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Bertrand Russel:
Mathematics possesses not only truth but supreme beauty, a beauty cold and austere,
like that of sculpture, sublimely pure and capable of a stern perfection, such as only the
greatest art can show.
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The Beauty in Numbers
Question 1.1. We often give our students various puzzles that slowly lead into algebraic explanations of the demonstrated phenomena. These interesting patterns may be familiar to you. Or not.
If they are, feel free to move forward.
1. Record squares of the following numbers: 1, 11, 111, 1111, 11111, . . . , 111111111.
2. Record the following sums: a · 8 + a0 s one’s digit, where a ∈ {1, 12, 123, 1234, . . . , 123456789}.
3. Record the following sums: a·9+(a0 s one’s digit+1), where a ∈ {0, 1, 12, 123, 1234, . . . , 12345678}.
Follow up by finding the following sums: a·9+b, where as a moves through {0, 9, 98, 987, 9876, . . . , 98765432},
b moves through {8, 7, 6, . . . , 0}. Then, of course, multiply the results with the same number
of digits.
4. Multiply 142857 by numbers 2 through 8.
5. Multiply 12345679 by the first nine multiples of 9. Do the same with 987654321.
6. Or maybe you should multiply 999999 by numbers 1 thorugh 10?
7. Or square numbers whose every digit is 9?
Question 1.2. Perform the following: Choose any three-digit number (where the units and hundreds digits are not the same). Reverse the digits of this number you have selected. Subtract
the two numbers (naturally, the larger minus the smaller). Once again, reverse the digits of this
difference. Now, add your last two numbers. What did you get?
Of course, you can investigate why this happens. Before you do so, inspect the following:
• Look at the first nine multiples of 1089. What do you notice?
• You’ve found that 1089 · 9 = 9801. Consider 10989 · 9, 109989 · 9, etc.
• Are there any other four-digit numbers whose multiple is equal to its reversal? Does the
inserting the nines phenomenon hold for that number as well?
• Start with any two-digit number, add the squares of its digits to form a new number, continue
the process. This sequence of numbers will eventually reach 1 or 89. Is that surprising?
• Explain the original puzzle.
• As a final partying from 1089: it appears that 1089 is both a square and difference of squares:
332 = 652 − 562 which is unique among two-digit numbers!
Question 1.3. You’ve considered sums of squares of digits of any given number. Let’s try to
consider sums of cubes of the number’s digits. What do you notice? Are there any numbers where
the sum of the cubes of its digits is the same as the original number?
Question 1.4. Let’s try another procedure:
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• Begin by selecting a four-digit number (except one that has all digits the same).
• Rearrange the digits of the number so that they form the largest number possible. Then
rearrange the digits of the number so that they form the smallest number possible.
• Subtract these two numbers (obviously, the smaller from the larger).
• Take this difference and continue the process, over and over and over, until you notice something disturbing happening. Dont give up before something unusual happens.
Question 1.5. Using any whole number as the first term, form a sequence using the following
rules:
If the number is odd, then multiply by 3 and add 1.
If the number is even, then divide by 2.
What happens?
Question 1.6. Which numbers can be expressed as the sum of consecutive integers? Can you
prove your claims?
Question 1.7. People love to play word games, so why not bring the word games into mathematics?
Palindromes are, for some reason, entertaining, and often found in dates. There were several
dates in October 2001 that appeared as palindromes when written in American style: 10/1/01,
10/22/01, and others. In February, Europeans had the ultimate palindromic moment at 8:02 p.m.
on February 20, 2002, since they would have written it as 20.02, 20-02-2002. What other numbers
are palindromic? Is there a way to generate a whole bunch?
Question 1.8. If you’re into weird shortcuts, you should think about speed multiplying:
11 To multiply a two dgit number by 11, just add the two digits and place the sum between the
two digits, or
21 Double the number, then multiply by 10 and add the original number.
31 Triple the number, then multiply by 10 and add the original number.
41 Quadruple the number, then multiply by 10 and add the original number.
Question 1.9. When Russian peasants multply numbers they use the method depicted in the
following table:
43
92
86
46
172 23
344 11
688
5
1376
2
2752 1
so 43 · 92 = 172 + 344 + 688 + 2752 = 3956. So all you ever need to know how to do is multiply
and divide by 2! Explain.
16 19 26 49
Question 1.10. Reduce to lowest terms the following fractions: 64
, 95 , 65 , 98 . Compare the results
to the original numbers. What do you notice? Can you find all the two digit numbers for which
this nifty reduction works? What about three digit numbers?
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Geometric Wonders
Question 2.1. The point P is any point on the circle with center O. Perpendicular lines are drawn
from P to perpendicular diameters, AB and CD, meeting them at points F and E, respectively.
If the diameter of the circle is 8, what is the length of EF ?
Question 2.2. In the figure shown below, point E lies on AB and point C lies on F G. The area
of parallelogram ABCD = 20 square units. Find the area of parallelogram EF GD.
Question 2.3. Compare these two problems:
1. Consider the globe of the earth with a rope wrapped tightly around the equator. The rope
will be about 24,900 miles long. We now lengthen the rope by exactly 1 yard. We position
this (now loose) rope around the equator so that it is uniformly spaced off the globe. Will a
mouse fit under the rope?
2. Suppose the world is a perfect sphere, and a rope is taut with the world along a great circle.
How much longer would the rope need to be to be lifted 10cm of the ground all along its
length?
Which question would you be more tempted to give to your students?
Question 2.4. Consider any two lines, each with three points located anywhere on the lines. Then
connect the points of the first line to those on the second line, but do not connect the corresponding
points. That is, don’t connect the rightmost point on one line to the rightmost point on the other
line or dont connect the two middle points. What can you say about the three intersection points
that were formed?
Question 2.5. In 1640, at the age of 16, the famous mathematician Blaise Pascal published a
one-page paper titled Essay Pour les Coniques, which presents us with a most insightful theorem.
What he called mysterium hexagrammicum states that the intersections of the opposite sides of a
hexagon inscribed in a conic section are collinear.
Question 2.6. In 1806, at the age of 21, a student at the cole Polytechnique, Charles Julien
Brianchon (17851864), published an article in the Journal de Lcole Polytechnique that was to
become one of the fundamental contributions to the study of conic sections in projective geometry.
His development led to a restatement of the somewhat forgotten theorem of Pascal and its extension,
after which Brianchon stated a new theorem, which later bore his name. Brianchons theorem, which
states “In any hexagon circumscribed about a conic section, the three diagonals cross each other
in the same point.”
Question 2.7. The feet of the perpendiculars drawn from any point on the circumcircle of a
triangle to the sides of the triangle are collinear.
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Question 2.8. Finally, fold some stuff: I”m sure you’ve worked with the golder ratio before. A
one dimensional version of the “beauty” is called the Golden Section. Simply, for the segment AB,
the point P partitions (or divides) it into two segments, AP and P B, such that
AP
PB
=
.
PB
AB
Take a strip of paper, say about 12 inches wide, and make a knot. Then very carefully flatten
the knot as shown in the next figure. Notice the resulting shape appears to be a regular pentagon,
that is, a pentagon with all angles congruent and all sides the same length.
Can you find golden sections? What might you want to call by a name of Golden Triangle?
Question 2.9. Let A and B be the midpoints of th sides EF and ED of an equilaterals triangle
DEF . Extend AB to mee the circumcircle of DEF at C. B divides ACaccording to the golden
section.
Question 2.10. Let c be the length of the hypotennuse of a rigth triangle whose other two sides
have lengths a and b. Prove that (no, of course not, we know that one):
√
a + b ≤ c 2.
When does the equality hold?
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