High School Mathematics-Physics SMILE Meeting
1997-2006 Academic Years
Mathematics: Puzzles
14 October 1997: Roy Coleman [Morgan Park High School]
and erase the
right line you get
(good come-back
for students that say "I only left off a little bit") 7641 Number not same digits, in decreasing order
1467 same digits, but in increasing order
----------- subtract
6174 This number, sometimes called Voodoo
number, will always occur, although it may
take up to 15 iterations to appear.
It is mapped into itself by the iteration.
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
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:
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:
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 ResearchFirst 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.
915 118th Avenue, SE
PO Box 96088
Bellevue WA 98009
http://www.ber.org/
Tel: 1 - 800 - 735-3503
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 79. Isn'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.
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 |
Subtraction: Original Problem Modulo 9
5273 --->
8 |
Multiplication: Original Problem Modulo 9
635 --->
5 |
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:Thank you for sharing that, Monica!The sum of the squaws on the two hides is equal to the squaw on the hippopotamus.
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:
|
... OR ... |
|
Note that, in each case, all numbers from 0 to 15 are present, and the rows, columns, and diagonals add up to 30. Fred 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":
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 |
| 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 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
| 3 | 6 | 9 | 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 |
| 1 | 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 | 3 | 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 | 3 | 1 | 8 | 6 | 4 | 2 | 9 |
| 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 9 |
| 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:
| one-sies | two-sies | three-sies | four-sies | five-sies | six-sies | seven-sies | eight-sies | nine-sies |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 4 | 6 | 8 | A | C | E | 10 | 12 |
| 3 | 6 | 9 | C | F | 12 | 15 | 18 | 1B |
| 4 | 8 | C | 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:
| one-sies | two-sies | three-sies | four-sies | five-sies | six-sies | seven-sies | eight-sies | nine-sies |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 4 | 6 | 8 | A | C | E | 1 | 3 |
| 3 | 6 | 9 | C | F | 3 | 6 | 9 | C |
| 4 | 8 | C | 1 | 5 | 9 | D | 2 | 6 |
| 5 | A | F | 5 | A | F | 5 | A | F |
| 6 | C | 3 | 9 | F | 6 | C | 3 | 9 |
| 7 | E | 6 | D | 5 | C | 4 | B | 3 |
| 8 | 1 | 9 | 2 | A | 3 | B | 4 | C |
| 9 | 3 | C | 6 | F | 9 | 3 | C | 6 |
| A | 5 | F | A | 5 | F | A | 5 | F |
| B | 7 | 3 | E | A | 6 | 2 | D | 9 |
| C | 9 | 6 | 3 | F | C | 9 | 6 | 3 |
| D | B | 9 | 7 | 5 | 3 | 1 | E | C |
| E | D | C | B | A | 9 | 8 | 7 | 6 |
| F | 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:
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:
| _ | 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 |
| 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 |
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 |
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:

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!Finally, Bill mentioned that the following trivia questions were answered in the book: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."
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 elements. Would the tables -- for multiplication and for addition -- be identical in form [isomorphic]? For additional information see the website Modular Arithmetic: http://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.Can anybody help Camille to solve this problem? [Note: there is a hint in the write-up of our 08 September 1998 meeting].
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
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: 0-7611-1826-8This reference was also mentioned by Roy Coleman in the HS Math-Physics SMILE meeting of 24 September 2002. Walter gave us the following question out of that book:
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:
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:
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:
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 |
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 |
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
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 Dummies: http://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.