4. MATHEMATICAL PROBLEM SOLVING
Historically, algebra grew out of arithmetic,
and it ought so to grow afresh for each
individual.
Mathematical Association, 1934, 5
Over the centuries mathematicians, both amateur and professional, have been
fascinated by questions and problems about integers. Attempts to solve these
problems have led to significant developments in various branches of mathematics.
In this section we consider number theoretic problems – these include multiples,
remainders, divisibility, primes, sequences and series, Fibonacci numbers, etc. We
look at mathematical problem solving, i.e. “abstract problems”, activating all our
knowledge of previous sections about induction and deduction, analysis and
synthesis and the logic of proof.
Here is an appetiser:
A little logic
PROBLEM 127:
Find the value of
…+
1
x3
+
1
x2
+
1
x
2
3
+1+x+x +x +…
The famous mathematician Leonhard Euler (1707-1783) solved it like this:
…+
= (… +
=
1
x3
1
x3
+
+
1
x2
1
x2
+
1
x
+
1
x
2
3
+1+x+x +x +…
2
3
+ 1) + ( x + x + x + …)
1
x
+
1
1− ( x ) 1− x
x
x
+
x −1 1− x
=0
=
Do you agree? Motivate!
PROBLEM 128:
1
1
1
1
1
Use induction to find a formula for Sn = 1 + 2 + 4 + 8 + 16 + 32 + …
Use your formula to find the values of S10 and S100.
Now prove that your formula is correct using deduction.
n
Use deduction to deduce the value of S∞, i.e. S∞ = Lim ∑
n →∞
186
i =1
1
2 n −1
PROBLEM 129:
Study this alternative approach to find S∞of the series in problem 128.
Comment on the validity of the argument and the truth of the conclusion:
1
1
1
1
1
1
1
S∞ = 1 + 2 + 4 + 8 + 16 + 32 + …
1
1
1
1
⇒ S∞ = 1 + 2 (1+ 2 + 4 + 8 + 16 + 32 + …)
1
⇒ S∞ = 1 + 2 S∞
1
⇒ 2 S∞ = 1
⇒ S∞ = 2
PROBLEM 130: Check this reasoning – comment on the validity of the
argument and the truth of the conclusion:
S∞ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + …
⇒ S∞ = 1 + 2(1 + 2 + 4+ 8 + 16 + 32 + 64 + …)
⇒ S∞ = 1 + 2S∞
⇒ S∞ = -1
PROBLEM 131:
Find the solution to the equation (2x + 4)2 = (x + 1) 2
Comment on the following proposed solution:
(2x + 4)2 = (x + 1)2
⇒ 2x + 4 = x + 1
⇒ x = -3
(Take the square root of both sides)
Consecutive numbers
PROBLEM 132: CONSECUTIVE NUMBERS
Investigate the sum of 3 consecutive natural numbers.
We will discuss this problem at length to reinforce our basic approach to investigative
mathematical activity.
An inductive attack may yield:
1+2+3=6
2+3+4=9
3 + 4 + 5 = 12
4 + 5 + 6 = 15
We easily recognise that the results are all multiples of 3, leading to the conjecture
that the result will always be a multiple of 3. However, realising the pitfalls of
induction, we should realise that it is necessary to prove the validity, or to try to
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explain why the result is a multiple of 3. Such a proof or explanation calls for the
introduction of a “generalised number” to cater for any natural number, indeed, for all
natural numbers. The sum of three consecutive natural numbers can then by
symbolised as follows:
n + (n + 1) + (n + 2)
The advantage of having first formulated an inductive conjecture is that we now know
what we want, therefore the next step seems obvious:
n + (n + 1) + (n + 2) = 3n + 3
n + (n + 1) + (n + 2) = 3(n + 1)
From this form, we can recognise that 3(n + 1) is a multiple of 3 for any value of n and
we have proved the validity of our conjecture through deduction.
We make a few further remarks:
First: In the inductive approach we formulated the conjecture that the result is always
a multiple of 3. Now, while looking at the general result 3(n + 1), in the stage Polya
calls looking back, if we analyse the meaning of the expression 3(n + 1), we realise
(discover, if you like) that interpreting n+1 as the “middle number”, the answer is
always 3 times the middle number. In other words, we do not only know that the result
is a multiple of 3, we have now through deduction discovered which multiple of 3.
Often, to understand the meaning and significance of such a generalisation, it is
necessary to illustrate it through specialisation, it is through looking at special or
specific cases. For example, if we look at 99 + 100 + 101, this new “theorem” says
that without adding, we can find the result as 3 × 100 = 300. (We note that in this
sense 3(n + 1) is a function rule, and specialisation involves finding function values,
one of our problem types!)
Second: The meaning and significance of manipulation in such generalised arithmetic
situations (mathematical problem solving or number theory), is different from that in
“applied problem solving” in the previous section. We recall that in applied problem
solving we constructed more convenient equivalent expressions to make finding of
output numbers, finding of input numbers and the analysis of the behaviour of function
values easier (see, for example, Activity 15, Scarce metal). Here the purpose of
manipulation is to construct an equivalent expression that we can readily recognise as
being of a certain form, for example as a multiple of 3 (the form is 3n), or as an odd
number (the form is 2n + 1), or as a remainder of 4 after division by 5 (the form is 5n +
4).
Third: With experience one learns that it is often more useful, and even more beautiful
to the mathematician to symbolise such situations as
(n – 1) + n + (n + 1) = 3n
We note that n has changed its meaning: n still represents any natural number, but in
this context it now refers to the middle number, and n + 1 now refers to the third number.
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PROBLEM 133:
1. The sum of three consecutive numbers is 86. Find the three numbers.
2. The sum of three consecutive numbers is 126. Find the three numbers.
One could pack up and leave it there. But that is not the mathematical attitude!
Looking back, the mathematical point of view immediately wants to extend this. We
vary the conditions, generalise, for example wondering what if we take four numbers,
or five …? Let’s jump into a daring guess:
We know that the sum of 3 consecutive numbers is a multiple of 3.
Conjecture: The sum of n consecutive numbers is a multiple of n.
Of course we understand what that means – through specialisation we mean that we
are simultaneously saying that the sum of 4 consecutive numbers is a multiple of 4,
the sum of 10 consecutive numbers is a multiple of 10, etc.
We have two tasks:
1. Is this conjecture true? Meaning is it always true?
2. If so, why? If not, why not?
To tackle the problem one would interplay between induction and deduction. I would
start with a little bit of specialisation, because, frankly, I am not sure how to set about
proving the general case! Specialisation will give me a feeling for the situation and
may give me some clues of how to tackle the general proof. However, checking some
specific cases of the conjecture will also give us an indication of whether we believe
the conjecture is true or not. It serves little purpose in trying to prove a conjecture that
turns out to be not true! We gain psychological conviction that a conjecture is true or
not through checking special cases.
For example, let’s test if it is true that the sum of four consecutive numbers is a
multiple of 4.
On the spur of the moment, and contradictory to the remark in the previous
paragraph, let’s try to give a general, deductive proof for the sum of any four
consecutive numbers. So let n be the first of the 4 numbers, then
S4 = n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3)
From the structure (form) of this result we can at most say that the sum is always
even (a multiple of 2). Actually, S4 can never be a multiple of 4, because the second
factor 2n + 3 is always odd.
Do you make sense of this compound statement? Let’s doodle on this for awhile.
There are three things to understand. First, the statement that 2(2n + 3) can never be
a multiple of 4. Do you agree? Does it make sense? Maybe you would like to check it
by specialising for n = 1, 2, 3, 4, …:
2 × 5, 2 × 7, 2 × 9, 2 × 11, … = 10, 14, 18, 22, …
None of the four examples are multiples of 4, so it seems correct. But are you sure that the
sequence 10, 14, 18, 22, … will never produce a multiple of 4? How can you be sure?
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Second, the statement: “the second factor 2n + 3 is always odd”. Are you convinced
that the statement is true? We can check 2n + 3 for n = 1, 2, 3, 4, … It is
5, 7, 9, 11 … Are you sure it can never generate a multiple of 2? That would be an
inductive generalisation, and we cannot be sure that the pattern is continued for ever.
However, we can reason it out logically by using the structure of the expression:
2n is an even number for all n ∈ N.
So 2n + 3 is an even number plus an odd number, which is an odd number.
So 2n + 3 cannot ever be even (a multiple of 2).
The third thing we must understand is the logical connection: because! The second
statement is presented as explanation or proof for the first. How does the statement
“2n + 3 is always odd” prove that “2(2n + 3) can never be a multiple of 4”? We should
be able to complete the logical reasoning: For 2(2n + 3) to be a multiple of 4, the
second factor (i.e. 2n + 3) must also be a multiple of 2 (4 = 2 × 2), which it is not. So
the argument follows the Modus Tollens structure: P ⇒ Q, Q is false, so P is false.
So, the deductive explanation is based on logic, using other general statements that
we accept as true, e.g. the sum of any even number and any odd number is always
an odd number.
So, we have given a deductive proof that the sum of four consecutive numbers is not a
multiple of 4. Of course, we could simply have checked a specific case as counterexample: 1 + 2 + 3 + 4 = 10, which is not a multiple of 4, and that disproves the conjecture.
This means that our conjecture that the sum of n consecutive numbers is a multiple of
n is in general false, because it is false for n = 4, which is therefore a counterexample disproving our conjecture. Just to rub it in, and we should probably have
done this first: the conjecture is not true for n = 2 either, as is illustrated by a specific
case, e.g. 1 + 2 = 3, which is not a multiple of 2.
PROBLEM 134:
We have shown that the sum of two consecutive numbers is not a multiple of 2.
The mathematical mind will always make a positive conjecture out of such a
negation: Prove that the sum of any two consecutive numbers is odd.
The art of “doing” mathematics and the power of mathematics, lies in the
mathematician's ability to continually look at familiar situations from a different
perspective, putting on different glasses and seeing the familiar differently. For
example, in Problem 124 we worked with two consecutive numbers. But in any two
consecutive numbers one number is even and the other is odd. So if we now do not
focus on the idea of consecutive numbers, but rather focus on this even and oddness
of the two numbers, the mathematician reasons: Is it a peculiarity of consecutive
numbers that the sum of an even number and an odd number is odd? The
mathematician makes the bold generalisation and conjectures: the sum of any even
number and any odd number is odd.
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PROBLEM 135:
Conjecture: The sum of any even number and any odd number is odd.
Why is this a generalisation of the conjecture in Problem 134?
Prove the conjecture!
Where do we stand now? The conjecture that the sum of n consecutive numbers is a
multiple of n is now thoroughly proven as untrue. One could pack up and leave it
there. But that is not the mathematical attitude! The conjecture is generally wrong,
yes, meaning it is not always true. But it also means that it is sometimes true! After all,
we know that it is true for n = 3. Shall we just see n = 3 as a special case, an
exception to the rule, and leave it?
Sometimes we do, for example 0 + 0 = 02 and 2 + 2 = 22 are the only two cases for
which x + x= x2, so we will not continue the investigation. Can you find other values of
x for which x + x= x2 is true?
Just because you cannot find any does not mean that there are not any!
Absence of evidence does not imply evidence of absence!
PROBLEM 136: x + x = x2
Prove that x = 0 and x = 2 are the only numbers that make x + x = x2 true.
To recap: We know that the sum of n consecutive numbers is a multiple of n for n = 3.
The mathematical mind will change the problem:
For which values of n is the sum of n consecutive numbers a multiple of n and for which not?
Let’s specialise by taking the very special case of n = 1, for which it obviously is true that
the sum of one number is a multiple of 1! Do you agree? Does it make sense to you?
So we have that our conjecture is true for n = 1 and n = 3 and not true for n = 4. Even
in this small database, the mathematical mind is all the time looking for patterns and
making conjectures, depending on what knowledge one brings to the situation: 1 and
3 are not just any numbers – they are odd numbers, and 4 is an even number. So the
mathematical mind jumps and makes a wild guess:
Conjecture: The sum of n consecutive numbers is multiple of n if n is odd.
How wild is this conjecture? Do you want to check it for n = 5? Try a special case like
1 + 2 + 3 + 4 + 5 = 15
which is a multiple of 5, so it looks promising, because our conjecture is now stronger:
It is true for n = 1 and 3 and now possibly also for 5. Why possibly? We also notice
from our example that our other result for n = 3 is also true, namely that the answer is
5 times the middle number. Shall we prove that it is true for any five consecutive
numbers?
191
Let the first number be n, then
S5 = n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 5 = 5(n + 1)
So that proves both our conjectures for n = 5. When will we be able to say it is true for
all odd numbers n? You should know the answer by now: After we have checked it for
all odd numbers n! You should now also know that it cannot be done by induction, but
only by deduction, which means that we must work with the general case, which
means we must call on the generality of algebraic symbolism.
If we want to prove the conjecture for any odd number, we have to go to a next level
of abstraction, working with two ideas of generality at the same time:
First, let’s reserve our symbol n for the odd number of terms (we want to prove that
the sum of n numbers is a multiple of n). Then, to work with any sequence of
consecutive numbers, we must introduce a different symbol, say m, for the first
number, i.e. m can be 1 or 2 or 5 or 10 or any natural number, but it is important that
we do not say which one. No matter what m is, the next natural number will be m +1,
etc. So we can start to express the sum as
Sn = m + (m + 1) + (m + 2) + (m + 3) + … to n terms, where n is odd.
It is not so easy to write down the nth term! We have to understand the structure in
the terms and generalise this structure. If you do not see it so easily, it will help if we
organise our information, for example:
T1
m
T2
m+1
T3
m+2
T4
m+3
T5
m+4
Tn
m+?
Sn = m + (m + 1) + (m + 2) + (m + 3) + … [m + (n – 1)] ………………………(1)
= (m + m + m + … to n terms) + [1 + 2 + 3 + 4 + 5 + … + to (n – 1) terms]
= mn +
n(n – 1)
2
Now we are in a completely different mode of mathematical thinking, namely in
manipulation of symbols, which, if we are fluent and dextrous, we should be able to
do automatically without thinking:
n(n – 1)
2
2mn + n(n – 1)
=
2
2mn + n2 – n
=
2
Sn = mn +
So we have simplified the fraction! But what does it mean? There is no such thing as
“simplest form” out of context. How we write it depends on what we want!
We want to show that n is a factor of Sn … We have to revisit what that means – it
means that n divides into Sn a whole number of times. If N = ab, then a and b are
factors of N. So we have a two-fold problem.
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First what does it mean – the logic is that
a+b
a
b
c =c +c
Can we say that 2 divides into n2 and 2 divides into n? No, we cannot, because,
remember, n is odd! We have to look with new eyes:
2mn + n2 – n
2
2mn + n(n – 1)
=
2
Sn =
How is this more useful? Well, it's about logic and reasoning of the type “If …then”: If
n is odd, then n – 1 is the previous number and therefore is even, so n – 1 and
therefore n(n – 1) and therefore Sn is divisible by 2. So we now know for sure that if
we divide Sn by n, the answer is a whole number! So n is a factor of Sn. That proves
our conjecture: The sum of an odd number of consecutive numbers is odd.
We can pack up and leave it. But the mathematical mind does not operate like that!
What about even values of n? Yes, we know that Sn is not divisible by n if n is even.
But maybe we can say something else?
Let’s follow exactly the same reasoning as above, but with n any even number:
2mn + n2 – n
and
2
n(2m + n – 1)
=
2
Sn =
n is an even number.
We can see that Sn is divisible by 2m + n – 1, but only if n is even, because it is only
n
when n is even that 2 and therefore Sn is a whole number! Now note that
2m + n – 1 = m + [m + (n – 1)]
which means that 2m + n – 1 is the sum of the first and last terms! This we can see
from examples like:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7 + (5 + 6)
= 11 + 11 + 11 + 11 to 5 terms
So our investigation has resulted in the following theorem.
The sum of n consecutive numbers is a multiple of n if n is odd and if n is even it is a
multiple of a + b where a and b are the first and the last terms.
One could pack up and leave it there. But that is not the mathematical attitude! Let’s
quote Polya (1957, 14-15):
Even fairly good students, when they have obtained the solution of the problem and
written down neatly the argument, shut their books and look for something else. Doing
so, they miss an important and instructive phase of the work. By looking back at the
completed solution, by reconsidering and reexamining the result and the path that led
193
to it, they could consolidate their knowledge and develop their ability to solve
problems. A good teacher should understand and impress on his students the view
that no problem whatever is completely exhausted. There remains always something
to do; … we could improve any solution, and, in any case, we can always improve our
understanding of the solution.
Our What if … questions could take us in at least two directions …
PROBLEM 137:
Looking back on the theorem The sum of n consecutive numbers is a multiple of
n if n is odd: What if we changed the type of number? What if it were not
consecutive natural numbers,
• but consecutive odd numbers … e.g. 5 + 7 + 9?
• but consecutive even numbers … e.g. 4 + 6 + 8?
• but consecutive multiples … e.g. 3 + 6 + 9 or 12 + 18 + 24?
• but consecutive terms of an arithmetic sequence … e.g. 2 + 5 + 8?
• but consecutive terms of a geometric sequence … e.g. 2 + 6 + 18?
We should realise that there are two approaches to Problem 137: You can first investigate
the special cases and then generalise to an arithmetic sequence. Or you can first prove a
conjecture for consecutive terms of any arithmetic sequence, and then specialise. Any
conjecture that is true for consecutive terms of an arithmetic sequence is immediately
without any further work through specialisation also true for consecutive multiples, for
consecutive even numbers, for consecutive odd numbers, and for consecutive natural
numbers, because they are all special cases of an arithmetic sequence.
So solving a general problem can be very economical and is therefore the way
finished mathematics (tidy mathematics) is often presented in a logical sequence. As
an example from school, take the theorems concerning quadrilaterals. Definitions are
carefully designed so that a square is a special rectangle and a rectangle and a rhombus
are special parallelograms. Then theorems are proven for a parallelogram and after that
immediately without further work, all the theorems that apply to a parallelogram by
specialisation also applies to a rhombus, a rectangle and a square …
In mathematics in the making (untidy mathematics), it is of course not always easy to
anticipate the more general conjecture when you are working on a specific case …
PROBLEM 138:
Looking back: The sum of n consecutive numbers is a multiple of n if n is odd.
What if we changed the operation? What if it were not the sum, but the product?
PROBLEM 139: SUMS OF SQUARES
We have found that the sum of three consecutive numbers is always a multiple
of 3. Now investigate the sum of the squares of three consecutive numbers …
194
PROBLEM 140:
Take any natural number and form a triplet (number, next number, sum), e.g.
(4, 5, 9). Prove that one and only one of the three numbers is a multiple of 3.
Note: This means that one of the three numbers is always a multiple of 3, but
only one of the three can be a multiple of 3, not two or three simultaneously.
PROBLEM 141:
Study this arrangement of the natural numbers:
2
6
1
3
Row 1
4
8
7
9
5
10 11 12 13 14 15 16
Row 2
Row 3
(a) How many numbers (or triangles) are there in Row 40? And in Row n?
(b) What are the first (left-most), middle, and last numbers in Row 40? And in Row n?
(c) Find the sum of all the numbers in Row 40. And in Row n?
(d) Find the sum of all the numbers in the first 40 rows. And in the first n rows?
(e) How many small triangles are there in total in the first 40 rows? And in the first n rows?
PROBLEM 142:
Take any 3 consecutive numbers, say 4, 5, 6.
Notice that 52 = 25 and 4 × 6 = 24 ...
Investigate, generalise and prove!
What if?
What if it were not consecutive numbers, but consecutive odd numbers?
Or consecutive even numbers?
Or consecutive multiples of 3 (e.g. 12, 15, 18)?
Or any arithmetic sequence?
Or …
PROBLEM 143:
Prove each of the following conjectures:
(a) If you take any multiple of 2, then add 1, then square, then add 3, the result is
always a multiple of 4.
(b) If you take any multiple of 4, then add 1, then square, then add 7, the result is
always a multiple of 8.
(c) If you take any multiple of 6, then add 1, then square, then add 11, the result is
always a multiple of 12.
(d) Write down the next conjecture with the structure of (a), (b) and (c), and prove it.
(e) Generalise!
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ACTIVITY 28:
Consider this arrangement of numbers:
Colomn 1 Colomn 2
↓
↓
0
1
7
8
14
15
21
22
28
29
...
2
9
16
23
30
3
10
17
24
31
4
11
18
25
32
Colomn 6 Colomn 7
↓
↓
5
6
12
13
19
20
26
27
33
34
←
←
←
←
←
Row 1
Row 2
Row 3
Row 4
Row 5
Now answer the following question. Justify your thinking by supplying complete
arguments (logical reasoning).
1. Write down the numbers in Row 100.
2. In what column and row will the number 4567 fall?
3. Find the 30th number in Column 6.
4. The vertical sequences in the table are all Arithmetic Sequences. These
sequences can be defined in two ways:
• they have a constant difference, or
• the have a general term Tn = an + b
Prove that these two definitions are equivalent.
5. Mandy says that in any 2 by 2 “square” (like the shaded one) the “cross sums” are
equal, e.g. 2 + 10 = 9 + 3. Is she correct? If so, explain why this should be true.
6. Andy says that in any 3 by 3 “square” the sum of the nine numbers is equal to 9
times the middle number. Is he correct? If so, can you explain why?
7. Show that, and explain why, the numbers in Column 1 is closed for addition and
for multiplication, but that the numbers in none of the other columns are closed for
addition or multiplication.
(Closed for addition means that the sum of any two numbers in the sequence is
also in the sequence.)
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ACTIVITY 29: THE MULTIPLICATION TABLE
There are many patterns in the multiplication table. A partially completed
multiplication table, up to 10 × 10, is shown below:
×
1
2
3
4
5
6
7
8
9
10
1
1
2
3
4
5
6
7
8
9
10
Row 1
2
2
4
6
8
10
12
14
16
18
20
Row 2
3
3
6
9
12
15
18
21
24
27
30
4
4
8
12
16
20
24
28
32
36
40
5
5
10
15
6
6
12
18
7
7
14
21
8
8
16
24
9
9
18
27
10
10
20
30
Diagonal 2
Diagonal 1
Column 1 Column 2
1. In the table, Diagonal 1 is a line of symmetry.
What does this mean? Why does the table have this line of symmetry?
2. Look at Row 7 (or Column 7), the multiples of 7: 7, 14, 21, 28, ...
Find the 100th multiple of 7.
Show that the sum of any two multiples of 7 is again a multiple of 7 (we say the
set is closed for addition).
Show that the product of any two multiples of 7 is again a multiple of 7 (we say
the set is closed for multiplication).
Is the set of multiples of 7 closed for the operations of subtraction and division?
3. Add the numbers in Row 2 to the numbers in Row 3, column by column:
Row 2:
2
4
6
8
10 …
Row 3:
3
6
9
12 15 …
Sum:
5
10 15 …
What patterns do you notice? Can you explain this?
Does the pattern occur for other rows? Generalise and explain.
4. Look at Diagonal 1, the perfect squares: 1, 4, 9, 16, 25, ...
Describe how this sequence of numbers is formed:
• by addition (recursive relationship)
• by multiplication (functional relationship)
Find the 30th number in the sequence.
Why do you think these numbers are called squares?
197
5. Find the following number sequences in the table and notice how they are
formed. Find the next three numbers and the 100th number in each sequence.
(a) 2, 6, 12, 20, 30, 42, 56, 72, 90; ... Is 420 a number in this sequence?
(b) 3, 8, 15, 24, 35, 48, 63, 80, ...
Is 960 a number in this sequence?
(c) 5, 12, 21, 32, 45, 60, ...
Is 720 a number in this sequence?
6. Look at Row 2 in the table, the even numbers: 2, 4, 6, 8, ...
(a) Find (1) the 50th even number
(2) the 100th even number.
(b) Find the sum of the first 50 even numbers, i.e.
2 + 4 + 6 + 8 + 10 + 12 + ... + 98 + 100
One way of doing this is to investigate the sum of a smaller number of even
numbers to see if you see a pattern to help you to predict the sum:
S(1):
2=2
S(2):
2+4=6
S(3): 2 + 4 + 6 = 12
Complete the table:
# even numbers added
Sum of the numbers
1
2
2
6
3
12
4
20
5
6
…
20
50
n
What other patterns can you use to find in the sum?
7. Investigate the sequence of the sums of the numbers in the “L” shapes in the
table on the left.
Remember, it is not about numerical answers! Here is the numerical answer:
13, 23, 33, 43, … The mathematical attitude is to make sure that it is always true,
and why it is true. And to do that you must investigate the structure …
×
1
2
3
4
5
×
1
2
3
4
5
1
1
2
3
4
5
1
1
2
3
4
5
2
2
4
6
8
10
2
2
4
6
8
10
3
3
6
9
12
15
3
3
6
9
12
15
4
4
8
12
16
20
4
4
8
12
16
20
5
5
10
15
5
5
10
15
9. Investigate the sequence of the sums of numbers in consecutive squares in the
table on the right. The numerical answer is 1, 9, 36, 100, …
10. If we combine the two results in 8 and 9, we have:
13 = 12
13 + 23 = 32
13 + 23 + 33 = 62
Generalise and prove …
198
Designing divisibility tests
Let’s develop a few divisibility tests. However, this is not at all about isolated pieces of
knowledge, but about the generic processes (i.e. the method can also be used
elsewhere) and about appreciating the power of algebra to aid our reasoning!
Let’s start with a divisibility test for 7. Let N be any whole number, then
N = 10a + b, a, b ∈ N, b ≤ 9
N = 7a + 3a + b
⇒ N is divisible by 7 if 3a + b is divisible by 7.
Example: For 91: 3a + b = 27 + 1 = 28, which is divisible by 7, so 91 is divisible by 7.
Example: For 266: 3a + b = 3×26 + 6 = 84, which is divisible by 7, so 91 is divisible by 7.
Example: For 83: 3a + b = 24 + 3 = 27, which is not divisible by 7, so 91 is not divisible by
7.
Or, here is another method:
2N = 20a + 2b = 21a + (2b – a)
⇒ 2N and therefore N is divisible by 7 if 2b – a is divisible by 7.
Example: For 49: 2b – a = 18 – 4 = 14, which is divisible by 7, so 49 is divisible by 7.
Or:
N = 100a + 10b + c, a, b, c ∈ N, b, c ≤ 9
N = (98a + 7b) + (2a + 3b + c)
N = 7(14 + b) + (2a + 3b + c)
⇒ N is divisible by 7 if 2a + 3b + c is divisible by 7.
Let’s design a divisibility test for 19:
N = 10a + b, a, b ∈ N, b ≤ 9
2N = 20a + 2b
=19a + a + 2b
⇒ 2N and therefore N is divisible by 19 if a + 2b is divisible by 19.
Example: For 95: a + 2b = 9 + 10 = 19, which is divisible by 19, so 95 is divisible by 19.
Or, 4N = 40a + 4b = 38a + (2a + 4b)
⇒ 4N and therefore N is divisible by 19 if 2a + 4b is divisible by 19.
Example: For 95: 2a + 4b = 18 + 20 = 38, which is divisible by 19, so 95 is divisible by
19.
PROBLEM 144:
Design several different divisibility tests for 9, 11, 13, 17, 37 …
Test them with numerical examples.
199
PROBLEM 145: abc abc
Choose any three-digit number, then repeat the digits to form a six-digit number,
e.g. 293293. Now divide the number by 7, then by 11, then by 13.
What do you notice?
Do it for other numbers. What do you notice?
Formulate a conjecture in words and in symbols.
Will this always happen? Are you sure? Can you explain why?
PROBLEM 146: an + bn
For which n∈N is each of the following expressions divisible by 10?
4n + 6n
3n + 7n
2n + 8n
1n + 9n
Is n = 10 a special case, or can the result be generalised?
For which values of n is an + bn divisible by a + b?
For which values of n is an – bn divisible by a – b?
PROBLEM 147: MIRROR IMAGE
Investigate the sum of a two-digit number and its mirror image (e.g. 47 + 74).
Formulate a conjecture and prove it.
What if? What if we investigate the sum of a three-digit number and its mirror
image (e.g. 372 + 273)? And four-digit numbers …?
Can you generalise? Can you prove your conjectures?
What if? Investigate the difference between mirror images (e.g. 724 – 427) …
PROBLEM 148: FRIENDLY NUMBERS
If a number is a multiple of the sum of its digits, we call the number a friendly
number. For example, 18 is friendly because the sum of its digits is 1 + 8 = 9
and 18 is a multiple of 9. 15 is not friendly, because 15 is not a multiple of 1 + 5= 6.
Find all the friendly numbers less than 100.
PROBLEM 149: SYMMETRICAL FRIENDLY NUMBERS
Prove: If 10a + b is a friendly number, then 10b + a is also a friendly number.
PROBLEM 150: GEOMETRIC SEQUENCES
In Problem 14 you proved that the sum of two Arithmetic Sequences is again an
Arithmetic Sequence.
Investigate analogous conjectures for a Geometric Sequence.
200
PROBLEM 151: GEOMETRIC SUMS
Look at any Geometric Sequences, for example:
3, 6, 12, 24, 48, …
(a) Prove that a Geometric Sequence is closed under multiplication, i.e. the product
of any two terms of the sequence is always also a term of the sequence.
(b) Form a new sequence
(i) by squaring the terms, e.g. 32, 62, 122, … = 9, 36, 144, …
(ii) by adding consecutive terms, e.g. 3 + 6, 6 + 12, 12 + 24, … = 9, 18, 36, …
(iii) by adding the squares of consecutive terms, e.g. 32+62, 62+122, 122+242, …
In each case: is the new sequence Geometric?
PROBLEM 152: POWERS
In the arrangement below, we start with the sequence of powers of 2 and then
form new sequences by adding the two terms above it:
1
2
3
4
32 64 …
48 96 …
9
18 36 72 144 …
27 54 108 216 …
6
8
12
16
24
Row 1
Row 2
Row 3
Row 4
Why is Row 2 = 3 × Row 1 and in general Row n = 3 × Row (n – 1)?
Why are all the horizontal sequences multiples of powers of two?
Why are the “vertical” sequences, e.g. 1, 3, 9, 27, … multiples of powers of 3?
Find the first term in Row 10.
Now add Row 1 and Row 2 term by term to form a new sequence: 1 + 3, 2 + 6, ...
What kind of sequence is this?
PROBLEM 153: THE FIBONNACI SEQUENCE
Look at these sequences. You can use this Fibonacci worksheet.
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … (The Fibonacci sequence)
3, 7, 10, 17, 27, 44, 71, 115, ...
These sequences T1; T2; T3; T4; ... are in general defined by the following two rules:
• T1 and T2 are any two natural numbers
• Tn+1 = Tn-1 + Tn with n > 1
(a) Show that for any such sequence the sum of the first 10 terms,
10
∑T
i =1
i
= S10 = 11× T7 .
(b) If T4 = 11 and T10 = 207, find T11.
(c) If T72 – T62 = 517 and T9 = 76, find T10.
(d) Prove that if Tk is a multiple of 5, then every 5th term after that is also a multiple of 5.
(e) Investigate: Tn + Tn+3 = 2Tn+2; Tn + Tn+4 = 3Tn+2; …
(f) Investigate: S3 = T1 + T2 + T3 = T5 – T2; S4 = T6 – T2; …
201
ACTIVITY 30: POLYGONAL NUMBERS
The ancient Greeks (300 BCE), used calculi (Greek for small stones or pebbles) as an
aid to “calculate”. They were also very interested in patterns in numbers, e.g. what
numbers could be arranged to form polygons, as shown here. The names of the
sequences speak for themselves!
Triangular numbers:
T1=1
T2 = 3
T3 = 6
T4 = 10
T5 = 15
Square numbers:
S1=1
S2 = 4
S3 = 9
S4 = 16
Pentagonal numbers:
P1=1
P2 = 5
P3 = 12
P4 = 22
Hexagonal numbers:
H1=1
H2 = 6
H3 = 15
H4 = 28
1. Can you explain the names of the sequences?
2. Complete this table. Look for patterns, describe your patterns …
n
1
2
3
4
5
6
7
8
Tn
1
3
6
10
15
21
28
36
Sn
1
4
9
16
25
36
Pn
1
5
12
22
Hn
1
6
15
28
7n
1
8n
1
202
10
30
n
3. Note that 1 and 36 are triangular numbers, but also square numbers. Can you find
other triangular numbers that are square?
4. Show that the product of any two square numbers is always again a square
number (we say the set of square numbers is closed under multiplication)
5. Prove that the sum of any two consecutive triangular numbers is a square.
6. Show that 8×Tn + 1 is a square number. Which square number, i.e. find the value
of m so that 8×Tn + 1 = Sm.
Use this relationship to easily calculate T30.
7. Note that H1, H2, H3, are also triangular numbers. Can we say that all hexagonal
numbers are also triangular numbers? How can we be sure? Convince yourself,
convince a friend, convince an enemy!
8. Look at the sequence of triangular numbers:
1; 3; 6; 10; . . .
Investigate the differences between squares of consecutive triangular numbers.
We can look at special cases:
(T2)2 – (T1)2 = 9 – 1 = 8
(T3)2 – (T2)2 = 36 – 9 = 27
(T4)2 – (T3)2 = 100 – 36 = 64
Do you recognise the answers as cubes?
(T2)2 – (T1)2 = 9 – 1 = 8 = 23
(T3)2 – (T2)2 = 36 – 9 = 27 = 33
(T4)2 – (T3)2 = 100 – 36 = 64 = 43
Can say that the difference is always a cube? Do you believe it? Prove it!
The wonderful thing is that we can find, say (T100)2 – (T99)2 = 1002, without
knowing what T100 or T99 is!
Calculate T100 – T99.
9. We have been able to find the general term of sequences like 1, 3, 6, 10, …
i.e. sequences that are not Arithmetic (constant difference) or Geometric (constant
ratio) sequences. These sequences can be defined in two ways:
• they have a constant second difference, or
• the have a general term Tn = an2 + bn + c
Prove that these two definitions are equivalent.
10. How will we find the sum of such a sequence, e.g.
Sn = 1 + 3 + 6 + 10 + to n terms? Investigate and invent a method!!
203
PROBLEM 154:COEFFICIENTS
1. Check the expansion of these binomials:
2
(x + 1)2 = x + 2x + 1
3
2
(x + 1)3 = x + 3x + 3x + 1
Extend the structure. Describe any patterns.
Now write down the expansion of (x + 1)10.
2. The sum of the coefficients of the expansion of (x + 1)2 is 1 + 2 + 1 = 4
Now find the sum of the coefficients of the expansion of (x + 1)37
PROBLEM 155: SEQUENCE OF ROOTS
Let a be one of the roots of x2 – x – 1 = 0
Now investigate the sequence 1, a, a2, a3, … Show that:
(a) a2 = a + 1
(b) a3 = 2a + 1
(c) a4 = 3a + 2
(d) a5 = 5a + 3
(e) a6 = 8a + 5
(f) an+2 = an+1 + an
A few prime problems
PROBLEM 156:
Show that any prime number can be expressed as 6n + 1 or 6n + 5
Let’s solve this problem here (See also page 6 and Problem 74):
If a number is divided by 6 it leaves a remainder of 0, 1, 2, 3, 4 or 5. Therefore any
number can be expressed as 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4 or 6n + 5.
If the number is of the form 6n it is divisible by 6 and therefore not prime.
If the number is of the form 6n + 2 = 2(3n + 1) it is divisible by 2 and thus not prime.
If the number is of the form 6n + 3 = 3(2n + 1) it is divisible by 3 and thus not prime.
If the number is of the form 6n + 4 = 2(3n + 2) it is divisible by 2 and thus not prime.
So only 6n + 1 and 6n + 5 do not have another factor and may be prime.
Again, to understand what this means, many of us would probably need to check
special cases. For example:
71 is prime. Can it be expressed as 6n + 1 or 6n + 5?
Looking back: Make sure of the meaning of n in the above reasoning! Should n ∈ N or
should n ∈ N0? Any other exceptions?
Looking back: Mathematicians prefer to rather say any prime number can be expressed
as 6n + 1 or 6n – 1. First: are they correct? Second: why is this the preferred form?
204
PROBLEM 157:
Prove the converse of 156: If a number can be written as 6n + 1 or 6n + 5, then it is prime.
PROBLEM 158:
Is 1 a prime number? If not, why not?
Note: The argument that the definition of a prime is “a number with only two
factors, namely 1 and itself”, therefore 1 cannot be prime because 1 and itself is
the same, is untenable! The meaning of “and” is inclusive: If each of these two
statements is true: “Ame is older than 10”; “Ame is younger than 20”, then the
combined statement “Ame is older than 10 and younger than 20” is true!
Click here for a discussion.
PROBLEM 159:
If we divide a whole number by 3 the possible remainders are 0, 1 or 2.
Now take any whole number, square it, then divide by 3.
Do you expect the possible remainders to be 0, 1, or 2?
Investigate and prove your findings!
PROBLEM 160:
A prime number greater than 2 is divided by 12. What are the possible
remainders?
A prime number greater than 2 is squared and then divided by 12. What are the
possible remainders?
PROBLEM 161:
Find all a, b∈ N so that
(b) a2 + b2 = 65
(a) a2 + b2 = 13
(e) a2 – b2 = 65
(d) a2 – b2 = 13
(c) a2 + b2 = 72
(f) a2 – b2 = 72
PROBLEM 162:
(a) 2, 3 are two primes that differ by 1. Find all prime pairs of the form p, p + 1.
(b) 3, 5 are two primes that differ by 2. Find all prime pairs of the form p, p + 2.
(c) 3, 5, 7 are three primes that differ by 2. Find all prime triplets of the form p, p + 2, p + 4.
(d) 2, 5 are two primes that differ by 3. Find all prime pairs of the form p, p + 3.
(e) 3, 7, 11 are three primes that differ by 4. Find all prime triplets of the form p, p + 4, p + 8.
(f) 5 is a prime that is 1 more than a square. Find all primes of the form n2 + 1.
(g) 3 is a prime that is 1 less than a square. Find all primes of the form n2 – 1.
205
PROBLEM 163: PRIME ARITHMETIC SEQUENCE
Look at the sequence
5; 11, 17, 23, 29
It is an Arithmetic Sequence. Check that each term is prime.
Let’s call it a Prime Arithmetic Sequence (PAS).
What is the longest PAS that you can find? Prove that it is the longest!
Remember: Mathematics is not a spectator sport! Make sure that you have really
invested much energy in the problem before continuing reading!
Do not read until you have solved the problem!
One would probably start with an inductive attack on the problem. This can be done
in two ways:
1. Write a computer program to generate primes (!), or find a list somewhere and then
look for common differences between the primes. For example, see this attached
list of the first 1000 primes.
2. Generate specific Arithmetic Sequences and then test whether the terms of the
sequence are prime.
This can be very frustrating, because the sequences keep breaking down early! You will
come under the impression that there is very little pattern in the distribution of primes!
On finding primes
Approach 2 is an example of practice in context: we can learn much in the process of
trying to solve the problem! For example, you will repeatedly have to test whether the
numbers are prime, and this gives practice in this method, as well as in divisibility tests,
etc.
Generating prime numbers is not that easy. Fact is that there is no polynomial formula
that generates only primes or generates all the primes. For example:
Suppose the quadratic formula f(n) = an2 + bn + c generates primes for n∈ N0.
Then, for f(c) = ac2 + bc + c = c(ac + b + 1) which is divisible by c, so f(c) is not prime.
So the longest sequence of primes that f may produce is n = c1.
However, the following theorem (Dirichlet's Theorem) assures us that there are
infinitely many primes in any Arithmetic Sequence:
If a and b are relatively prime2 positive integers, then the arithmetic progression
a, a+b, a+2b, a+3b, ... contains infinitely many primes.
For example, look at this Arithmetic Sequence with formula 5 + 6n, n∈ N0:
5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125,
131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227,
233, 239, 245, 251, 257, 263, 269, 275, 281, 281, ...
1
2
Compare section 2.3: n2 – n + 11 is prime for n = 0 to 10, and n2 – n + 41 for n = 0 to 40.
Two integers are called relatively prime if they have no common factor except 1, so 4 and 9 are relatively prime.
206
There is an excellent online calculator on the WWW Interactive Mathematics Server3
generating primes in an arithmetic sequence. You may browse through the list looking
for consecutive terms!
To find small primes the most efficient method is the Sieve of Eratosthenes4 (ca 240 BC):
Make a list of all the integers less than or equal to n (greater than one) and strike
out the multiples of all primes less than or equal to the square root of n, then the
numbers that are left are the primes
To find individual small primes trial division works okay. Just divide by the primes, e.g.
to show 211 is prime, just show that 2, 3, 5, 7, 11, 13 are not factors of 2115.
When investigating special cases, one makes observations leading to conjectures,
which one then tests against further special cases. Or one can try to prove your
conjectures deductively, with the main aim that any “theorem” can help to make the
inductive search easier or to explain why the search is unsuccessful. I briefly show a
few pieces of deduction based on inductive conjectures.
We are in general working with the Arithmetic Sequence p, p + d, p + 2d, ... p + (n – 1)d
where p is the first term, d is the common difference and all terms are prime.
Let’s first investigate the influence of d.
Theorem 1: 2, 3 is the only PAS with d = 1.
Proof: If p, p + 1 is any other PAS, then p is prime, therefore odd, then obviously
p + 1 is even, and therefore not prime.
Note: This proof is based on the Modus Tollens structure: P ⇒ Q, Q is false, so P is false!
Theorem 2: If d is odd, p = 2.
What the theorem says that there is no PAS with d odd except 2, 3; 2, 5; 2, 7; ...
Proof: If p, p + d is a PAS with p ≠ 2 and d odd, then p is prime, therefore p is odd and
d odd, then p + d is even, and therefore not prime.
Deduction: If d is odd, the longest PAS has length 2! So we need consider only
even values of d! This already helps the search!
Theorem 3: 3, 5, 7 is the longest PAS with d = 2
Proof: Let p, p + 2, p + 4 be any other PAS.
Now look at the sequence of consecutive numbers: p, p + 1, p + 2, p + 3, p + 4
If p is not 3, then one of p + 1, p + 2 is a multiple of 3
If p + 1 is a multiple of 3, we can write
p + 1 = 3k where k ∈ N
⇒ T3 = p + 4 = 3k + 3 = 3(k + 1) is a multiple of 3 and therefore not prime
If p + 2 is a multiple of 3, then T2 is not prime.
3
The URL is: http://wims.unice.fr/~wims/ Open Calculators and plotters, then the program Primes.
See http://www.faust.fr.bw.schule.de/mhb/eratosiv.htm for an interactive Sieve!
5
Interesting observations here are
a) Why only primes – why not check whether 4, 6, … are factors? Can you explain why?
b) One has to search only for prime factors less than the square root of the number! Can you explain why?
4
207
Note: We are here using a very important property of natural numbers, namely:
In any sequence of n consecutive natural numbers, one of the numbers is a multiple of n6
Convince yourself (you can use this Excel worksheet):
In any 2 consecutive numbers, one of them is a multiple of 2
In any 3 consecutive numbers, one of them is a multiple of 3, etc.
Theorem 4: 3; 7, 11 is the longest PAS with d = 4
Proof: Let p, p + 4, p + 8 be any other PAS with d = 4
If p is not a multiple of 3 (except 3 itself), then one of p + 1, p + 2 is a multiple of 3.
Consider them one at a time:
If p + 1 = 3k where k ∈ N
⇒ T2 = p + 4 = 3k + 3 = 3(k + 1) is a multiple of 3 and therefore not prime
If p + 2 = 3k where k ∈ N
⇒ T3 = p + 8 = 3k + 6 = 3(k + 2) is a multiple of 3 and therefore not prime
Theorem 5: 5; 11, 17, 23, 29 is the longest PAS with d = 6
Proof: Let p, p + 6, p + 12, p + 18, p + 24 be any other PAS with d = 6
If p is not a multiple of 5 (except 5 itself), then one of p + 1, p + 2, p + 3, p + 4 is a
multiple of 5. Consider them one at a time:
If p + 1 = 5k where k ∈ N
⇒ T2 = p + 6 = 5k + 5 = 5(k + 1) is a multiple of 5 and therefore not prime
If p + 2 = 5k where k ∈ N
⇒ T3 = p + 12 = 5k + 10 = 5(k + 2) is a multiple of 5 and therefore not prime
If p + 3 = 5k where k ∈ N
⇒ T4 = p + 18 = 5k + 15 = 5(k + 3) is a multiple of 5 and therefore not prime
If p + 4 = 5k where k ∈ N
⇒ T5 = p + 24 = 5k + 20 = 5(k + 4) is a multiple of 5 and therefore not prime
Show that if d = 8 , longest PAS has length 2! And if d = 10?
Absence of evidence does not mean there is evidence of absence!!!
PROBLEM 164:
Theorem 6: The longest PAS with d = kp (a multiple of p) is 2.
Note: Illustrate and prove the theorem!
We can easily determine how the first term determines the maximum length of a PAS:
Theorem 7: The longest PAS with p = 3 is 3
Proof: Let 3, 3 + d, 3 + 2d, 3 + 3d be a sequence of length 4.
This is not possible, because the fourth term 3 + 3d = 3(1 + d) is a multiple of 3 and
therefore not prime.
6
Also note that we can express this in the following way:
Any natural number can be written as either 2n or 2n + 1 (leaves a remainder of 0 or 1 when divided by 2).
Any natural number can be written as either 3n or 3n + 1 or 3n + 2 (leaves a remainder of 0, 1 or 2 when
divided by 3).
208
This can now easily be generalised:
Theorem 8: The longest PAS with first term p has length p.
Proof: Let the sequence be p, p + d, p + 2d, p + 3d, . . . .
Then Tp+1 = p + pd = p(1 + d), which is a multiple of p and therefore not prime.
So the longest that the PAS can be is p.
Remark: The promise that starting with a relative large p, say p = 31 will give a long
PAS of 31 terms is foiled by the influence of d. For example, let’s start with any p and
d = 30:
Theorem 9: The longest PAS with d = 30 has length 6.
Proof: Let the sequence be p, p + 30, p + 60, p + 90, p + 120, p + 150, . . .
If p = 7 we have 7, 37, 67, 97, 127, 157 which is a PAS with length 6 (187 = 11 × 17)
If p is not a multiple of 7 (except 7 itself), then one of p + 1, p + 2, p + 3, p + 4, p + 5,
p + 6 is a multiple of 7. So:
If p + 1 = 7k where k ∈ N
⇒ T5 = p + 120 = 7k + 119 = 7(k + 19) is a multiple of 7 and therefore not prime
If p + 2 = 7k where k ∈ N
⇒ T2 = p + 30 = 7k + 28 = 7(k + 4) is a multiple of 7 and therefore not prime
If p + 3 = 7k where k ∈ N
⇒ T6 = p + 150 = 7k + 147 = 7(k + 21) is a multiple of 7 and therefore not prime
If p + 4 = 7k where k ∈ N
⇒ T3 = p + 60 = 7k + 56 = 7(k + 8) is a multiple of 7 and therefore not prime
If p + 5 = 7k where k ∈ N
⇒ T7 = p + 180 = 7k + 175 = 7(k + 25) is a multiple of 7 and therefore not prime
If p + 6 = 7k where k ∈ N
⇒ T4 = p + 90 = 7k + 84 = 7(k + 12) is a multiple of 7 and therefore not prime
This means that even with p = 29 or 71, which promises potential lengths of 29 and
71, with d = 30 the longest possible PAS has length 6.
Lesson: Deduction can save us from fruitless inductive searches!
Theorem 10: If d = 20, then the longest PAS has length 2
Proof: Let the sequence be p, p + 20, p + 40, . . .
If p is not a multiple of 3 (except 3 itself), then one of p + 1, p + 2 is a multiple of 3.
If p + 1 = 3k where k ∈ N
⇒ T3 = p + 40 = 3k + 39 = 3(k + 13) is a multiple of 3 and therefore not prime
If p + 2 = 3k where k ∈ N
⇒ T2 = p + 20 = 3k + 18 = 3(k + 6) is a multiple of 7 and therefore not prime
There are two frustrations here:
Even if we prove a potential length of, say 6 in theorem 9 by considering multiples of
7, then as the terms in the sequence become larger, larger factors like 11, 13, …
sabotage the sequence! For example 23, 53, 83, 113, 143 (= 11 × 13)!
209
If we work deductively with a formula like 29 + 60k, the only thing we can say is that if
k = 29 or a multiple of 29, the number will be divisible by 29, so we have a promise of
a length of maybe 29! But there “always” are other, smaller factors. That happens
a+b a b
because our analysis is based on the fact that if
= + we can say that a + b is
c
c c
divisible by c if a is divisible by c and b is divisible by c. But a + b can be divisible by c
even if neither of a and b is divisible by c, e.g. 5 + 9 is divisible by 7 although neither
of 5 nor 9 is divisible by 7.
The longest known PAS
You would not have much success with an inductive attack! As a consolation, here is
a 10-term arithmetic sequence of primes with common difference 210.
199, 409, 619, 829, 1 039, 1 249, 1 459, 1 669, 1 879, 2 089
The record PAS7 (at least when I last checked!) has length 22, given by
11 410 337 850 553 + 4 609 098 694 200×n, 0 ≤ n ≤ 21
I used the Primes search tool on the WWW Interactive Mathematics Server8 to
generate these 22 primes in arithmetic sequence:
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
n=
0:
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
15:
16:
17:
18:
19:
20:
21:
p = 11 410 337 850 553
p = 16 019 436 544 753
p = 20 628 535 238 953
p = 25 237 633 933 153
p = 29 846 732 627 353
p = 34 455 831 321 553
p = 39 064 930 015 753
p = 43 674 028 709 953
p = 48 283 127 404 153
p = 52 892 226 098 353
p = 57 501 324 792 553
p = 62 110 423 486 753
p = 66 719 522 180 953
p = 71 328 620 875 153
p = 75 937 719 569 353
p = 80 546 818 263 553
p = 85 155 916 957 753
p = 89 765 015 651 953
p = 94 374 114 346 153
p = 98 983 213 040 353
p = 103 592 311 734 553
p = 108 201 410 428 753
I used the Factoris tool on the WWW Interactive Mathematics Server to check that
for n = 22, p = 112 810 509 122 953 = 61 × 107 × 1 907 × 9 063 277 is not prime!
7
Source: Pritchard, P. A.; Moran, A.; and Thyssen, A. (1995). Twenty-Two Primes in Arithmetic Progression.
Math. Comput. 64, 1337-1339.
8
The URL is: http://wims.unice.fr/~wims/ Open Calculators and plotters, …
210
THE FIRST 1 000 PRIMES (the 1 000th is 7919)
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211
223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347
349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467
479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617
619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761
769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919
929 937 941 947 953 967 971 977 983 991 997
1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 1087 1091 1093 1097 1103
1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223 1229 1231
1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367
1373 1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487
1489 1493 1499 1511 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607
1609 1613 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733
1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 1871 1873
1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 1993 1997 1999
2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 2099 2111 2113 2129
2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 2221 2237 2239 2243 2251 2267 2269
2273 2281 2287 2293 2297 2309 2311 2333 2339 2341 2347 2351 2357 2371 2377 2381 2383
2389 2393 2399 2411 2417 2423 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539
2543 2549 2551 2557 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677
2683 2687 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789
2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 2909 2917
2927 2939 2953 2957 2963 2969 2971 2999
3001 3011 3019 3023 3037 3041 3049 3061 3067 3079 3083 3089 3109 3119 3121 3137 3163
3167 3169 3181 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301
3307 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 3433
3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 3541 3547 3557 3559
3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 3659 3671 3673 3677 3691 3697 3701
3709 3719 3727 3733 3739 3761 3767 3769 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853
3863 3877 3881 3889 3907 3911 3917 3919 3923 3929 3931 3943 3947 3967 3989
4001 4003 4007 4013 4019 4021 4027 4049 4051 4057 4073 4079 4091 4093 4099 4111 4127
4129 4133 4139 4153 4157 4159 4177 4201 4211 4217 4219 4229 4231 4241 4243 4253 4259
4261 4271 4273 4283 4289 4297 4327 4337 4339 4349 4357 4363 4373 4391 4397 4409 4421
4423 4441 4447 4451 4457 4463 4481 4483 4493 4507 4513 4517 4519 4523 4547 4549 4561
4567 4583 4591 4597 4603 4621 4637 4639 4643 4649 4651 4657 4663 4673 4679 4691 4703
4721 4723 4729 4733 4751 4759 4783 4787 4789 4793 4799 4801 4813 4817 4831 4861 4871
4877 4889 4903 4909 4919 4931 4933 4937 4943 4951 4957 4967 4969 4973 4987 4993 4999
5003 5009 5011 5021 5023 5039 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147
5153 5167 5171 5179 5189 5197 5209 5227 5231 5233 5237 5261 5273 5279 5281 5297 5303
5309 5323 5333 5347 5351 5381 5387 5393 5399 5407 5413 5417 5419 5431 5437 5441 5443
5449 5471 5477 5479 5483 5501 5503 5507 5519 5521 5527 5531 5557 5563 5569 5573 5581
5591 5623 5639 5641 5647 5651 5653 5657 5659 5669 5683 5689 5693 5701 5711 5717 5737
5741 5743 5749 5779 5783 5791 5801 5807 5813 5821 5827 5839 5843 5849 5851 5857 5861
5867 5869 5879 5881 5897 5903 5923 5927 5939 5953 5981 5987 6007 6011 6029 6037 6043
6047 6053 6067 6073 6079 6089 6091 6101 6113 6121 6131 6133 6143 6151 6163 6173 6197
6199 6203 6211 6217 6221 6229 6247 6257 6263 6269 6271 6277 6287 6299 6301 6311 6317
6323 6329 6337 6343 6353 6359 6361 6367 6373 6379 6389 6397 6421 6427 6449 6451 6469
6473 6481 6491 6521 6529 6547 6551 6553 6563 6569 6571 6577 6581 6599 6607 6619 6637
6653 6659 6661 6673 6679 6689 6691 6701 6703 6709 6719 6733 6737 6761 6763 6779 6781
6791 6793 6803 6823 6827 6829 6833 6841 6857 6863 6869 6871 6883 6899 6907 6911 6917
6947 6949 6959 6961 6967 6971 6977 6983 6991 6997
7001 7013 7019 7027 7039 7043 7057 7069 7079 7103 7109 7121 7127 7129 7151 7159 7177
7187 7193 7207 7211 7213 7219 7229 7237 7243 7247 7253 7283 7297 7307 7309 7321 7331
7333 7349 7351 7369 7393 7411 7417 7433 7451 7457 7459 7477 7481 7487 7489 7499 7507
7517 7523 7529 7537 7541 7547 7549 7559 7561 7573 7577 7583 7589 7591 7603 7607 7621
7639 7643 7649 7669 7673 7681 7687 7691 7699 7703 7717 7723 7727 7741 7753 7757 7759
7789 7793 7817 7823 7829 7841 7853 7867 7873 7877 7879 7883 7901 7907 7919 END
211
REFERENCES
Bell, A. W. (1976). A Study of Pupils' Proof-Explanations in Mathematical Situations.
Educational Studies in Mathematics, 7, 23-40.
Brown, S. I. & Walter, M. I. (1983). The Art of Problem Posing. Philadelpia, PA:
Franklin Institute Press.
Davis, P. & Hersh, R. (1981). The Mathematical Experience. Boston: Houghton- Mifflin.
Department of Educaton (2001). National Strategy for Mathematics, Science and
Technology Education in General and Further Education and Training.
Department of Education (2002). Revised National Curriculum Statement for Grades
R-9 (Schools) – Mathematics. Government Gazette, Vol. 443, No. 23406. Pretoria.
http://education.pwv.gov.za/content/documents/11.pdf
De Villiers, M. D. (1987). Elementary Logic. Stellenbosch: Research Unit for
Mathematics Education.
De Villiers, M. (1990). The role and function of proof in mathematics. Pythagoras, 24,
17 - 24.
Fennema, E. & Romberg, T. A. (Eds.) (1999). Mathematics Classrooms that Promote
Understanding. London: Lawrence Erlbaum.
Griffiths, H. B. (1978). The structure of pure mathematics. In G. T. Wain (Ed.),
Mathematical Education, 8-24. New York: Van Nostrand Reinhold Company.
Hiebert, J., Carpenter, T., Fennema, E., Fuson, K., Human, P., Murray, H., Olivier, A.
& Wearne, D. (1996). Problem-solving as a basis for curriculum development and
instruction: The Case of Mathematics. Educational Researcher, 25(4), 12-21.
Hiebert, J., Carpenter, T.P., Fennema, E., Fuson, K., Human, P., Murray, H., Olivier, A.
& Wearne, D. (1997). Making sense: teaching and learning mathematics with
understanding. Portsmouth, New Hampshire: Heinemann Educational Books.
Jacobs, H. R. (1982). Mathematics. A Human Endeaver. New York: W. H. Freeman
and Company.
Lakatos, I. (1976). Proofs and Refutations: The Logic of Mathematical Discovery.
New York: Cambridge University Press.
Mathematical Association (1934). The Teaching of Algebra in Schools. London: G.
Bell & Sons.
Mason, J., Burton, L. & Stacey, K. (1985). Thinking Mathematically. New York:
Addison Wesley.
Murray, H., Olivier, A. & Human, P. (1998). Learning through problem solving. In
A. Olivier & K. Newstead (Eds.), Proceedings of the Twenty-second International
Conference for the Psychology of Mathematics Education: Vol. 1, 169-185.
Stellenbosch, South Africa.
National Council of Teachers of Mathematics. (1989). Curriculum and Evaluation
Standards for School Mathematics . Reston, VA: NCTM.
National Research Council (1989).
Academy Press.
Everybody Counts. Washington: National
212
Oakes, J. (1985). Keeping Track: How Schools Structure Inequality. New Haven,
Connecticut: Yale University Press.
Olivier, A. (1988). The construction of an algebraic concept through conflict.
In A. Borbás (Ed.), Proceedings of the Twelfth International Conference for the
Psychology of Mathematics Education, 511-518. Veszprém, Hungary.
Olivier, A. (1992). Handling pupils' misconceptions. In M. Moodley, R.A. Njisani &
N. Presmeg (Eds.), Mathematics Education for Pre-Service and In-Service, 193-209.
Pietermaritzburg: Shuter & Shooter.
Olivier, A. (1992). Computation via Problem Solving. In T. Marsh (Ed.), Proceedings
of the Eleventh National Congress on Mathematics Education, 173-189. Port
Elizabeth: Mathematical Association of Southern Africa.
Orton, A. (Ed.) (1999). Patterns in the Teaching and Learning of Mathematics.
London: Cassell.
Pólya, G. (1954). Mathematics and plausible reasoning (Volume 1: Induction and
Analogy in Mathematics; Volume 2: Patterns of Plausible Inference ). Princeton:
Princeton University Press.
Pólya, G. (1957). How to Solve It. Princeton University Press, Princeton, New Jersey.
Pólya, G. (1962,1965/1981). Mathematical Discovery (Volume 1, 1962; Volume 2,
1965). Princeton: Princeton University Press. Combined paperback edition, 1981.
New York: Wiley.
Resnick, L. (1989). Treating mathematics as an ill-structured discipline. In R. Charles &
E. Silver (Eds.), The Teaching and Assessing of Mathematical Problem Solving, 32-60.
Reston, VA: National Council of Teachers of Mathematics.
Schoenfeld, A. H. (1985). Mathematical Problem Solving. Orlando, FL: Academic Press.
Schoenfeld, A. H. (1989). Problem solving in context(s). In R. Charles & E. Silver
(Eds.), The Teaching and Assessing of Mathematical Problem Solving, 82-92.
Reston, VA: National Council of teachers of Mathematics.
Schoenfeld, A. H. (1992). Learning to think mathematically: Problem solving,
metacognition, and sense-making in mathematics. In D. Grouws (Ed.), Handbook for
Research on Mathematics Teaching and Learning, 334-370. New York: MacMillan.
http://www-gse.berkeley.edu/Faculty/aschoenfeld/LearningToThink/Learning_to_think_Math.html
Shell Centre for Mathematical Education. (1984). Problems with Patterns and
Numbers. Nottingham, England: Shell Centre.
Silver, E. A. (1994). On mathematical problem posing. For the Learning of
Mathematics, 14(1).
Taylor, N. R. & Vinjevold, P. (Eds.) (1999). Getting Learning Right. Report of the
President's Education Initiative Research Project. Johannesburg: Maskew Miller Longman.
213
APPENDIX: Is 1 a prime?9
Like any good question, this can be answered at several levels. The first simple
answer is that the generally accepted definition of a prime number is “a positive
integer greater than 1 which has no divisors other than 1 and itself.”
An equivalent definition is to specify that a prime number is a positive integer which
has exactly two divisors. Since 1 has exactly one divisor, it is not prime.
A composite number is then defined to be a positive integer with three or more
divisors.
If, on the other hand, you define a prime number as a positive integer which has no
divisors other than 1 and itself, then 1 must be regarded as a prime number.
Which is the correct definition? Both have been used over the years, and it really
depends on how you want to develop the discussion of prime numbers.
Mathematicians are nowadays in general agreement that it is more convenient to use
a definition which excludes 1 from the set of prime numbers.
The prime reason (if you will pardon the pun) is that one of the first results in number
theory is The Fundamental Theorem of Arithmetic, which states:
Every positive integer greater than 1 is expressible as the product of prime
numbers, and, except for the order of the factors in the product, such an
expression is unique.
Thus 1001 = 7 x 11 x 13, and there is no other way of expressing 1001 as a product
of prime numbers, except by changing the order of the factors. However, if you allow
your definition of prime to include the number 1, you could write 1001 in many
different ways:
1001 = 1 x 7 x 11 x 13 = 1 x 1 x 7 x 11 x 13, etc.
So the Fundamental Theorem of Arithmetic would have to be changed if your
definition of prime included the number 1, to:
Every positive integer greater than 1 can be expressed as a product of prime
numbers greater than 1, and, except for the order of factors in the product,
such an expression is unique.
Thus, if you allow 1 to be a prime number, you must immediately exclude it from your
first major theorem about prime numbers. That is enough justification for agreeing not
to consider 1 to be a prime. Note that it is not wrong, in any moral, legal or
mathematical respect, to regard 1 as a prime number. But if you use your own
personal definition for a concept which is not how others view that concept, you are
liable to be misunderstood.
9
See Mathematical Digest, University of Cape Town
214