Question 992526
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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;28 &nbsp;and &nbsp;29 &nbsp;of February of a leap-year 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;(28 + 29 + 1 + 2 + 3 = 63). 


<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 in 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;63&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 that the dates &nbsp;28, &nbsp;29, &nbsp;1, &nbsp;2 &nbsp;and 3 &nbsp;satisfy the condition &nbsp;28 + 29 + 1 + 2 + 3 = 63. 

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



This problem was just offered and solved in this forum one-two weeks ago (# 990684, the link 

http://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.990684.html ).


The problem and the solution were also placed in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/word/misc/Spending-money-according-to-dates.lesson?content_action=show_source>Spending money according to dates</A>&nbsp; in this site.