Question 987601
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<B>Answer</B>. &nbsp;There is only one solution:

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;27 and 28 &nbsp;of February and &nbsp;1, 2, &nbsp;and &nbsp;3 &nbsp;of March.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(27 + 28 + 1 + 2 + 3 = 61). 


<B>Solution</B>


If these days would be inside one month, &nbsp;then the dates are &nbsp;(x-2), &nbsp;(x-1), &nbsp;x, &nbsp;(x+1) &nbsp;and &nbsp;(x+2), &nbsp;where x is the date of the middle of &nbsp;5 days. 

Then the sum must be multiple of &nbsp;5, &nbsp;since 


(x-2) + (x-1) + x + (x+1) + (x+2) = 5x.


But the integer &nbsp;61&nbsp; is not multiple of &nbsp;5. &nbsp;Contradiction. 


Hence, &nbsp;the dates are partly the end of some month and the beginning of the next month. 


Then, &nbsp;it is easy to check the the dates &nbsp;27, &nbsp;28, &nbsp;1, &nbsp;2 &nbsp;and 3 &nbsp;satisfy the condition &nbsp;27 + 28 + 1 + 2 + 3 = 61. 

Next, &nbsp;it is easy to see that there is no other solution.