Download Some Math Club Experiences Shailesh Shirali 5–7 April, 2012

Some Math Club Experiences
Shailesh Shirali
Rishi Valley School, Rishi Valley, A.P., India
Community Math Centre, Rishi Valley School
5–7 April, 2012
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Apr 2012
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No one would dispute the importance of having a forum like a Math Club
in a school. (Or, similarly, having a Math Circle in one’s locality.)
In this brief talk I will describe some activities I have conducted at a Math
Club for upper primary and secondary level students.
Essentially we solved interesting problems. In some cases I followed this by
developing some theory based on the problems. On other days we did some
investigations. Typically, all these were in the areas of Elementary Number
Theory (ENT), Combinatorics, Number Games, Board Games and so on.
I was often struck by the sophistication of children’s reasoning.
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Apr 2012
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Partial list of topics
1
Problems on divisibility
2
Constructing third order magic squares
3
Solving cryptarithms
4
Problems in combinatorics
(Other topics covered: Triangular Numbers; Number Sequences; Method of
Differences; Fibonacci Numbers; Enumeration; Graph Theory; . . . .)
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Apr 2012
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Part I
Problems on Divisibility
Is it true that . . .
1
. . . the sum of three consecutive numbers is always a multiple of 3?
2
. . . the sum of two consecutive odd numbers is always a multiple of 4?
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Apr 2012
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Sum of three consecutive numbers
Example of pre-algebra reasoning:
The middle number is 1 less than the previous number and 1 less
than the next number. Therefore the sum is the same as 3 times
the middle number.
This is a kind of verbal proof. But it shows a clear understanding of the
underlying algebra.
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Apr 2012
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Sum of two consecutive odd numbers
Example of pre-algebra reasoning:
If you add two consecutive odd numbers, it is the same as
doubling the number in-between. This number is even, so the
sum is divisible by 4.
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Apr 2012
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Proof by induction!
Some students gave what is unmistakably the seed of a proof by induction.
Thus, one student wrote:
1 + 3 = 4, every time you add 2 to each, therefore you add 4 more.
In essence this is an inductive proof.
This experience suggests that the inductive principle is implicit in the way
we internalize the number system.
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Apr 2012
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A corollary
One student even noticed a corollary:
The sum of 4 consecutive numbers is never a multiple of 4.
This comes in the wake of:
• The sum of 2 consecutive numbers is never a multiple of 2.
• The sum of 3 consecutive numbers is always a multiple of 3.
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Apr 2012
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Here’s how the student argued:
The sum of four consecutive numbers is twice the sum of the two
in-between numbers, and this sum is always odd. Twice an odd
number is not a multiple of 4. Hence . . . .
In the same way one can say: The sum of 6 consecutive numbers is always
a multiple of 3. But it is odd, being a sum of three odd numbers. Hence it
is never a multiple of 6. Thus, it is 3 more than a multiple of 6.
Generalization: The sum of k consecutive numbers is a multiple of k if k is
odd, and 21 k more than a multiple of k if k is even.
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Apr 2012
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Investigations
1
Can every positive integer be expressed as the difference of two perfect
squares?
2
Can every positive integer be expressed as the sum of two or more
consecutive positive integers?
3
Is it true that there is no perfect square whose Tens and Units digits
are both odd?
4
Is it true that from any collection of five integers, one can select three
integers whose sum is a multiple of 3?
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Apr 2012
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Difference of two squares
We shall use the known fact that the sum of the first n odd numbers is n2
(i.e., 1 + 3 + · · · + (2n − 1) = n2 ). So: a difference of two squares is the
sum of consecutive odd numbers. Example:
52 − 22 = (1 + 3 + 5 + 7 + 9) − (1 + 3) = 5 + 7 + 9.
If the number of odd numbers is odd, then the sum is odd.
If the number of odd numbers is even, then the sum must be a multiple of
4, since the sum of a consecutive pair of odd numbers is a multiple of 4.
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Apr 2012
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A corollary
So a sum of consecutive odd numbers is either odd or a multiple of 4.
Corollary
A number of the form 4k + 2 cannot be written as a difference of two
squares.
This does not fully answer the question — we still have to prove that every
odd number and every multiple of 4 can be expressed as a difference of two
squares. But that is easily answered. (Any odd number can be written as
n2 − (n − 1)2 . Now multiply by 4 as many times as needed . . . .)
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Apr 2012
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In the end we get a pretty result! —
Theorem
A positive integer n can be written as a difference of two perfect squares if
and only if it is not of the form 4k + 2.
It is striking to see how ideas from different strands of mathematics hang
together . . . .
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Apr 2012
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Part II
Third Order Magic Squares
Magic squares are a favourite topic for most audiences.
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The task
Fill the 9 cells with the numbers from 1 to
9, using each number just once, so that the
rows, columns and diagonals all have the
same sum.
The common sum of the rows, columns and diagonals is the magic sum of
the magic square.
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The magic sum
The first question is:
What is the value of the magic sum?
The answer comes forth soon enough: The sum of all the 9 numbers is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45,
so the magic sum must be 45 ÷ 3 = 15.
This part comes easily enough from the children, though not instantly.
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Apr 2012
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The central number
The next question is:
What number occupies the central cell?
This is much less obvious!
A common response: The central number must be 5, because it is the
central number in the set of the digits from 1 to 9.
However, they are unable to justify this meta-mathematical logic! (Not
very surprising — it isn’t logical at all!)
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But children sometimes come up with surprising solutions . . . .
Let e be the number in the central square, e ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}.
We consider the different possibilities for e, and then eliminate them one
by one.
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If e ∈ {9, 8, 7, 6} . . .
What could be the neighbours of the central cell?
8
9
8
7
9
Remember: the
9
magic sum is 15.
Hence: e 6= 9,
9
8
9
Shailesh Shirali (RVS)
9
6
Math Club
9
e 6= 8, e 6= 7,
e 6= 6.
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If e ∈ {1, 2, 3, 4} . . .
What could be the neighbours of the central cell?
2
1
2
3
1
Remember: the
1
magic sum is 15.
Hence: e 6= 1,
1
2
1
Shailesh Shirali (RVS)
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4
Math Club
1
e 6= 2, e 6= 3,
e 6= 4.
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The Sherlock Holmes principle
Therefore, e 6= 9, 8, 7, 6, and also e 6= 1, 2, 3, 4. Hence e = 5.
When you have eliminated the impossible, whatever remains,
however improbable, must be the truth.
There is just one possible occupant for the central square!
Having got the central number and knowing the magic sum, it is easy to fill
in the remaining entries. (But even here there is scope for working out the
solution.)
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Apr 2012
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8
1
6
3
5
7
4
9
2
Children naturally get different versions of this basic square, and one can
bring out ideas of reflection and rotational symmetry, invariance, etc.
(The account I have presented is actually how it happened.)
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Apr 2012
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Another proof
Actually, a stronger result is possible.
x
In a 3 × 3 magic square with magic sum S, if the
number in the central square is x , then S = 3x .
First, a trivial observation: In a 3 × 3 magic square with magic sum S, the
sum of all the numbers in the square is 3S.
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Draw four ‘lines’ as shown, through the central
cell. Add the numbers along each line. As each
x
line has sum S, the four lines yield a sum 4S.
The four lines cover all the cells in the square.
Every cell gets covered just once — except the
central cell which gets covered four times.
This implies that 4S = 3S + 3x , and so S = 3x .
Hence the average of the numbers in each line is x .
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Corollary
In a 3 × 3 magic square, the numbers in each ‘line’ through the central
square are in arithmetic progression.
Following this, one can proceed to study 4 × 4 magic squares, which too
have some nice laws.
But a very nice way of ending this unit is to draw attention to the fact that
the above corollary and the observation that S = 3x are proved in the first
chapter of a very famous book. (Can you guess which one?)
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Part III
Cryptarithms
Coded arithmetic problems always go down well with children at
this age.
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Problem. If A, B, C , D, E represent the digits 1, 2, 3, 4, 5 in some order,
and
AB × C = DE ,
find which letter represents which digit. (Here, ‘AB’ refers to the two digit
number with tens digit A and units digit B. Likewise for ‘DE ’.)
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
 {A, B, C , D, E } = {1, 2, 3, 4, 5},

AB × C
= DE .
Solution. Children quickly realize that 5 is the ‘key’ digit. Question.
Which letter represents 5?
They see that C 6= 5; next, B 6= 5 and also E 6= 5.
Then they see that A 6= 5. Hence D = 5, and DE ∈ {51, 52, 53, 54}.
The factorizations 51 = 17 × 3, 53 = 53 × 1, 54 = 18 × 3 = 27 × 2
show that DE = 52. Now they have the answer: 13 × 4 = 52.
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More problems of this genre . . .
1
Which four digit numbers have the property that under multiplication
by 4, the digits get reversed? —
ABCD × 4 = DCBA.
2
Solve this cryptarithm: PAPA × 2 = MAMAN.
3
Solve this cryptarithm: ABCDEF × 5 = FABCDE .
4
Solve this cryptarithm: TWO × TWO = THREE .
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Part IV
Combinatorics
A somewhat unfamiliar subject at the school level, the problem
solving opportunities that combinatorics offers is enormous. Two
examples:
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Diktat from a classteacher
A class has 25 children seated in a square
array. The teacher asks them to change
their places as follows: Each child must
shift to a desk immediately in front of her,
behind her, to her left, or to her right.
Can the children do what they have been asked?
Hint. Examine the cycles created by the movements.
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A party game
Take any number n. Replace it by any two smaller numbers, say a and b,
whose sum is n. Record their product ab.
Repeat the same step for each of the two smaller numbers, and iteratively
continue this till all the numbers left are 1s.
Now compute the sum of all the products. Whichever route you take to
reach the final string of 1s, you get the same sum in the end!
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For example, starting with n = 6 one may trace the following trajectory:
Step
Partition
Product
1
4+2
8
2
(3 + 1) + 2
3
3
(3 + 1) + (1 + 1)
1
4
((2 + 1) + 1) + (1 + 1)
2
5
(((1 + 1) + 1) + 1) + (1 + 1)
1
The sum of the products is 8 + 3 + 1 + 2 + 1 = 15.
Check that you get the same sum whichever trajectory you take.
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Part V
The Burden of Proof
Traditionally, geometry has been seen as the stage where students learn the
notion of formal proof. There are good reasons for this.
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1
2
Setting is concrete, the objects
Suitable for developing the skill
can be ‘seen’.
of framing and dealing with
Assertions made are easy to
intermediate objectives.
understand.
3
4
5
Allows the possibility of
Involves contemplating many
significant mathematical
statements together, drawing
learning without having to
conclusions from the collection
grapple with the complexities of
as a whole.
axiom systems.
McClure, J.E. (2000). Start where they are: geometry as an introduction
to proof. American Mathematical Monthly, 107, 44–52.
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Also, in geometry there is the situation where a small number of plausible
assumptions lead to a large number of surprising and beautiful results.
Isaac Newton expressed this point beautifully, in his inimitable style, when
he wrote in the Preface of Principia Mathematica:
It is the glory of geometry that from so few principles, fetched from
without, it is able to accomplish so much.
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But similar things can be said about elementary number theory (ENT)!
ENT may not have the incredible richness of plane geometry.
But it is a field where exploration and experimentation are easier to do.
Many results can be empirically stumbled upon by students.
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And then there are these inspiring words from Prof G H Hardy:
The elementary theory of numbers should be one of the very best
subjects for early mathematical instruction. It demands very little
previous knowledge; its subject matter is tangible and familiar;
the processes of reasoning which it employs are simple, general
and few; and it is unique among the mathematical sciences in its
appeal to natural human curiosity.
A month’s intelligent instruction in the theory of numbers ought
to be twice as instructive, twice as useful, and at least ten times
as entertaining as the same amount of ‘calculus for engineers.’
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