Question 1162656
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            Here is another  ("elegant")  solution,  which tutor @greenestamps wants to see

            (and you, probably,  too).



The base for this solution is the same rule of divisibility by  9:


<pre>
    The number is divisible by 9 if and only if the sum of its digits is divisible by 9.
</pre>

The more general form of this rule is


<pre>
    For any number, the remainder of dividing by 9 is the same as the remainder of dividing by 9 the sum of its digits.
</pre>

For both forms, &nbsp;see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-9-rule.lesson>Divisibility by 9 rule</A>  

in this site.


<pre>
It follows immediately from these rules, that the remainder of division the number L(m) by 9 
is equal to the remainder of divisibility by 9 the number


    1 + 2 + 3 + . . . + 9 + 10 + 11 + 12 + . . . + 99 + 100 + 101 + 102 + . . . 999 + 1000 + 1001 + 1002 + . . . + the last 4-digit number m = 

    = the sum of the first "m" natural numbers = {{{(m*(m+1))/2}}}.



So, all we need is to check the remainder of dividing by 9 of several numbers of the form  {{{(m*(m+1))/2}}},
starting from m = 2021, 2022, 2023, 2024 . . . 



The table below (made using the function r = mod(N,9) of Excel) shows these remainders


    m                              2021    2022   2023  2024

    r = {{{mod((m*(m+1))/2,9)}}}          6       3      1     0



Thus that minimal value of "m" the problems is asking for is  2024.    <U>ANSWER</U>
</pre>

Solved.