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

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Puzzles and Curious Problems by Henry Ernest Dudeney: $53$

Finding a Birthday
A correspondent informs us incidentally that on Armistice Day (Nov. 11, 1928)
he would have lived as long in the $20$th century as he lived in the $19$th.
This tempted us to work out the day of his birth.
Perhaps the reader may like to do the same.


Solution

Midday on $\text {19}$ Feb $\text {1873}$.


Proof

We calculate which day of the year (that is, which by index from $1$ to $366$) is $\text {11}$ Nov $\text {1928}$, 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 $\text {1}$ Jan $\text {1901}$, 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 $\text {31}$ Dec $\text {1873}$.

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

Day $49$ of $1873$ is $\text {18}$ Feb $\text {1873}$.

Half a day after that brings us to midday on $\text {19}$ Feb $\text {1873}$.

$\blacksquare$


Sources

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