10-Minute Math

I did a search for “volume formulas” and the second image returned was this one.  I almost provided these posters to a student until I noticed two mistakes…

With a new year comes a new Doomsday to keep in mind.  No no no, this doesn’t have to do with the world ending in 2012 – it’s Conway’s Doomsday Algorithm, of course!

Dr. Conway developed a sweet method to determine the day of the week for any day in history.  It takes some practice.  Dealing just with the current year is pretty simple, provided you know the Doomsday. 

Doomsday 2012 is Wednesday*.

• This means that the following dates are all Wednesday: 4/4, 6/6, 8/8, 10/10, 12/12. 
  What day of the week is Christmas 2012?  13 days past a Wednesday… must be 1 day before a Wednesday.
• If you work nine to five at the local 7-11, you can remember four more Wednesdays: 9/5, 7/11, 5/9, 11/7.
• Einstein’s birthday = Pi Day = 3/14 = Wednesday.
• The last day of February (2/29 this year) is a Wednesday.
• Lastly, 1/4 is Wednesday (3 out of 4 years it’s the 3rd… but on leap years it’s the 4th).

What day of the week is your birthday? Halloween?  Independence Day? 

*In case you’re interested, the link up top does a thorough job of teaching the method, but here’s the quick version. Start with the century day.  For 2000 it is Tuesday (2).  Then look at the 2-digit year, 12, and ask how many 12’s are in it. (1). What is the remainder? (0) How many 4’s are in this number? (0). Add up (2)+(1)+(0)+(0) = 3 = 3rd day of the week = Wednesday. 

Craziness? How about one more example from the future: 2055.  Century day is still (2).  How many 12’s in 55? (4). What is the remainder? (7).  How many 4’s in this number? (1).  Add q‘em up: (2)+(4)+(7)+(1) = 14.  Take out multiples of 7 and you get 0.  The 0th day of the week is Sunday. Now you can find the day of the week for any date in 2055.

We noticed a prominence of powers of two in our family this year.  My daughter will turn , my son will turn , my wife and I will turn and celebrate anniversary number .  (This may just be a fluke of the Mayan calendar).

Given this information, will there be another numerically interesting year for us, or is this it?  We can’t look forward to all being powers of 2 again, but perhaps there’s something else in store for us?

I was cutting out some cookies the other day, and noticed that my method wasn’t really that efficient.

So I started using this pattern, and challenge you to find an even more efficient way to pack stars in a plane.

Step 1: Make a thing. 

I went with a leaf shape.

Step 2: Effect > Distort & Transform > Transform.  Make sure to preview, then make some copies of that leaf and scale ‘em down.  Add a reflect X, scale dimensions down to 90%, and put some Move Vertical, and with some tweaking you should get a fern-looking thing.

Why does this remind me of the geometric series elves?

Step 3: Object > Expand Appearance so that the effect becomes materialized.  Select all, right-click and group that sucker.

Step 4: Now for the fractal part: iterate!  I rotated my tree-looking thing 90º clockwise so it resembled the initial leaf.  Then I repeated the transformation to make a bunch of these tree things with a scale factor of 90%. 

With 50 copies of a shape made of 50 parts, you’ll notice your computer start to slow down a bit.  I decided to stop while I was ahead… not a perfect fractal, but good enough for your Christmas card.

There’s a nice feature in Adobe Illustrator that makes it easy to visualize exponential change.  Here’s a little example to show your students how

Start by drawing something… in this case, an elf.  Say he has a height of one unit.  Select your drawing, right-click, and group it.

Now hit up the Effects > Distort & Transform > Transform

Here’s where it gets cool.  First click Preview, then put a number in the Copies box (I used 100 to approximate infinity).  This will create a bunch of copies of the original, but since they’re all the same size, you can’t see them.  Change the horizontal and vertical scales to 50%, and space out the elves by increasing the Move>Horizontal slider.  Drop the vertical slider so that the elves are on the same baseline.

You can now see that each elf is half the height of the previous.  This is a geometric sequence.  To find the value of the series, we need to add the heights of the elves, which we can do by adjusting the horizontal and vertical sliders to stack the elves on top of each other (make sure to have students make a prediction first).

All that’s left is to see how tall that stack of elves on top of the first one is.  Remember that the big elf is one unit, so let’s use him as the measuring stick.  Copy and paste the big elf, then turn off his transformation effect and drag him up to measure the series.

Students should see that the sum of the heights of all the little elves is equal to 1 big elf, which is to say 1/2 + 1/4 + 1/8 + … = 1.