Question 1119742
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<pre>
If you write all 4-digit numbers from 0 to 9999 in column

    0000
    0001
    0002
    0003

    . . . 

    9999


replacing/filling by zeroes the leading positions in 1-digit, 2-digit and 3-digit numbers,

you will have 10000 = {{{10^4}}} numbers (= lines), each containing 4 digits.


So, the total number of positions (place-holders) for digits in these {{{10^4}}} lines is  {{{4*10^4}}}.


Now, it is OBVIOUS ( ! <U>attention</U> ! - <U>It is the key idea of the solution</U> !), each digit from 0 to 9 does appear equal number of times 

in these place-holders.


Therefore, each digit (including '1') appears  {{{(4*10^4)/10}}} = {{{4*10^3}}} = 4000 times.


<U>Answer</U>.  The digit '1' appears 4000 times if we write all whole numbers from 1 to 9999.
</pre>

Solved.


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It is a nice Math circle level problem.