Question 990684
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Let me re-formulate the problem in more clear way:


Mr. Z had a habit of spending money according to dates. i. e. If date is 19 he was spending Rs.19 and if date is 15, he was spending Rs. 15. 
One night he calculated total spending of 5 consecutive days, and he found that he spent Rs. 63 in 5 days. 
So, identify the dates.


<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 &nbsp;(a leap-year)&nbsp; 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 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;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 the 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 see that there is no other solution.