Solution

Answer: 210

We construct the irrational decimal

$$0.12345678910111213141516\cdots$$

by concatenating the positive integers in order.

Let $d_n$ denote the $n$-th digit after the decimal point.

We must compute

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

1. Mathematical analysis

The key idea is to locate which integer contains the $n$-th digit.

We group integers by their number of digits.

Step 1: Count digits contributed by each block

1-digit numbers

Numbers:

$$1,2,\dots,9$$

There are $9$ numbers, each contributing $1$ digit.

Total digits:

$$9 \times 1 = 9$$

2-digit numbers

Numbers:

$$10,11,\dots,99$$

There are

$$90$$

numbers, each contributing $2$ digits.

Total digits:

$$90 \times 2 = 180$$

Cumulative total:

$$9 + 180 = 189$$

3-digit numbers

$$100 \text{ to } 999$$

Count:

$$900$$

Digits contributed:

$$900 \times 3 = 2700$$

Cumulative:

$$2889$$

In general:

$k$-digit numbers contribute

$$9 \cdot 10^{k-1} \cdot k$$

digits.

Finding $d_n$

To determine $d_n$:

Determine which digit-length block contains position $n$.
Determine which exact number inside that block.
Determine which digit of that number.

Example: $d_{12}$

The first 9 digits come from $1$-digit numbers.

So position $12$ lies among the $2$-digit numbers.

Relative position inside the 2-digit block:

$$12 - 9 = 3$$

Each number contributes 2 digits.

$$(3-1)//2 = 1$$

So we are at the second 2-digit number after 10:

$$11$$

The digit index:

$$(3-1)\bmod 2 = 0$$

which is the first digit of $11$, namely:

$$1$$

So $d_{12}=1$, matching the statement.

Compute the required digits

We now determine:

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

The resulting digits are:

$$1,\ 1,\ 5,\ 3,\ 7,\ 2,\ 1$$

Thus the product is

$$1 \times 1 \times 5 \times 3 \times 7 \times 2 \times 1$$

$$= 210$$

2. Python implementation

def champernowne_digit(n):
    """
    Return the nth digit of Champernowne's constant:
    0.123456789101112...
    """

    digit_length = 1
    count = 9
    start = 1

    # Find which block contains the nth digit
    while n > digit_length * count:
        n -= digit_length * count
        digit_length += 1
        count *= 10
        start *= 10

    # Find the actual number containing the digit
    number = start + (n - 1) // digit_length

    # Find the digit inside that number
    digit_index = (n - 1) % digit_length

    return int(str(number)[digit_index])

# Required positions
positions = [1, 10, 100, 1000, 10000, 100000, 1000000]

product = 1

for p in positions:
    product *= champernowne_digit(p)

print(product)

3. Code walkthrough

Function definition

def champernowne_digit(n):

Computes the $n$-th digit.

Initialize block information

digit_length = 1
count = 9
start = 1

These represent:

current digit size,
how many numbers in the block,
first number in the block.

Initially:

1-digit numbers,
9 such numbers,
starting at 1.

Locate the correct block

while n > digit_length * count:

If $n$ exceeds the digits in the current block, subtract them and move on.

Example:

2-digit block contributes $2 \times 90 = 180$.

Find the exact number

number = start + (n - 1) // digit_length

This computes which number within the block contains the target digit.

Find the digit position

digit_index = (n - 1) % digit_length

Determines which digit inside the number is needed.

Return the digit

return int(str(number)[digit_index])

Convert to string and extract the digit.

Small-example verification

For $n=12$:

after removing the first 9 digits,

remaining position is $3$ inside the 2-digit block.

second 2-digit number is $11$,
first digit is $1$.

Correct.

4. Final verification

The digits found are:

$$d_1=1,\quad d_{10}=1,\quad d_{100}=5,\quad d_{1000}=3,\quad d_{10000}=7,\quad d_{100000}=2,\quad d_{1000000}=1$$

Their product:

$$1 \cdot 1 \cdot 5 \cdot 3 \cdot 7 \cdot 2 \cdot 1 = 210$$

The value is small and reasonable since many extracted digits are small.

Answer: 210

Project Euler Problem 40