mathematics - Odd birthdate surprise

Last year, just before my birthday in February, I noticed something unsual. The last two digits of my birth year were the same as my age. How old was I then and was this an unusual occurrence?

Answer

Generally, people born in $1900 + n$ will have this birthday in the year $1900 + 2n$, when they are $n$ years old. The same can be applied to any century. You'll notice this is an even number; the year in the question is odd because you noticed this just before your birthday. That is, the Beddian birthday (see Kate Gregory's link in the quetion) was in 2012, but you noticed the phenomenon in the next year, 2013.

Working backwards, if this year is $c + m$ (where $c$ is a century such as 1900 or 2000 and $m$ is even and less than 100), those with a Beddian birthday this year were born in $c + \frac{m}{2}$ or $(c - 100) + \frac{m + 100}{2} = (c - 100) + \frac{m}{2} + 50 = c + \frac{m}{2} - 50$.

You can obtain the second option by "going back" a century, making $c - 100$ the new century part and $m + 100$ the new "$2n$" part from the first paragraph. Since $m < 100$, $\frac{m}{2} + 50 < 100$, and so the year is valid. You can't go back another century, or else the "$m$" portion will be more than two digits, which wouldn't make any sense.

Thus, for the year 2012: $$c = 2000, m = 12\\ \text{birth year is } \boldsymbol{6}=\frac{m}{2} \text{or } \boldsymbol{56}=\frac{m}{2} + 50$$

As Rob Watts notes, you're more likely to have been 56 than 6. Additionally, this happens every even year for those born turning $(year ~mod~ 100)/2$ and the same age plus or minus 50.