Question 993570
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It is about this problem again and again:


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Mr. &nbsp;<B>X</B>&nbsp; had a habit of spending money according to dates. &nbsp;For example, &nbsp;if date was &nbsp;19&nbsp; he was spending &nbsp;19 rupees, &nbsp;and if date was &nbsp;15&nbsp; he was spending &nbsp;15 rupees. 
One night he calculated total spending of &nbsp;5&nbsp; consecutive days - Monday to Friday, &nbsp;and he found that he spent &nbsp;63 rupees in &nbsp;5&nbsp; days. 
So, &nbsp;identify the dates. &nbsp;(Assume that everyday spending is integer number of rupees). 
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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 and 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.



The solution of this problem was placed in this forum one week ago 
http://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.992162.html .


The solution for the similar problem was placed in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/word/misc/Spending-money-according-to-dates.lesson>Spending money according to dates</A>&nbsp; in this site couple of weeks ago.