Question 1162751
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            The method of  Mathematical  Induction is  IRRELEVANT  to this problem.


            The proof is constructed based on other principles.




&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>THE PROOF</U>



Let an arbitrary set of &nbsp;37 positive integer numbers is given.


I organize &nbsp;7 boxes numbered from &nbsp;0 &nbsp;to &nbsp;6.  &nbsp;So, &nbsp;the boxes are numbered &nbsp;0, 1, 2, 3, 4, 5 and 6.


I distribute all these &nbsp;37 numbers in these &nbsp;7 boxes according their remainders modulo 7.



If some box has at least &nbsp;7 numbers, &nbsp;then these &nbsp;7 numbers provide the required sum.


 
If there is no a box with at least &nbsp;7 numbers, &nbsp;it means that each box has no more than &nbsp;6 numbers and all boxes have no more than &nbsp;6 numbers.


If all boxes have at least one number, &nbsp;then their sum  &nbsp;0 + 1 + 2 + 3 + 4 + 5 +  6 = 21 &nbsp;&nbsp;(mod7) &nbsp;&nbsp;is divisible by &nbsp;7.


If not all boxes have at least one number, &nbsp;it means that at least one box is empty.


In this case, &nbsp;we have &nbsp;6 boxes with no more than &nbsp;6 numbers in each, &nbsp;which gives &nbsp;6*6 = 36 numbers.


But then the &nbsp;37-th number breaks this scheme.



So, &nbsp;having &nbsp;37 numbers in &nbsp;7 boxes, &nbsp;we &nbsp;EITHER &nbsp;can find a box with at least &nbsp;7 numbers in it 


<pre>
     and then this box provides the required sum of 7 numbers,
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OR, &nbsp;otherwise, &nbsp;all &nbsp;7 boxes have at least one number each,


<pre>
    and then we can form the sum  0 + 1 + 2 + 3 + 4 + 5 + 6 = 21 (mod7)

    of numbers from these boxes which is divisible by 7.
</pre>


The proof is completed.



The key to the proof &nbsp;(and to the statement itself) &nbsp;is this equality &nbsp;&nbsp;37 = 6*6 + 1.



Done.