Henry Ernest Dudeney/Puzzles and Curious Problems/53 - Finding a Birthday/Solution

by : $53$

 * Finding a Birthday

Solution

 * Midday on.

Proof
We calculate which day of the year (that is, which by index from $1$ to $366$) is, which is a leap year: $28 = 7 \times 4$:


 * $31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 10 + \tfrac 1 2 = 315 \tfrac 1 2$

the $\tfrac 1 2$ because at noon we are halfway through the day.

Between $1901$ and $1927$ there are:
 * $6$ leap years of $366$ days

leaving:
 * $21$ years of $365$ days.

This adds up to $10 \, 176 \tfrac 1 2$ days since, which is where the century started.

It remains to determine the date which is the day $10 \, 176 \tfrac 1 2$ days before this date.

Some $28$ years before $1901$ is somewhere in February $1873$.

We need to narrow that down.

Between $1874$ and $1900$ there are again $6$ leap years of $366$ days, and $21$ years of $365$ days, so:
 * the number of days between the start of $1874$ and the end of $1900$

is exactly the same as:
 * the number of days between the start of $1901$ and the end of $1927$

so we do not have to adjust for that.

Instead we know that we want the date of day $315 \tfrac 1 2$ counting backwards from.


 * $365 - 315 \tfrac 1 2 = 49 \tfrac 1 2$

Day $49$ of $1873$ is.

Half a day after that brings us to midday on.