You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

Number theory : A set of friendly examples

Even and odd numbers

Two whole numbers are added together. If their sum is odd, which statements below are
always true? Which are always false? Which are sometimes true and sometimes false?
1 Their quotient is not a whole number.
2 Their product is even.
3 Their difference is even.
4 Their product is more than their sum.
5 If 1 is added to one of the numbers and the product is found, it will be even.
The Collatz conjecture
Choose any whole number to start with.

If it is odd, multiply it by 3 and then add 1.
If it is even, divide it by 2. Then repeat this process on the number just obtained. Keep repeating the procedure.

For example, if you start with 58, the resulting chain of numbers is
58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, …

The Collatz conjecture, made by Lothar Collatz in 1937, claims that, if you repeat this
process over and over, starting with any whole number greater than zero, eventually you will finish up with the sequence … 4, 2, 1, 4, 2, 1.

A conjecture is a statement that is thought to be true but has not been proved mathematically to be true for all cases. Although the Collatz conjecture has been shown to work − often very quickly − for many whole numbers, there are some quite small numbers that take a very long time to come down to … 4, 2, 1, 4, 2, 1.Apply this process to all the whole numbers greater than zero and less than or equal to 30.
For each one, find:
• how many steps it takes to reach 1 the frst time
• the largest number in the sequence. (For the sequence above, 58 takes 19 steps and reaches a maximum of 88.)
Look for shortcuts and work with a partner if you like.

Long division
Here is a way to check how good your long division skills are. If you are able to follow it
through and get to the end without making a mistake, you can consider yourself a qualiꏨed long division champion.
• Start with any two-digit number (for example, 58). Write it three times so that a six-digit  number is formed (585 858).

  • Divide this number by 21. There should not be any remainder. If there is, try and find out where you made your mistake and fix it.
    • Now divide this new four- or possibly five-digit number by 37. Once again, there should be no remainder.
    • Finally, divide this number − which should by now have only three or four digits − by 13.
    You will know if you got it right by looking at the number you are left with.
    Explain why this exercise works.
    (Doing any of this exercise on a calculator is still interesting but is de뀠nitely wimping out!)

Totient numbers

1 A totient number is the number of fractions between 0 and 1 (not including 0 or 1) for
a given denominator that cannot be reduced to a simpler equivalent fraction. The totient
number of 2 is 1, since we have \frac{1}{2}; of 3 it is 2, since we have \frac{1}{3} and \frac{2}{3}; and of 4 it is also 2, since we have \frac{1}{4}
and \frac{3}{4} (\frac{2}{4} can be reduced to \frac{1}{2}). The totient number of 5 is 4, since we have \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; and of 6 it is 2, since we have \frac{1}{5} and
\frac{5}{6}. Find the totient numbers forall denominators up to 12.

2 For any denominator n, there are n fractions between 0 and 1 (including 0 but not 1). Of
these fractions, some will be counted towards the totient number of n, but others will
cancel down and count towards the totient number of one of the factors of n. Using this
information and the totient numbers from the previous question, calculate the totient
numbers for 15, 18, 20 and 24.

3 The totient number is related to the prime factors of the original number, since these will
determine which fractions can be cancelled. Using this information, calculate the totient numbers of 72, 81, 98 and 100.

\bf{Last \hspace{0.1in}Digits \hspace{0.1in} of \hspace{0.1in}powers}
\bf{Square \hspace{0.1in}Numbers}

Without using a calculator, can you say which of this set of numbers could not be square numbers?

8116801, 251301659, 3186842, 20720704.

You can just by checking the last digit (units digit) of each number.

Do a bit of experimentation with a calculator and find the four digits that square numbers end in. (This eliminates the third number in this set).

Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set).

Complete these sentences with what you have discovered:

* In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______.
* The second last digit (tens place) of an odd square number is always _______.

\bf{Cube \hspace {0.1in} Numbers}

Cube numbers behave rather differently.

A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed.

Complete this table:
\left| \begin{array}{cc}     \mbox {if a number ends in} & \mbox{its cube will end in}\\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ 8 & \\ 9 & \end{array}\right|

\bf{Fourth \hspace{0.1in}Powers}

Fourth powers are in fact just square numbers that have been squared. For example, 7^{4}=7^{2} \times 7^{2}= 49^{2}=2401.

Since 4^{2}=16 and 9^{2}=81, the last digit of a fourth power can only be 0, 1, 5 or 6.

\bf{Fifth \hspace{0.1in}powers}

Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you create tables of second powers, third powers, fourth powers and fifth powers of all numbers from zero to 20. Check, compare…actually, it is fun to “compare rate of growth of powers with increasing integers”…this idea involves rudimentary ideas of calculus…

\bf{Obstinate \hspace{0.1in} numbers}

An odd number can usually be written as a sum of a prime number and another number, which is a power of two. This is true for all odd numbers greater than one but less than 100.

For example, if we choose 23, we can say that it is equal to 23=19 + 2^{2}. There is one more way to represent 23: it is 7 + 2^{4}. So, there are two ways to represent 23 as a sum of a prime number and a power of two. But, 21+2^{1} and 15+2^{3} do not work as both 21 and 15 are not prime numbers.

Some odd numbers like this can be expressed in many ways.

Try to find various representations as sum(s) of prime number and a power(s) of two of the following integers: 45, 29, 59, 95.

If you are adventurous or courageous, try to find such representations of all odd numbers lying from 1 to 100. You need a lot of patience and stamina and grit…but you will develop an “intuitive feel or tactile feel for numbers”…that’s the way math begins…

There are some odd numbers which cannot be expressed as a sum of a prime number and a power of two. Such numbers are called \bf{obstinate \hspace{0.1in} numbers}.

An example of an obstinate number is \bf{251} as the working below shows:

251-2^{1}=249=3 \times 83

251-2^{2}=247=13 \times 19

251-2^{3}=243=3 \times 81

251-2^{4}=235= 5 \times 47

251-2^{5}=219= 3 \times 73

251-2^{6}= 187= 11 \times 17

251-2^{7}=123=3 \times 41

The next power of 2 is 2^{8}=256, which is clearly greater than 251. Hence, 251 is an obstinate number.

In fact, 251 is the third obstinate number. The first two lie between 100 and 150. Find these two odd numbers keeping track of how you eliminated the other twenty three odd numbers between 100 and 150.

\it{Remember \hspace{0.1in} to \hspace{0.1in} be \hspace{0.1in} systematic \hspace{0.1in}}.

Making a list of the powers of two up to 2^{8} might be a good place to start with. Look for short cuts and patterns as you proceed further.

Have fun with numbers !!

Regards,
Nalin Pithwa.

Axiomatic Method : A little explanation

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : a \in A ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

  1. If to equals we add equals, we get equals.
  2. If from equals we take equals, the remainders are equal.
  3. If equals are multiplied by equals, the products are equal.
  4. If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If n=1, the result is evident.

Step 2: By the induction hypothesis the result is true when n=k; we must prove that it is correct when n=k+1. Let S be any set containing exactly k+1 real numbers and denote these real numbers by a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}. If we omit a_{k+1} from this list, we obtain exactly k numbers a_{1}, a_{2}, \ldots, a_{k}; by induction hypothesis these numbers are all equal:

a_{1}=a_{2}= \ldots = a_{k}.

If we omit a_{1} from the list of numbers in S, we again obtain exactly k numbers a_{2}, \ldots, a_{k}, a_{k+1}; by the induction hypothesis these numbers are all equal:

a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}.

It follows easily that all k+1 numbers in S are equal.

*************************************************************************************

Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

Check your talent: are you ready for math or mathematical sciences or engineering

At the outset, let me put a little sweetener also: All I want to do is draw attention to the importance of symbolic manipulation. If you can solve this tutorial easily or with only a little bit of help, I would strongly feel that you can make a good career in math or applied math or mathematical sciences or engineering.

On the other hand, this tutorial can be useful as a “miscellaneous or logical type of problems” for the ensuing RMO 2019.

I) Let S be a set having an operation * which assigns an element a*b of S for any a,b \in S. Let us assume that the following two rules hold:

i) If a, b are any objects in S, then a*b=a

ii) If a, b are any objects in S, then a*b=b*a

Show that S can have at most one object.

II) Let S be the set of all integers. For a, b in S define * by a*b=a-b. Verify the following:

a) a*b \neq b*a unless a=b.

b) (a*b)*c \neq a*{b*c} in general. Under what conditions on a, b, c is a*(b*c)=(a*b)*c?

c) The integer 0 has the property that a*0=a for every a in S.

d) For a in S, a*a=0

III) Let S consist of two objects \square and \triangle. We define the operation * on S by subjecting \square and \triangle to the following condittions:

i) \square * \triangle=\triangle = \triangle * \square

ii) \square * \square = \square

iii) \triangle * \triangle = \square

Verify by explicit calculation that if a, b, c are any elements of S (that is, a, b and c can be any of \square or \triangle) then:

i) a*b \in S

ii) (a*b)*c = a*(b*c)

iii) a*b=b*a

iv) There is a particular a in S such that a*b=b*a=b for all b in S

v) Given b \in S, then b*b=a, where a is the particular element in (iv) above.

This will be your own self-appraisal !!

Regards,

Nalin Pithwa