$d_{1000000}$ = 1

Thus the product is 210

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Project Euler : Finding The nth Digit Of The Fractional Part Of Irrational Number

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