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Note: This question is inspired from this post.

The number of days since I was born is approximately the number of days in (cube root of year) years (ignoring leap years), and today is the (cube root of year)$^{2}$th day of the year.

It is also worth to mention that the date format of my DOB (Date Of Birth), written in MM/DD/YYYY is that DD is the $n^{th}$ Fibonacci Number, $n$ is the value of MM, and YYYY is the product of MM, DD, and the cube root of the year.

Question:

What date is today and what day I was born?

Hint:

Today's date falls one day earlier than the day of the week I was born.

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    $\begingroup$ Concerning the Fibonacci sequence, "Many writers begin the sequence with $0$ and $1$, although some authors start it from $1$ and $1$ and some (as did Fibonacci) from $1$ and $2$". Which of these three enumerations do you follow? $\endgroup$ Commented Dec 6 at 7:58
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    $\begingroup$ @AlexRavsky 0 and 1. $\endgroup$ Commented Dec 6 at 8:40
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    $\begingroup$ And does "cube root of (the) year" always refer to the current given year, or depend on the context? $\endgroup$
    – Nautilus
    Commented Dec 6 at 8:50
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    $\begingroup$ @AlexRavsky Consider all years between having only 365 days. As an example, there are only 4015 days from 2024 to 2035 if we ignore leap years. $\endgroup$ Commented Dec 6 at 9:24
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    $\begingroup$ So both the current year and your birth year are perfect cubes? $\endgroup$ Commented Dec 6 at 9:30

1 Answer 1

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Today’s date is 6/18/2197 and the day you were born was 8/21/2184.

I utilized excel formulas to quickly guess and check different cube years for today. See below for the formulas and solve process. The main part to check was the factors of born yyyy/C, as those needed to be Fibonacci n and nth number, and multiply back to born yyyy. You can narrow the search by knowing the Fibonacci numbers are constrained to properly formatted dates. Excel diagram of solve process

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  • $\begingroup$ @Aero You’re almost correct. The day I was born was correct, but not the today’s date. $\endgroup$ Commented Dec 6 at 21:49
  • $\begingroup$ @ThirdyYabata I think I have corrected. Used Wolfram to double check the C squared day of the year instead of my faulty arithmetic. $\endgroup$
    – Aard
    Commented Dec 7 at 0:26

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