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Happy 10/10/10!

Published on Sunday, October 10, 2010 in , ,

Happy 10/10/10! It's one of those silly number holidays, much like last year's 9/9/09. What can we do for 10/10/10?

I think it's a perfect time to examine the powers of 10. That's never been done better than in Charles and Ray Eames' classic educational film, Powers of 10:

Did anybody else ever notice that the narrator specifically mentions that the picnic is happening in October? Did they choose this because of the 10th month? He does say it's taking place “early one October“, so maybe it's a subtle 10/10 joke.

If the film got you thinking, you can experience it online in a more hands on manner, at powersod10.com. This site is very well developed, and really helps bring the experience to life.

On a geekier note, if you use binary, 101010 equals 42. This is silly, fun, and geeky, which seems to go perfectly with the day. As a matter of fact, many of us geeks are celebrating today as 42 Day.

This seems like the perfect time to bring up what I believe should be called Douglas Adams' identity: 1337% of Pi = 42.

Are you doing anything silly or special for 10/10/10? Let's hear about it in the comments!

1 Response to Happy 10/10/10!

12:02 AM

Happy 10/10/10 to all!

10/10/10 represents Sunday, October 10, 2010.
We write:
101010 or in this form: 1,0,1,0,1,0
Either way, the number is a repeating pattern, in which the unit of repeat has length 2
We see this the days of the week, or a tessellation.

It's in the form of ababab or (a,b,a,b,a,b)
ababab = ab * 10101 = ab*3*7*13*37

Let's play with these numbers. For example:

Consider the sequence: 10, 1010, 101010

which is the beginning of the sequence A163662:

http://www.research.att.com/~njas/sequences/A163662

A020988(n) written in base 2 : 0, 2, 10, 42,... [ (2/3)*(4^n-1) ]

http://www.research.att.com/~njas/sequences/A020988

And also :
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10/10/10 is also interesting because the first two numbers added together equal
the first two digits of the year: 10/10/2010 -> 10 + 10 = 20.

Such dates occur every 201 years, the previous one was 09/09/1809 and
the next one will be 11/11/2211.

I'll share another problem related to this:
Let's write the date in the "mm-dd-yy" format
Let's consider another sum, in the example above:

Case 1: the first two numbers added together equal
to the first two digits of the year

It's easy to find other such dates

Case 2: the sum of the first two numbers added together equal to the LAST two digits of the year.

For example, January 1, 2002 or 01/01/02, where 1+1 = 2.

Would anyone like to find out the total number of sum-days in previous centuries?
and from January 1, 2002 to December 31, 2100 inclusively?