To save money, a meticulous person had a oddly unique way of splurging money.

Each day the person would splurge an amount equal to the month-digits + day dollars.

So for the first of April it would be 4+1 = 5 Dollars, while on first of June it would be 6+1=7 dollars. On the 10th of April it would be 4+10 = 14 dollars.

One day the person counted a total splurge of 97 dollars for the last 5 consecutive odd numbered days including this day.

What date can it be?


2 Answers 2


The following answer looks much more complicated than it actually is. But it dictates a more "logical" way to solve this.

There are two cases:

  • All five dates are in the same month

    Let the middle date be $d/m$. Then the spending is $5m+5d$ which is divisible by 5, hence cannot be $97$.

  • Five dates are in two different months

    • The first month has $29/30$ days.
      The five days are in an arithmetic progression mod 30 hence mod 5. Taking mod $5$ of the five days, we see that each days must be distinct modulo 5, hence the sum of days (not months) is divisible by 5. Therefore the sum of the five months must be 2 mod 5. It follows that there must be three dates on the first months and two on the second. So, the dates are 25,27,29,1,3 which sum to $85$. So, the month has to be $\lfloor {97-85 \over 5}\rfloor $ $=2$, ie. February. This has only 28 days. However, on leap years this works and leads to the solution-

      25 Feb, 27 Feb, 29 Feb, 1 Mar, 3 Mar

    • The first month has $31$ days.
      We can divide in four cases according as how many days in the first month.

      25,27,29,31,1 -> sum exceeds 97
      27,29,31,1,3 -> m = -0.2, not an integer
      29,31,1,3,5 -> m = 5 = May
      31,1,3,5,7 -> m = 7.2, not an integer

      This leads to the solution-

      29 May, 31 May, 1 June, 3 June, 5 June

    • The first month has $28$ days, ie. February on a non leap year.
      There are only four cases according as how many days fall in February. None gives a solution.


It is

June, 5th (#1) or in a leap year it can be March 3rd (#2) too

Calculation #1:

29+5 + 31+5 + 1+6 + 3+6 + 5+6 = 97

Calculation #2:

25+2 + 27+2 + 29+2 + 1+3 + 3+3 = 97

How I solved it?
I made an excel sheet with all odd days in a year and a second column for the sum of the last 5 days.

  • $\begingroup$ You got it right...That was quick. Hail MS Excel !!! $\endgroup$
    – kanchirk
    Commented Sep 22, 2015 at 12:14
  • 2
    $\begingroup$ I was about to post the answer, when I got the "new answer has been posted" notification appeared. Same answer, but I've done it on a piece of paper :D. Nice thinking with Excel, well done! $\endgroup$ Commented Sep 22, 2015 at 12:24
  • $\begingroup$ I was just about to do that. It was more of a math problem than a puzzle. $\endgroup$
    – user14492
    Commented Sep 22, 2015 at 12:26

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