High School Mathematics-Physics SMILE Meeting
1997-2006 Academic Years
Mathematics: Puzzles

14 October 1997: Roy Coleman [Morgan Park High School]

Porter:
Take a 3 digit number with distinct digits, reverse the digits, and subtract the smaller from the larger. Then take the new number, and reverse its digits, and add the two together. The answer is always 1089.

                  htu                       865
                100h + 10/t + u          -  568     reverse digits
                100u + 10/t + h           ---------
              __________________            297
     
                                          + 792     reverse digits
           99h - 99u = 99(h-u)           ---------
            Cast out 9's                   1089

08 September 1998: Bill Colson, Moderator [Morgan Park HS]
He showed us his favorite Math Word Problem that appears to provide too little information to determine the children's ages. The person asked the question cannot solve it with the information given, but would know the answer if told that the oldest child has red hair [and thus that there is an oldest child, and not two of the same age].  See the website http://www.varatek.com/scott/family_math.html.

He recommended a book called The Chicken from Minsk and 99 Other Infuriatingly Challenging Brain Teasers from the Great Russian Tradition of Math and Science, by Yuri B. Chernyak and Robert M. Rose [out of print, according to http://www.amazon.com ---PJ].

29 September 1998:  Alan Tobecksen [Richards Voc HS]
He showed a rope that was tied to each hand and between two people...and the objective was separate from one other. He left the resolution of the problem till next week. 13 October 1998: Al  showed the solution to the rope puzzle of the last meeting. The trick is to put a loop in the rope, and then feed it through the rope around one the hand of the other person. You either become free in the process, or else become more tightly enmeshed in the process.

24 November 1998: Al Tobecksen [Richards Vocational HS]
He proposed a game based upon removing objects from rows, with the rows initially containing 1, 3, 5, 7 objects. Two people take turns removing some or all objects from only one of the rows, and the person who makes the last move loses the game. To stimulate student interest and excitement concerning his potential humiliation, he promised an A for the grading period to any student that beat him in the game. What is the winning strategy? [more about this in the future!]

06 April 1999: A riddle from Lilla Green

The first digit was your original number [# times you want to eat out each week]. The second two digits are your age.

07 December 1999: Porter Johnson (IIT)
put us to work cutting out a paper pattern to construct an icosahedron [http://www.geom.umn.edu/docs/education/build-icos/]. When some of us had completed it, we had what could pass for a nifty and science-based Christmas tree ornament. Porter gave us  mathematical and other insights into this shape. The icosahedron is one of the 5 Platonic Solids; see http://library.thinkquest.org/22584/image/soilds.jpg.  Thanks, Porter!

01 February 2000: Earl Zwicker (IIT)
showed us a topological puzzle. He cut a sheet of paper (8.5 ´ 11) into the shape of a plus sign (+), after first folding it in half one way, then the other. The folds were used as a visual guide, and Earl cut a good half inch on each side of each fold, leaving a + shaped piece of paper with each leg a good inch wide. He then folded the vertical legs to meet in a loop, and taped them together. Then the horizontal legs to make another loop (taped) lying below the first (& naturally perpendicular to the first loop). Next, he used a scissors and cut each loop in half "longways" - along its circumference. Behold! The result was a rectangle, like a picture frame!

Next, he held up what appeared to be two intertwined hearts. This was the result of making a half-twist before taping the legs together to make a Möbius Strip of each loop. Then cutting as before along the "circumference" of each. Try it!  See the website http://www.scidiv.bcc.ctc.edu/Math/Mobius.html.

28 March 2000: Sally Hill (Clemente HS)
put us through a paper-and-scissors exercise to construct a geodesic dome. A template, which consists of 21 equilateral triangles, is given on the website http://hilaroad.com/camp/projects/dome/dome.html. We crowded around the table and carefully cut out section after section from the pattern copied onto many pieces of paper. Fitting and gluing (use stick glue to make it easy) 5 sections together will do it!

Sally, thanks for introducing us to this useful idea and website resource!

28 March 2000: Ed Robinson (Collins HS)
posed the following challenge to us. We are given twelve coins, one of which is either heavier or lighter than the others. Using a balance, how - with no more than three weighings - can we determine which coin is the odd one? He noted that 1990s pennies have a zinc cladding, and therefore weigh different from a 1972 penny, which is made of solid copper. This would make it possible for us to do this experimentally on the balance he had placed on the table.

Ed proceeded to show us a theoretical strategy to do this, starting with just 6 coins, and 9 coins---namely, split the coins into 3 sets, and go from there. A guiding principle is that you can tell which coin of 3 is bad by making one weighing. One can expand the same strategy to 12 coins. Ed actually carried out the experiment with a set of coins, and it worked! He also handed out the written solution to the problem: for details click here:

PJ Comment: Also, check these websites:

You got us thinking, Ed! Thanks!

28 March 2000: Porter Johnson (IIT Physics)
passed out copies of "Weighing Problem," a write-up from five years ago, which is similar to Ed's approach, with differences in discussion.  For details, click here:

10 October 2000 Ed Robinson (De LaCruz School)
gave each of us a sheet of blank paper, then challenged us to find a pattern in playing a game that he called "Nim Mod." The first step was to sketch a rectangle and divide it into 4 boxes; (N = 4). The game is played similarly to tic-tac-toe, with one player making Xs and the other making Os. The loser of the game is always the player who is forced to fill in the last empty box, because none other is left. But each player, on his turn, may fill in 1, 2, or 3 boxes with his X (or O). With N = 4, it is clear that the Starter player (S) can fill in three boxes, leaving only one box empty, and forcing the second player to fill in the last box to become the loser of that particular game:

X X X O
Thus, we can designate that the Starter (S) would be the winner (unless he is makes a mistake!) for N = 4. Ed had us continue the game for N = 5, and then he worked it at the board, showing us that S may easily be forced to lose the game by his opponent, so for N = 5, S loses. He put us through these games:
N = 6, S wins
N = 7, S wins
...
N= 9, S ??
and so on. No matter how high N becomes, if the problem is divided into sets of N = 4, then we can find if S is a winner or loser by dividing N by 4, and finding the Remainder, R.
If R = 0, S wins, if R = 1, S loses, if R = 2, etc. 
(Try it and see for yourself!) His students enjoy the math and trying to figure out the pattern. Thanks, Ed! We learned another new idea today!

05 December 2000 Sally Hill (Clemente HS)
had two identical Tower of Hanoi puzzles on the table, and she invited a teacher and a student to come up front to solve the puzzles. Each puzzle is made from three dowel rods held vertically on a wooden base. The rods are in line and the distance between the 1st and 2nd is the same as between the 2nd and 3rd. (Draw a picture!) The 1st rod is surrounded by (sticks up through the holes in the center of) a set of 5 wooden rings which lie on top of each other and are at rest on the base. The bottom ring has the largest diameter, and each successively higher ring has a smaller diameter than the ring below it. The object is to move all the rings from the 1st rod to the 3rd rod. However, you may move only one ring at a time from any rod to another, and you may never place a larger ring on top of a smaller ring while making these moves. After the teacher and student had made some attempts, Sally gave a hint for the first move. It wasn't long and the student had out-paced the teacher! Isn't that the way it always works?! Anyhow, on the board, Sally placed hints for moves that looked like this:

                            number           differences
                            of moves         in column                        
              1 ring          1
                                                2
              2 rings         3
                                                4
              3 rings         7
                                                8
              4 rings        15
                                               16
              5 rings        31
Porter Johnson said a variation on this puzzle is the Chinese ring puzzle, or Cardan's Rings. [see an image of Cardan's Rings, http://www-groups.dcs.st-and.ac.uk/~history/Diagrams/Cardan's_rings.jpeg, and a brief history of math puzzles http://www.mefferts-puzzles.com/history.html.]

Sally also gave us a handout of showing an X-shaped pattern of 9 empty boxes:

                     0       0
                       0   0
                         0
                       0   0
                    0        0

Can you make the sum of each diagonal and the corners equal to 26? Use the digits 1 through 9 only once.

Thanks, Sally! (Got it yet?)

14 March 2001 Bill Colson
handed out copies of an article Test Yourself [SAT I questions from recent tests designed to measure verbal and math skills] from the 12 March 2001 issue of Time Magazine: [http://www.time.com/time/].

10 April 2001 Bill Colson (Morgan Park HS, Math)
showed a workbook, Stretching Your Math Students' Achievement, Motivation, and Involvement: Grades 7 - 12 Resource Handbook by Irv Lubliner, recently obtained from the following source:

Bureau of Education and Research
915 118th Avenue, SE
PO Box 96088
Bellevue WA 98009
http://www.ber.org/
Tel: 1 - 800 - 735-3503
First he showed a clear exposition from that book of the rope trick (topological puzzle) that has been shown several times in SMILE, such as 13 October 1998.

Then he showed us how to play a game called MAXIT, illustrating the point with a 4 ´ 4 square lattice. He put an ´ into one location, and had us to call out numbers between -10 and +10 for the other locations, with the result as shown:

5 8 6 2
3 ´ -7 6
-1 -3 0 1
4 -9 9 9

People on one side of the room were the UP'S AND DOWN'S, whereas those on the other side were the RIGHT'S AND LEFT'S. After flipping a coin to see who moves first, the winner was allowed to move the ´ --- either up or down, or right or left, respectively, to another location. The number in that location is replaced by an ´, and they get the number of points corresponding to that number, and you cannot move into the location of an ´. The last team to be able to move ends the game---and the team with the largest point total wins.

The next game involved the creation of a magic square, such as the following one:

14 20  3 12  7
32 38 21 30 25
19 25  8 17 17
11 17  0  9  4
17 23 6 15 16

The numbers in this table may look unrelated, but they have not been randomly chosen, because if you pick five numbers, each from a different row and a different column, and take the sum, you will get the total 79. For instance, the five numbers shown in bold give 14 + 25 + 25 + 0 + 15 = 79.  We have, in fact, generated 120 different combinations of numbers adding up to the total of 79Isn't that remarkable? Surprisingly, there is nothing unique about the number 79, and you can see how the table was made by adding another row and another column to it:

** 11 17  0  9  4
 3 14 20  3 12  7
21 32 38 21 30 25
 8 19 25  8 17 17
 0 11 17  0  9  4
 6 17 23 6 15 16

The inner numbers are generated by taking the sum of the corresponding numbers in the first row and first column. e.g. 21 + 17 = 38. The "magic number" 

79---which is merely the sum of the ten numbers in the first row and the first column---can by changed by changing those 10 "seed numbers".

01 May 2001 Roy Coleman [Morgan Park HS] Puzzle
handed out the following pattern on a sheet of paper:

 
What should go in the region in which ?? is located?  The answer is based upon the pattern for displaying numbers on hand-held calculators, which involves a display on which any of seven lines are present. The number 8 involves all seven lines whereas the number 4 has four lines and the number 7 three lines, as shown:

Roy's pattern shows a complementary image, in which only the "unlit" lines are shown. Let us put in the lighted lines by using RED lines:
 

Therefore the missing pattern is given by the two GREEN segments.

Capice?

25 September 2001 Fred Schaal (Lane Tech HS, Mathematics) 
He commented about his presentation of the 11 September 2001 SMILE meeting, that none of his students had read or heard of My Friend Flicka, and thus did not identify Flicka as a horse.  Therefore, he suggested the following syllogism:

He also gave these statements in conditional form:

One might question whether the second statement is actually true, and thus the conclusion, which may seem perfectly reasonable, cannot be made upon the basis of this "false syllogism".

23 October 2001 Monica Seelman (ST James School)  Casting out Nines
Monica
pointed out that you can check arithmetical operations by calculating the entries modulo base 9, and then checking the arithmetical operations modulo 9.  This check on arithmetic would work on any base, but it is especially convenient using modulo 9, since you get the number mod 9 by repeatedly summing the digits.  For example, 1285 ---> 1+2+8+5=16 ---> 1+6 = 7.  She did the following sample problems

Addition:

Original Problem    Modulo 9

      362        --->          2
      256        --->          4
      147        --->          3
 ----------                 -------
      765        --->          9

Subtraction:

Original Problem    Modulo 9

     5273        --->          8
  -   189        --->           0
 ----------                 -------
     5084        --->          8

Multiplication:

Original Problem     Modulo 9

      635     --->             5
   ´   24     --->         ´  6
 ----------                 -------
  15240     --->             3

Fred Schaal mentioned that in hexadecimal notation, in which the counting sequence is

1  2  3  4  5  6  7  8  9 A  B  C  D  E  F  10 ...

he just turned the age of 3F, and next year would become age 40.  Dream on about hex code and remember what the Beatles said [http://www2.uol.com.br/cante/lyrics/Beatles_-_When_I_am_64.htm], Fred!

23 October 2001 Fred Schaal (Lane Tech HS, Mathematics) 9 ´ 9 Magic Squares
Fred handed out a sheet containing the following empty lattice:

                                            
                 
                 
                 
                 
                 
                 
                 
                 

Fred then asked for a start number (we chose 11), as well as an add number (we chose 17).  Next, he put 11 into the middle square on the top row.  The idea is to implement "toroidal topology" with periodic boundary conditions, and to add 17 sequentially to the 11, and the total placed one square above and to the right of the previous element.  The first few numbers are shown below:

                    11                    
      147           
    130            
  113              
96                
                79
              62  
            45    
          28      

At this point we do not put 164 (147 + 17) into the location already occupied by 11; instead we put the 164 under the 147, and continue until we hit the next snag:

                      11 198                
      147 181        
    130 164          
  113 300            
  96 283              
266                 79
                62 249
              45 232  
           28 215    

once again, we proceed by putting the next number, 317, under the 300, and continue the procedure. to get

  793   980 1167 1354     11   198    385   572   759
  963 1150 1337   147   181   368   555   742   776
1133 1320   130   164   351   538   725   912   946
1303   113   300   334   521   708   895   929 1116
    96   283   317   504   691   878 1065 1099 1286
  266   453   487   674   861 1048 1082 1269     79
  436   470   657   844 1031 1218 1252     62   249
  606   640   817 1014 1201 1235   45   232   419
  623   810   997 1184 1371     28 215   402   589

The sum of every row, every column, every diagonal, and every set of numbers symmetrically placed about the diagonal is 6219, which is 9 multiplied by the central element, 691.  Why?

Porter Johnson suggested subtracting the start number 11 from each element, and then dividing each element by the add number 17, to obtain the following array:

 46 57  68 79   0  11 22 33 44 
56 67 78   8 10 21 32 43 45
66 77   7   9 20 31 42 53 55
76   6 17 19 30 41 52 54 65
  5 16 18 29 40 51 62 64 75
15 26 28   3 50 61 63 74   4
25 27 38 49 60 71 73  3 14
35 37 48 59 70 72   2 13 24
36 47 58 69 80   1 12 23 34

The sums are equal to 9 multiplied by the central number 40, or 360.  This property remains valid if you make the following operations:

Very interesting, Fred!

05 March 2002: Fred Schaal (Lane Tech HS Math) -- Reports of a "bad number" in the Magic Square/
Fred
pointed out that the 9 ´ 9 Magic Square Table in the SMILE notes of 23 November 2001 contained an error.  Can you find it?

 

  793   980 1167 1354     11   198    385   572   759
  963 1150 1337   147   181   368   555   742   776
1133 1320   130   164   351   538   725   912   946
1303   113   300   334   521   708   895   929 1116
    96   283   317   504   691   878 1065 1099 1286
  266   453   487   674   861 1048 1082 1269     79
  436   470   657   844 1031 1218 1252     62   249
  606   640   817 1014 1201 1235   45   232   419
  623   810   997 1184 1371     28 215   402   589

"It's all my fault!" -PJ

Very keen powers of observation, Fred!

11 December 2001: Monica Seelman (ST James School): Repeating Decimals
Monica
showed us that certain fractions, when expressed in base 10 decimal form, result in repeating decimals.  For example:

 

1/3 .3333 ...
1/9 .11111 ...
13/99 .131313 ...

She pointed out that the repeating decimal 0.c1c2c3...cn can be expressed as the fractional ratio of two integers:  c1c2c3...cn / 999...9. Here are some other examples

1 / 7 .142857 ... = 142857 / 999999
1 / 11 .090909 ...  = 9 / 99
13 / 99 .131313 ...

Interesting examples, Monica.

She ended the presentation with the following anecdote:

Three native American squaws wanted to be certain to have boy babies, and they sought their tribe's medicine man  for advice. The first one was told to sleep on a buffalo hide, and, indeed she did have a son. The second one was told to sleep on a horse side, and she had twin sons. The third one was told to sleep on a hippopotamus, and she had triplet sons. This proves the following:
The sum of the squaws on the two hides is equal to the squaw on the hippopotamus.
Thank you for sharing that, Monica!

19 March 2002: Fred Schaal (Lane Tech HS Mathematics) -- Even Magic Squares Fred suggested an extension of his Magic Square demonstration in the SMILE class of 23 November 2001, in which he was to make a magic square with an even number of squares, such as 4 ´ 4.   One of his students  found solutions at the website http://mathforum.org/alejandre/magic.square/adler/adler4.html:

0 7 11 12
13 10 6 1
14 9 5 2
3 4 8 15
  ... OR ...  
0 13 7 10
14 3 9 4
11 6 12 1
5 8 2 15

Note that, in each case, all numbers from 0 to 15 are present, and the rows, columns, and diagonals add up to 30Fred will discuss these examples, and consider the algorithm for generating this square and squares of higher order in the future.  Can you make it work for  2 ´ 2 squares, Fred?

Fred also asked about two keys, ITC and SLP, that were present on his old TI-35X calculator.  It was suggested that these keys represent "intercept" and "slope" for entered data.  Compare your calculator with the one on the website, http://www.datamath.org/Sci/Modern/TI-35X_1.htm.  Good luck on your quest for the meaning of keys, Fred.

24 October 2000 Don Kanner (Lane Tech HS)
told us about a puzzle he had thought of which seemed to tie physics and math together. (Handout). Given n identical resistors, if they are connected in various series/parallel combinations, [all planar, non-intersecting, irreducible] is there some sort of regular sequence of total resistance values that might be generated? For example, if each resistor has a value of 10 Ohms

:
number   series parallel other combinations

1

10 Ohms

-

-

2

20   "

5

-

3

30   "

3.3

15 6.7

4

40   "

2.5

16.7 13.3 ...

5

50   "

2.0

...

Don said the real difficulties begin with n = 5. Equations used by electrical engineers called the Y-delta and delta-Y Transformation Equations would be needed to eliminate circuits with the same resistance. The number of possible combinations increases rapidly with n, making it difficult to investigate. 

Don next asked us the following trivia question:

Given the following sequence of numbers

777       _ _ _ _ _       999999,

what numbers belong in the blanks?

Interesting questions, Don. Answers, anyone?

 21 November 2000 Don Kanner (Lane Tech HS)
He first announced the answer to a question previously posed: for n equal resistors combined in irreducible planar networks, the number of different total resistances is 2n-1.

11 September 2001 Don Kanner (Lane Tech HS, Physics)  200 Puzzling Physics Problems with hints and solutions
by Peter Gnadig, G. Honyek, K. F. Riley [Cambridge 2001] ISBN 0-521-77480-3
Don found the book interesting and stimulating for the most part, but felt that some of the problems "needed work":

  1. Q:  A "black box" has two sets of terminals on each side.  When a battery of voltage V is connected to one side, a voltage V/2 is read across the other side. When the battery is hooked across the other side, a voltage V is read across the first side What is inside?
    A: (But, what about internal resistance of the voltmeter?)
    
    
  2. Q:  If you push an object and it goes off the table [a distance of 1 meter], does it have wheels?  A:  Yes [Don pushed an empty plastic cola bottle off the table to show the absurdity of this question.]
  3. Q:  Why do bubbles in a glass of champagne accelerate on the way out of the glass?  A:  Allegedly, the force of gravity is balanced by the buoyant force.  But then, why would they accelerate at all?

Here is a review from the website http://www.amazon.com.

The problems are generally at the level of International Physics Olympiad competitions or higher, and are very suitable for freshman/sophomore honors physics courses. The problems are tough and very interesting, with many unique and unusual twists, guaranteed to challenge even the best students preparing for physics competitions at the senior high school and undergraduate levels. I would guess that those sitting for PhD qualifying exams will have difficulties with some of the problems. The surprise is that the solutions require no more than elementary calculus, although lots of original critical thinking and insight are necessary. 

Gnadig and Honyek are both leading the Hungarian physics olympiad teams for many years, while Riley is a Fellow of Clare College, Cambridge University. Riley had a previous equally interesting but slightly less difficult problem book called "Problems for Physics Students," covering quite a few interesting Cambridge University Natural Sciences Tripos and entrance exam type questions for the best UK high school students.

If you are a physics buff, you will derive countless hours of enjoyment (and frustration - if you resist the temptation of looking at the provided solutions too soon), plus it will bring your understanding of classical physics to a deeper level. 

A few other worthy physics problem books include Thomas & Raine's "Physics to a Degree" for undergraduates, and Dendy's "Cambridge Problems in Physics" for high school students aiming for Cambridge University entrance. The Russian books "Problems in General Physics" by Irodov and "Problems in Elementary Physics" by Bukhovtsev are very good too but they may be harder to find. An upcoming excellent (and tough) classical mechanics problem book is David Morin's Physics 16 course text at Harvard - still in its draft form but downloadable from Harvard website. 

02 April 2002: Don Kanner (Lane Tech HS Physics) -- ¿¿Magic Triangles??
Don
suggested a  variation of the Magic Square problem, in which we put numbers inside a triangular lattice, and seek solutions in which all external line entries and all angle bisector entries add to the same number, such as the following:

              /\
             /  \
            /____\
           /\    /\
          /  \  /  \
         /____\/____\
        /\    /\    /\
       /  \  /  \  /  \ 
      /____\/____\/____\  
For this triangular lattice, ten entries are required---one inside each triangle. In the following case, all numbers add to a total of 36:
              /\
             /18\
            /____\
           /\    /\
          / 8\ 3/ 7\
         /____\/____\
        /\ 19 /\ 17 /\
       /10\  /15\  /11\ 
      /____\/____\/____\  
Can one solve this problem with nine consecutive numbers; say, 1-9? You have definitely put down the gauntlet, Don!

02 April 2002: Fred Schaal (Lane Tech HS Mathematics) -- Even Magic Squares
Fred
showed the solution for the 4 ´ 4 magic square, as given on the website http://mathforum.org/alejandre/magic.square/adler/adler4.html.

 .    .    .    .  
.  .  .   . 
 .   .   . 
 .  .  .   . 

The idea is to put one of the numbers 1-16 into each of the 16 small squares, with no duplications, in such a way that the sums of rows, columns, and diagonal elements is the same --- 34, in this case.  The construction proceeds in two steps.  First, we count the squares across rows, starting across the top, and insert the number for all "diagonal" squares, as shown:

  1           4  
.   6  7  . 
.  10  11   . 
13   . .  16 
Second, pass by the unfilled squares in the same order and fill them with "missing numbers", starting with the largest and going down as unfilled squares are encountered:
 1 15   14    4  
12  6   7   9 
  8 10  11   5 
 13   3  2  16 

Congratulations; your magic square is completed!  But, this provides us with no insight for solving Mr 6 ´ 6.This way of constructing a 4 ´ 4 magic square can be found in a book by Jerome S. Meyer, Fun With Mathematics (Cleveland: World Publishing Company, 1952). Comment by PJ:  For information on magic squares, see http://en.wikipedia.org/wiki/Magic_square. For the example given below, the numbers all add to a total of 111.

 
   1  35   33     4   32     6 
 30   29  10    9  26    7
 18   20   22  21  17  13 
 19   14  16  15   23  24 
 12   11  27  28    8  25
 31    2    3  34    5  36

23 April 2002: Monica Seelman (St James) -- Digital Numbers, Geometrical Shapes, and Pouring Water from a Coffee Pot
Monica
handed out copies of an article: The Wonderful World of Digital Sums by M V Bonsangue, G E Gannon, and K L Watson, which appeared in the January 2000 issue of the periodical Teaching Children Mathematics, and discussed some  interesting applications.  First she made put a multiplication table on the board [PJ: here is a 9 ´ 9 version, like the ones on spiral notebooks in schools a few decades ago, which some of us remember very well.]

one-sies two-sies three-sies four-sies five-sies six-sies seven-sies eight-sies nine-sies
2 3 4 5 6 7 8 9
2 4 6 8 10 12 14 16 18
3 6 12 15 18 21 24 27
4 8 12  16 20 24 28 32 36
5 10 15 20 25 30 35 40 45
6 12 18  24 30 36 42 48 54
7 14 21  28 35 42 49 56 63
8 16 24 32 40 48 56 64 72
9 18 27  36 45 54 63 72 81

Monica patiently explained that a digital number is simply the sum of digits of a number, as illustrated here

Number Digital Number
 6   ® 6
13  ® 1 + 3 = 4
27  ® 2 + 7 = 9
38   ® 3 + 8 = 11®1 + 1 = 2

In other words, a digital number is just the sum of the digits of a number, which becomes a number: 1, 2, ... , or 9 [more technically, the number modulo 9 + 1]. Here is our 9 ´ 9 multiplication table , expressed in terms of digital numbers

one-sies two-sies three-sies four-sies five-sies six-sies seven-sies eight-sies nine-sies
2 3 4 5 6 7 8 9
2 4 6 8 1 3 5 7 9
3 6 9 3 6 9 3 6 9
4 8 7 2 6 1 5 9
5 1 6 2 7 3 8 4 9
6 3  9 6 3 9 6 3 9
7 5 1 8 6 4 2 9
8 7 6 5 4 3 2 1 9
9 9 9 9 9 9 9 9

Monica then used these columns of digital numbers to specify the order of connecting the vertices of a regular nine-sided polygon (nonagon), which was made by drawing a circle and then dividing it into nine equal segments (arcs), as shown:

Monica next showed us how to connect the points in the order given by the columns in the table.  We get the following types of figures in the nine cases:

Sequence Name

Sequence

Figure Name
One-sies:  1 ® 2 ® 3 ® 4 ® 5 ® 6 ® 7 ® 8 ® 9  Regular Nonagon
Two-sies 2 ® 4 ® 6 ® 8 ® 1®  3 ® 5 ® 7 ® 9 2-star Nonagon
Three-sies   3 ® 6 ® 9 ® 3 ® 6 ® 9 ® 3 ® 6 ® 9  Triangle
Four-sies 4 ® 8 ® 3 ® 7 ® 2 ® 6 ® 1 ® 5 ® 9 4-star Nonagon
Five-sies 5 ® 1 ® 6 ® 2 ® 7 ® 3 ® 8 ® 4 ® 9 4-star Nonagon
Six-sies 6 ® 3 ® 9 ® 6 ® 3 ® 9 ® 6 ® 3 ® 9 Triangle
Seven-sies 7 ® 5 ® 3 ® 1 ® 8 ® 6 ® 4 ® 2 ® 9 2-star Nonagon
Eight-sies 8 ® 7 ® 6 ® 5 ® 4 ® 3 ® 2 ® 1 ® 9 Regular Nonagon
Nine-sies 9 ® 9 ® 9 ® 9 ® 9 ® 9 ® 9 ® 9 ® 9 POINT

Comment by PJ:  Note that the Nine-sies are all nines [one point in the  figure], whereas the Three-sies and Six-sies involve the same 3 numbers, forming an equilateral triangle, but with the process of connection being done in opposite directions.  Similarly, the One-sies and Eight-sies form the same regular nonagon, but involve tracing it in opposite directions.  The Two-sies and Seven-sies trace the same 2-star nonagon, which hits alternate numbers and makes two revolutions before closing.  Finally, the Four-sies and Five-sies produce the same 4-star nonagon, which closes on itself after  making four revolutions.

Monica poured water from a coffee pot and raised the question as to why the coffee stream forms a twisting spiral when it comes off the lip of the coffee pot.  We will investigate this more in a future meeting! 

Fascinating stuff, Monica!

07 May 2002: Fred Schaal (Lane Tech HS, Mathematics) -- Hexcode Digital Numbers  
Fred
presented an extension of the the lesson on Digital Numbers by Monica Seelman at the 23 April 2002 SMILE meeting.  Monica converted multiplication tables [from 1 ´ 1 to 9 ´ 9] into tables of Digital Numbers, where digital decimal numbers are obtained by adding the digits of a number sequentially until a number between 1 and 9 is obtained [for example; 78 ® 7 + 8 = 15 ® 1+5  = 6].  With unerring Mathematical Logic, Fred decided that the same procedure would work with Hexcode Numbers [that is, hexadecimal numbers -- base 16].  First he reminded us how to count in base 10 as well as in base 16:

Base 10:  0   1   2   3   4   5   6   7   8   9   10   11   12   13   14   15   16  17  18  19   ... 
Base 16:    0  1  2  3  4  5  6  7  8  9   A   B   C   D   E   F  10  11  12  13  ... 

The new symbols A, B, C, D, E, and F represent the numbers between 9 and 16, whereas in base 16 the number 16 = 161+ 0 is written as [10]hex, etc.  He next made a multiplication table for base 16, a portion of which is given here:

Base 16 Multiplication Table
one-sies two-sies three-sies four-sies five-sies six-sies seven-sies eight-sies nine-sies
2 3 4 5 6 7 8 9
2 4 6 8 A C E 10 12
3 6 C F 12 15 18 1B
4 8 10 14 18 1C 20 24
5 A F 14 19 1E 23 28 2D
6 C 12  18 1E 24 2A 30 36
7 E 15  1C 23 2A 31 38 3F
8 10 18 20 28 30 38 40 48
9 12 1B  24 2D 36 3F 48 51
A 14 1E  28 32 3C 46 50 5A
B 16 21  2C 37 42 4D 58 63
C 18 24  30 3C 48 54 60 6C
D 1A 27  34 41 4E 5B 68 75
E 1C 2A  38 46 54 62 70 7E
F 1E 2D  3C 4B 5A 69 78 87

Now, the plan is to convert these numbers to Hexa-decimal Digital Numbers [for example; 7Ehex ® 7hex + Ehex = 15hex ® 1hex + 5hex  = 6hex]. When we do so, we get the following table:

Base 16 Multiplication Table Using Digital Numbers
one-sies two-sies three-sies four-sies five-sies six-sies seven-sies eight-sies nine-sies
2 3 4 5 6 7 8 9
2 4 6 8 A C E 1 3
3 6 C F 3 6 9 C
4 8 1 5 9 D 2 6
5 A F 5 A F 5 A F
6 C 9 F 6 C 3 9
7 E D 5 C 4 B 3
8 1 9 2 A 3 B 4 C
9 3 6 F 9 3 C 6
A 5 A 5 F A 5 F
B 7 E A 6 2 D 9
C 9 3 F C 9 6 3
D B 7 5 3 1 E C
E D B A 9 8 7 6
F F F F F F F F

 Now, divide a circle into 15 equal arcs, number the vertices using base 16 symbols [1 ... E], and connect the vertices in the order given in one of the columns.  You will produce a polygonal figure that eventually closes, as before.  The structure of the figures  [quite different in appearance from last time in the base 10 case]  is summarized here:

Sequence Name

Sequence

Figure Name
One-sies:  1 2 3  4 5 6 7 8 9 A B C D E F Regular 15-agon
Two-sies 2 4 6 8 A C E 1 3 5 7 9 B D F 2-star 15-agon
Three-sies   3 6 9 C F 3 6 9 C F 3 6 9 C F  Pentagon
Four-sies 4 8 C 1 5 9 D 2 6 A E 3 7 B F 4-star 15-agon
Five-sies 5 A F 5 A F 5 A F 5 A F 5 A F  Triangle
Six-sies 6 C 3 9 F 6 C 3 9 F 6 C 3 9 F 2-star Pentagon
Seven-sies 7 E 6 D 5 C 4 B 3 A 2 9 1 8 F 7-star 15-agon
Eight-sies 8 1 9 2 A 3 B 4 C 5 D 6 E 7 F 7-star 15-agon
Nine-sies 9 3 C 6 F 9 3 C 6 F 9 3 C 6 F 2-star Pentagon
Ten-sies A 5 F A 5 F A 5 F A 5 F A 5 F Triangle
Eleven-sies B 7 3 E A 6 2 D 9 5 1 C 8 4 F 4-star 15-agon
Twelve-sies C 9 6 3 F C 9 6 3 F C 9 6 3 F Pentagon
Thirteen-sies D B 9 7 5 3 1 E C A 8 6 4 2 F 2-star 15-agon
Fourteen-sies E D C B A 9 8 7 65 4 3 2 1 F Regular 15-agon
Fifteen-sies F F F F F F F F F F F F F F F Point

Interestingly, you get a 15-sided polygonal figure only for rows 1, 2, 4, 7, 8, 11, 13, and 14.  It is significant that these numbers, and only these numbers, have no common factors with the number 15.  The regular 15-agon is called a pentadecagon; for additional information see the website http://mathworld.wolfram.com/Pentadecagon.html. Here is a note of historical interest:

In Book IV, Euclid’s main achievement was to construct a regular pentagon in a circle. By combining this construction with that of an equilateral triangle, he was then able to construct a regular 15-sided polygon. In the 1790s, Gauss extended this idea to determine exactly which regular polygons can be constructed – they are ones based on the numbers 3, 5, 17, 257 and 65537 – the so-called Fermat primes.

08 October 2002: Monica Seelman [St James Elem]     Shoelaces, Bows, Knots, and Topology
Monica
taught us how to tie double and triple knots that can be untied by pulling the cord at one end, and she tried to figure out a pattern for such knots. She passed out a piece of cardboard rolled and taped into a cylindrical shape, which had two holes punched in it at one end, to simulate a shoe.  Also, she gave us black and white shoelaces that had been cut in half and tied together, so that each lace has a black half and a white half.  She gave us these methods for making double and triple knots:

She tried without success to tie quadruple knots that could be undone with just one string pull. Can anybody else figure out how to do it? Monica and Earl Zwicker showed various knots that could be untied by merely pulling an end of the string. Comment by PJ: Knot theory is a branch of topology, and knots that can be undone by pulling are not considered "knots" in the strict topological sense. It was a very clear presentation about a very knotty problem, Monica intervals.  To show us how simple this is, she played her recorded image back to us on our large TV Monitor, with impressive results. It was suggested that she could relate the "frame rep rate" to real time by recording the image of the second hand of a clock with her camera, either separately or as part of the apparatus.  You showed us how it really should be done,  Betty!

19 November 2002: Fred Schaal [Lane Tech High School, Mathematics]      Coding / Decoding and Matrices
Fred
showed us how to code and decode a message, based upon matrix multiplication.  He illustrated the procedure for encoding and decoding, using the following highly significant message:

GO LANE HOBBLE THE MUSTANGS
This simple message consists entirely of letters (to form words) and spaces (between words), without punctuation, capital letters, numbers, or other symbols.  Fred first encoded the message by identifying the space [_] and the 26 letters of the alphabet with the numbers 0 through 26, as follows:
_ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
The coded message can thus be converted into a list of numbers:
G O _ L A N E _ H O B B L E _ T H E _ M U S T A N G S _
 7 15 0 12 1 14 5 0  8 15 2 2 12 5 0 20  8 5 O 13 22 19 20 1 14  7 19 0
The next step is to parse the numbers in pairs, to obtain the following sequence of pairs:
| 7 15  |  0 12 |  1 14  | 5 0  |  8 15  |  2 2  | 12 5  | 0 20  |  8 5  | 0 13  | 22 19  | 20 1  | 14 7  | 19 0  |
Fred then asked us to invent a 2 dimensional matrix B of positive integers, with non-zero determinant. We came up with the following modest example:
    B =   | 6  3 |
          | 5  2 |
Then we multiplied each of the parsed pairs in the message [treated as a rows] by this matrix, to obtain new pairs. For example:
    [ 7 15 ] | 6  3 |  =  [ 117 51]   AND  [ 0 12 ]  | 6  3 | = [ 60 24]
             | 5  2 |                                | 5  2 |
We continue with to multiply to obtain the coded message [first two words shown]: 
Original Alphabet Message G O | _ L | A N | E _ First Row
Original Message as Numbers    7 15 | 0 12 | 1 14 | 5 0    Second Row
Coded Message 117 51 | 60 24 | 76 31 | 30 15 Third Row
How do we decipher the coded message contained in the third row? To do so, we first construct the inverse of the matrix B, which is
A =  | -2/3   1  | = B-1  
     |  5/3  -2  |
Note that A B-1 = I, the 2-dimensional identity or unit matrix.  To decipher the message, multiply the matrix A by the each of coded column pairs. For example
    [ 117 51 ] | -2/3  1 |  =   [7 15]     AND  [ 60 24 ] |-2/3  1 | = [ 0 12]
               |  5/3 -2 |                                | 5/3 -2 
In summary, we recover the decoded numbers on the second row, and then may convert it back to the letters of the first row. Fred said that it would be unwise for potential spies to use anything below a 5-dimensional matrix for transmitting coded messages. Here, as in many contexts, bigger is better. This is a standard "matrix code" which is simple to decode if you know the matrix, but takes some time to crack for a large matrix. The matrix must frequently be changed, of course, to guard against cracking. Relatively primitive codes of this type were used for communications in World War II.

Porter Johnson mentioned the book Between Silk and Cyanide: A Codemaker's War 1941-1945 by Leo Marks [Free Press 1999]  ISBN 0-6848-64223, which describes the experiences of a British cryptographer.  His codes for communicating with operatives behind Axis lines, based upon limericks and poems, were printed on parachute silk (You can guess what the cyanide pills were for!).  A great way to motivate students to learn matrix operations, Fred!

22 April 2003: Bill Colson [Morgan Park HS, Mathematics]        Geometry Puzzle
Bill
[passed around the drawing of an 8 unit ´ 8 unit square that had been divided into four pieces -- A, B, C, D --, as shown:

Bill cut out the pieces from the drawing and rearranged them into a 5 unit ´ 13 unit rectangle, as shown:

Note that we have been able to create a rectangle of area 65 units from a square of area 64 units!  How come? The pieces fit together remarkably well,  -- at least as well as in a jigsaw puzzle -- and there were no evident gaps.  And yet, we decided that there was something seriously wrong with the second picture.  The total area of pieces A + B+ C+ D is 64 units, as before, but the area of the rectangle is 65 units.  Bill said that there were several ways to explain the difficulty. Probably the easiest method is to note that the triangles formed by piece D and pieces D + A should be similar, so that 3 / 8 = 5 / 13! From this relation one could cross-multiply so that 39 = 40 --- which is ridiculous to everybody, except perhaps a Jack Benny aficionado.  Porter said that the diagonal line of the rectangle could not be straight, since its total length must be Ö [132 + 52]  = 13.9284... , whereas the diagonal's two segments have lengths of Ö[52+22]  = 5.3852... and Ö[32+82]  = 8.5440... , respectively.  Their total, 13. 9292... , is slightly greater than the diagonal's length!

Bill had seen this problem, as well as a number of other interesting mathematical puzzles and quandaries, in the book One Equals Zero and Other Mathematical Surprises by Nitsa Movshovitz-Hadar and John Webb [Key Curriculum Press 1998] ISBN 1-559530309-0.  [http://www.keypress.com/x6049.xml]

"One equals zero! Every number is greater than itself! All triangles are isosceles! Surprised? Welcome to the world of One Equals Zero and Other Mathematical Surprises. In this book of blackline activity masters, all men are bald, mistakes are lucky, and teachers can never spring surprise tests on their students!

The paradoxes and problems in each One Equals Zero activity will perplex your students, arouse their curiosity, and challenge their intellect. Each counterintuitive result, false analogy, and answer that defies expectation will encourage students to look at familiar mathematical situations in a new light. By solving the paradoxes, your students will come to better understand both the possibilities and the limitations of mathematics."

Finally, Bill mentioned that the following trivia questions were answered in the book:
  1. For what physical characteristic is almost everybody "above average"?  [number of fingers]
  2. For what physical characteristic is everybody "average"? [number of heads]

So that's where the missing square went to, eh Podner! Very slick, Bill!

06 May 2003: Monica Seelman [ST James School]      Psychic Puzzle
Monica presented her solution of the puzzle that appears on the Psychic Powers website:  http://mr-31238.mr.valuehost.co.uk/assets/Flash/psychic.swf (link is inactive).  On that website, you are to pick a number, such as 47, say.  Then, calculate the sum of the digits, 4 + 7 = 11.  Next, subtract the sum of the digits, 11, from the original number, 47.  The answer is 36.  In fact, the answer will always be a multiple of 9:  9 - 18  - 27 - 36 - 45 - 54 - 63 - 72 - 81.  This fact provides the explanation for the apparent demonstration of ESP on that website, since the images associated with multiples of 9 are always identical.  Check it out for yourself!  [Similar explanations were provided by Art DiVito and Ken Schug via handout.]

Monica next handed out a multiplication table, which would have looked like this, if only the one's place had been written, with any numbers in the ten's place left out, so that 7 ´ 7 = 9, etc.

´  :   0  1  2  3  4  5  6  7  8  9  0
 -      -   -  -  -  -  -  -  -  -  -
 0  :   0  0  0  0  0  0  0  0  0  0  0
 1  :   0  1  2  3  4  5  6  7  8  9  0
 2  :   0  2  4  6  8  0  2  4  6  8  0
 3  :   0  3  6  9  2  5  8  1  4   7   0 
 4  :   0  4  8  2  6  0  4  8  2  6  0
 5  :   0  5  0  5  0  5  0  5  0  5  0
 6  :   0  6  2  8  4  0  6  2  8  4  0
 7  :   0  7  4  1  8  5  2   9  6  3  0
 8  :   0  8  6  4  2  0  8  6   4  2  0
 9  :   0  9  8  7  6  5  4  3  2  1  0
 0  :   0  0  0  0  0  0  0  0  0  0  0

The following symmetries of this table almost appear to leap out of the page:

Porter Johnson pointed out that this multiplication procedure is called multiplication modulo 10, and that we have noticed some of the symmetry properties of this group of ten elements.  MODULO 10 ADDITION can also be performed (7 + 6 = 13 ® 3, etc) -- leading to a different table --  which also defines a group of 10 elementsWould the tables -- for multiplication and for addition -- be identical in form [isomorphic]?  For additional information see the website Modular Arithmetichttp://www.cut-the-knot.org/blue/Modulo.shtml.  Mathematics is about discovering patterns in things around us, and sometimes [as in playing chess, GO,  or bridge] the specific numbers are either irrelevant,  or play only an incidental role.

So that's how you do ESP with numerology!  Very nice, Monica!!

24 February 2004: Camille Gales [Coles Elementary School,  Math]           A question from "The Chicken from Minsk ... "
Camille passed out the following problem, which was taken from an out-of-print (?) book mentioned by Bill Colson at the High School Physics SMILE meeting of  08 September 1998.  Here is the question:

" ... Two mathematicians, Igor and Pavel, meet on the street. 'How is your family', says Igor. 'As I recall, you have three sons, but I don't remember their ages.' 'That's easy', says Pavel.  'The product of their ages is 36.'  Igor still looks confused, so Pavel points across the street.  'See that building?  The sum of their ages is the same as the number of windows facing us.'  Igor thinks for a minute and says, 'I still don't know their ages'.  'I'm sorry, says Pavel, 'I forgot to tell you:  my oldest son has red hair'.  'Now I know their ages', he says.
Do you? ... "
Source:  The Chicken from Minsk and 99 Other Infuriatingly Challenging Brain Teasers from the Great Russian Tradition of Math and Science, by Yuri B. Chernyak and Robert M. Rose
Can anybody help Camille to solve this problem? [Note:  there is a hint in the write-up of our 08 September 1998 meeting].
Thanks, Camille.

14 September 2004:  Walter McDonald [CPS Substitute]           Nines
Walter presented a riddle taken from the following source:

Title: 1000 PLAYTHINKS  Games of Science, Art, & Mathematics by Ivan Moscovich; Workman Publishing [http://www.workman.com/] 2002;  ISBN: 0-7611-1826-8 
This reference was also mentioned by Roy Coleman in the HS Math-Physics SMILE meeting of 24 September 2002Walter gave us the following question out of that book:
Can you find a way to express the number 100 using six 9's?
We found several ways of doing this, using six 9's and no other numerals.  Here are some of them:
9 ´ 9 + 9 + 9 + 9 / 9 = 100
(999 - 99) / 9 = 100
(999)(9.99) = 100
99 + 99 /99 = 100
(9 + 9 / 9) ´ (9 + 9 / 9 = 100
... and  ... using only four nines ...
99 + 9 /9 = 100
99 / .99 = 100
(9 / .9) ´ (9 / .9) = 100
Several variations of this question were suggested. Thanks for sharing this, Walter!

12 October 2004:  Don Kanner [Lane Tech HS,  Physics]           In-Service Meeting for Mathematics and Science Teachers 
In a weak moment Don agreed  to direct an in-service meeting of teachers in the region at Lane Tech HS on (?Saturday? 29 January 2005 -- or "whenever").  He asked our group for help in assembling written materials for reproduction (before 05 Nov 2004!), as well as suggestions as to what Physics teachers would like their students to know.  We came up with the following suggestions, in short order:

Good luck on your courageous undertaking, Don! 

09 November 2004: Riddle by Larry Alofs.

I'm thinking of a 5 digit number.  When I put a "1" after it, the result is 3 times as large as when I put a "1" in front of it.  What is the number?

07 December 2004: Porter Johnson reminded us of this question posed by Larry Alofs at the Math-Phys SMILE meeting of 09 November 2004.

"I'm thinking of a 5 digit number.  When I put a "1" after it, the result is 3 times as large as when I put a "1" in front of it.  What is the number?"
We may represent the original five-digit number as just "x", as well as  "D:abcde" in decimal form. According to the problem we need:
(expressed in decimal form):
D:abcde1 = 3 ´ (D:1abcde)
(expressed in terms of x):
10 * x + 1 = 3 * ( 105 + x)
... or ...     7 * x = 3 * 105 - 1
solve:            x = 299999 / 7 = 42857
checking:           3 ´ 142857 = 428571
... ok! ...
Note that the solution of the algebraic equation 7 x = 299999 comes out as an integer, even though we must divide by 7. Ever in search of a mathematical quandary, Fred Schaal asked whether the problem has a solution for an N-digit number for any value other than N = 5. In that more general case, one must solve the equation
7 * x  = 3 *  10N  - 1
A unique real solution x of this algebraic equation exists for any N, but x is an integer only for certain special choices:  N = 5, 11, 17, 23, ... .  For the case N = 11, the solution is x = 42857,142857 (for emphasis, commas have been inserted every 6 decimal places), which satisfies the check 
´  142857,142857  =  428571,428571
Can you find the solution for N = 17, 23, ... ?

Equations that must be satisfied for integers are called Diophantine equations.  Our equation for x is a Diophantine equation, and Fermat's Last Theorem involves surely the most famous Diophantine equation:

x N + y N = z N
This equation has solutions for integers (x, y, z) only for N = 2.

09 November 2004: Monica Seelman [ST James Elementary School]           Divisibility Test for 7 
Monica began by reviewing these simple rules to determine whether a given number is divisible by 2, 3, 4, 5, 6, 8, 9, and 10:

Number Divisibility Criterion
2 Last digit 2, 4, 6, 8, or 0
3 Sum of digits divisible by 3
4 Last two digits divisible by 4
5 Last digit 0 or 5
6 (Divisible by 2 and 3)
7 What about divisibility by 7?
8 Last three digits divisible by 8  
9 Sum of digits divisible by 9
10 Last digit 0
To consider divisibility by 7, Monica wrote down the number 1862. First, take the final digit (2), double it (4), and write it down under the second digit (6), and subtract that from the number with that final digit excised: 186 - 4 = 182. Then, repeat the process, taking the new final digit (2), doubling it (4), and subtracting that from the number with the final digit excised: 18 - 4 = 14. The number 14 is divisible by 7, so the original number is, as well. How can such a process possibly work? Let us write it out in detail:
 
               1862                          1862
              -  4             OR           -  42
               ---                           ----
               182                           1820 
              - 4                           - 420 
               --                            ----
               14                            1400

                      OR
                      
                1862 = 42 + 420 + 1400
                     = 7 ( 6 + 60 + 200 )
                     = 7 ( 266 )
We are subtracting a multiple of 7 from the original number at each stage. If we end up with a result divisible by 7, then the original numbers is, as well. Simple, non?  Now, how do we decide whether a number is divisible by 11?

Fascinating, Monica!

07 December 2004: Monica Seelman [ST James Elementary School]           Testing for Divisibility by 7: Follow-up 
Monica  reminded us of  the test described at the last MP SMILE meeting using the number 2164 --- which is not divisible by 7.  We begin by taking the last digit (4), doubling it, writing both numbers down (84), subtracting that from the test number (2164), and dropping the trailing zero.  We repeat the procedure until we cannot continue.  Is the remaining number divisible by 7?  If so, then so is the original number.  If not, then the original number is not divisible by 7.  Here is the example:

                PARTIAL               FULL   
               2  1  6  4          2  1  6  4
             -       8  4         -      8  4  
               ----------          ----------
               2  0  8             2  0  8  0 
             - 1  6  8            -1  6  8  0
               -------             ----------
                  4                   4  0  0

   4 IS NOT DIVISIBLE BY 7   --  AND NEITHER IS 400
                         STOP
Monica pointed out that the following numbers occur on the subtraction line:
Last Digit: 1 2 3 4 5 6 7 8 9
Subtraction Line:     21  42  63  84  105  126  147  168  189
All numbers on the subtraction line are divisible by 7 --- in fact they are divisible by 21.   Thus, we are always subtracting a multiple of 7, even after the zeros are included. 
Ain't Mathematics Wonderful?? Thanks, Monica.

15 November 2005: Larry Alofs  (Kenwood HS, retired)           LED's, etc.
Larry
handed out copies of  a Bill Amend classic "Foxtrot" comic strip from November 6, 2005, in which Jason presented a numerical word search (or nerdsearch): http://www.livejournal.com/community/comic_foxtrot/66515.html?mode=reply. The clues were various terms, integrals, derivatives, sums, etc. One of the questions involved the sum of squares of the first 47 integers, which is a special case of the expression

S(n) = 12 + 22 + 32 + . . . + n2
for n=47. One can show by induction that
S(n) = n (n + 1) (2 n + 1) / 6
The answer is correct for n=1; S(1) = 1. Assume that the answer is correct for the integer n. Then
S(n+1) = S(n) + (n+1)2 = (n+1) ( n (2n+1)/6 + (n+1)) = (n+1)/6 ( 2 n2 + n + 6 n + 6)
= (n+1)/6 ( 2 n2 + 7 n + 6) = (n+1)/6 (n+2) (2n+3).
Thus the result is true for n+1. For the specific example we have
S(47) = 35720 . . . S(48) = 35720 + 482 =38024
This is a really neat way for students to hone their skills in various areas of mathematics to which with they may have been exposed. 

Larry showed us some LED flashlights, a type of flashlight for which he is an enthusiastic proponent. White LED's need enough cells to provide more than 3.5 Volts (eg, three individual cells are required). Larry then showed a flashlight with a white LED that works with a single cell (1.5 volts). So how do they get the voltage high enough to run the flashlight?! Larry was not sure, but one possibility is a Voltage Multiplier circuit (see http://www.voltagemultipliers.com/html/multcircuit.html), which uses capacitors and diodes in series.  Larry made an analogy with the movement of water up a pipe against gravity (increasing its potential energy) using one-way valves separating segments of a pipe and a device that can simultaneously squeeze a section of the pipe and expand the section of the pipe just above the squeezed section. Very slick! Thanks, Larry!

24 January 2006: Dr Russell Bonham, research faculty member in Chemistry at IIT, provided this word problem, which had been given to an 8th grade public school class in Barcelona, Spain:

"A Duke arrives at a train station every day at 5 PM. He is met by his driver and taken home. One day the Duke catches an earlier train and arrives at the station at 4 PM and starts walking towards his home. The driver picks him up on the road and he arrives home 20 minutes sooner than usual. Assume the car and the man travel at constant speeds. For how long did the Duke walk?"
Can your students solve this problem?

21 February 2006: Walter McDonald (CPS substitute teacher and radiation technologist at the VA)                    Sangaku
Walter
told us about “sangaku”, Japanese temple geometry, which was a very popular pastime in Japan in the Edo Period (when there were Samurai), 1603 - 1867. See  http://www.wasan.jp/english/.  The participants figured out solutions to geometrical problems and puzzles and then recorded the solutions on beautiful wooden tablets. These tablets were sometimes hung under the roofs of shrines and temples. Walter got the example from the book Play Thinks by Ivan Moscovich  [http://www.amazon.com/1000-Play-Thinks-Paradoxes-Illusions/dp/0761118268].  Walter gave us one such geometrical problem as an example. It is a problem similar in spirit to those given in high school geometry. See the Mathworld web page http://mathworld.wolfram.com/CircleInscribing.html for a discussion of this problem.

21 March 2006: Ann Brandon (Joliet West, retired)                    Math Puzzle (Kakuro) and Paper Clip Energy
Ann
had us each take a paper clip and bend it into a triangle with one side of the triangle being formed from the two overlapping ends of the clip and held together by friction. When you drop it and the ends come apart, the potential energy stored in this configuration may be  released when it it hits the ground, and the triangle may jump back up. We didn't have a lot of success in our attempts to launch this device!

Ann then showed us two examples of a Kakuro puzzle, sort of like a crossword puzzle but with arithmetic sums (across and down) instead of words. You are not allowed to use the same numeral (1-9) in the same row or column of a given block.  Her puzzle was taken from the book Kakuro presented by New York Times puzzle editor Will Shortz --- one hundred addictive puzzles, St Martins Press 2006 (ISBN: 0-312-36042-8):  http://www.amazon.com/Kakuro-Presented-Will-Shortz-Addictive/dp/0312360428.  See also the book Kakuro for Dummieshttp://www.amazon.com/Kakuro-Dummies-Andrew-Heron/dp/047002822X.  These puzzles can be quite difficult, since they cannot be completed one piece at a time!
Rather puzzling, Ann.