Compatible Numbers

By Chandrajeet K


• The word "compatible" means "well-matched". Compatible numbers are numbers that are friendly with each other.

For example:

15 and 5 are compatible numbers, because, 5 goes into 15 evenly ---- 15/5=3
So are 15 and 3, because, 3 goes into 15 evenly ---- 15/3=5

• Compatible numbers make estimation or mental calculation easier. They are useful in estimating the sum, difference, product, or quotient. Compatible numbers are close in value to the actual numbers given in the problem. They often end in 0 or 5.

Let’s look at a few examples.

Example 1

Estimate 33 + 28.

Look for close numbers that are easier to work with. Multiples of 10 are easier to work with.

28 is closer to 30.

Adding 33 and 30 is easy.

So,

33 + 28 is approximately equal to 33 + 30 = 63.


Example 2

Estimate 12 + 59 + 38.

Now, look for numbers that can be added together to make a 10 or multiples of 10.

12 + 59 + 38

= 12 + 38 + 59 ----- Swap the positions of 59 and 38

= 50 + 59 ------------ The 2 and the 8 are compatible. They add to make a 10.

= 109


Example 3

Estimate 52 – 37.

Look for close numbers that are easier to work with. Multiples of 10 are easier to work with.

37 is closer to 40.

Subtracting 40 from 52 is easy.

So,

52 – 37 is approximately equal to 52 – 40 = 12.


Example 4

Estimate the value of 61 × 5.8.

Compatible numbers for 61 and 5.8 are 60 and 6 respectively.

So, 61 × 5.8 is approximately equal to 60 × 6 = 360.


Example 5

Estimate 33/8.

To get a good estimate, round the dividend to the nearest multiple of the divisor.

Look for a number ‘close to 33 and at the same time is divisible by 8’. In other words, try to find a multiple of 8 that is close to 33.

32 is the right choice.

So,

33/8 is approximately equal to 32/8 = 4.


Example 6

Estimate 29/6.5.

29 and 6.5 are not friendly with each other. Try to find a pair of compatible numbers one of which is close to 29 and the other is near 6.5.

Round the numbers 29 and 6.5.

30, 6 is the ideal pair of compatible numbers with 30 close to 29 and 6 close to 6.5.

Therefore, 29 ÷ 6.5 is approximately equal to 30 ÷ 6 = 5.

Our estimate is 5. The actual quotient is 4.46. We are not for away from the actual answer.


Sometimes we should be careful about our choice of compatible numbers. It is important to choose compatible numbers that are appropriate to a given situation.

Let’s look at an example.

83 apples have to be packed in boxes. Each box holds 10 apples. About how many boxes will you need?

In order to find the number of boxes needed to pack ALL the apples, we have to divide 83 by 10. But…83 and 10 do not go well together.

Can you think of a number that goes well with 10 and at the same time is closer to 83?

Let’s try 80.

Well…80 is indeed friendly with 10, since 10 goes into 80 evenly.
But REMEMBER! We have to pack ALL 83 apples. 83 is greater than 80. If we consider 80, then we’ll have 3 apples left unpacked.

Think of another number…90?

Hooray!

90 is close to 83 and is friendly with 10 since 90 is a multiple of 10.

83/10 is approximately equal to 90/10 = 9.

Therefore you’ll need about 9 boxes.

Amicable Numbers

By Chandrajeet K

The term “amicable” means “friendly”.
Amicable numbers are friendly numbers.

Amicable numbers are a pair of numbers such that the sum of the proper divisors of each number equals the other number in the pair.

Okay…

What are proper divisors?

Proper divisors are all the numbers that divide evenly into a number, including 1 but excluding the number itself.

Example

1, 2, 3, 4, 6, and, 12 are divisors of the number 12. They divide evenly into 12 without a reminder.
But…the PROPER divisors of 12 are: 1, 2, 3, 4, and 6; we should NOT include 12 here.



Let’s go back to “Amicable Numbers”.
220 and 284 are the smallest pair of amicable numbers.

The proper divisors of 220 are:
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110

The sum of all these proper divisors is:
1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284

Wow! That’s the other number in the pair.

Let’s now find the proper divisors of 284.
1, 2, 4, 71, and 142

Add them up.
1 + 2 + 4 + 71 + 142 = 220

Excellent! Indeed they are friendly numbers!

Easier said than done… It’s a lot of work to find divisors of big numbers.
Here are a few tips to help you finding the divisors and sum of the divisors of a number.

# 1

The quickest way to find the divisors of a number could be to obtain the prime factors of the number and then combine those factors in all possible ways.

Let’s look at an easy example.

You can use a factor tree to find the prime factors.

The prime factors of 24 are: 2, 2, 2, and 3
Let’s combine the factors in all possible ways to find the divisors of 24.
There are three 2’s and one 3 in 24.
Let’s first combine the 2’s.

2 is divisor of 24
2 x 2 = 4 is a divisor of 24
2 x 2 x 2 = 8 is a divisor of 24

Let’s now combine the 2’s and the 3.

3 is a divisor of 24
2 x 3 = 6 is a divisor of 24
2 x 2 x 3 = 12 is a divisor of 24
2 x 2 x 2 x 3 = 24 is a divisor of 24

So, the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
We include 1 among the divisors, because ‘1 is a divisor of every number’.
Leaving out the 24 we get the PROPER divisors of 24.
The next step is to find the sum of proper factors of 24 which is:

1 + 2 + 3 + 4 + 6 + 8 + 12 = 36

Phew, that’s tedious! There’s got to be a better way.

What if we can get the SUM OF THE PROPER DIVISORS straight away?

Well…there’s a formula available for the sum of the divisors. This might come in handy when you got to work with pairs of really big amicable numbers.

# 2

Let ‘n’ be a positive integer.
n = (p^a) x (q^b) x (r^c) x … represents the prime factorization of ‘n’.

Then the sum of ALL the divisors of ‘n’ is:
{[p^(a+1)] – 1}/(p – 1) x {[q^(b+1)] – 1}/(q – 1) x {[r^(c+1)] – 1}/(r – 1) x …

Let’s try using this formula with the SAME SIMPLE example (24).

The prime factors of 24 are: 2, 2, 2, and 3
The prime factorization of 24 is (2^3) x (3^1).

So, the sum of ALL the divisors of 24 is:

{[2^ (3+1)] – 1}/(2 – 1) x {[3^(1+1)] – 1}/(3 – 1)
= {[2^4] – 1}/(1) x {[3^2] – 1}/(2)
= {[16] – 1}/(1) x {[9] – 1}/(2)
= {15}/(1) x {8}/(2)
= 15 x 4
= 60

REMEMBER 60 is the sum of ALL the divisors of 24. But what we need is the sum of the PROPER divisors of 24. So, SUBTRACT 24 from 60, we get:
60 – 24 = 36

Hooray! It works! It matches with our answer!

Hope this helps!

The Counting Principle

By Chandrajeet K

The counting principle works for two or more events occurring in series. It is used to find the total number of possible outcomes when two or more events occur together.

The counting principle states that:

If a first event has ‘p’ possible outcomes followed by a second event that has ‘q’ possible outcomes followed by a third event that has ‘r’ possible outcomes and so on…, then there are a total of ‘p × q × r × …’ possible outcomes for the series of events.

We simply multiply the number of possible outcomes of each of the events in the series to get the total number of outcomes for the series.

Here are some examples using the counting principle:


Example 1

Helen has 4 blouses and 3 skirts. How many different outfits consisting of one blouse and one skirt are possible?

Solution

Helen has:

4 blouses – B1, B2, B3, B4
3 skirts – S1, S2, S3

The different possible outfits consisting of one blouse and one skirt are:

1. B1, S1
2. B1, S2
3. B1, S3
4. B2, S1
5. B2, S2
6. B2, S3
7. B3, S1
8. B3, S2
9. B3, S3
10. B4, S1
11. B4, S2
12. B4, S3

So:

Helen can make 12 different outfits consisting of one blouse and one skirt.
This can easily be calculated using the counting principle as 4 × 3 = 12.


Example 2

A restaurant offers a choice of 2 salads, 4 main courses, and 3 desserts. Find the number of possible choices when you choose one item from each category.

Solution

You have a choice of 2 salads – S1, S2
You have a choice of 4 main courses – M1, M2, M3, M4
You have a choice of 3 desserts – D1, D2, D3
As in the previous example, our list looks something like this…

The different possible choices consisting of a salad, a main course, and a dessert are:

1. S1, M1, D1
2. S1, M1, D2
3. S1, M1, D3
4. S1, M2, D1
5. S1, M2, D2,
6. S1, M2, D3
7. …
8. …
9. …

I'm not going to list all of them! I'll let you write them down.

By the counting principle we know there are a total of 2 × 4 × 3 = 24 possible choices of a 3-course meal.


Example 3

A state’s license plates have 3 letters followed by 3 digits.

(a) How many different license plates are possible if letters and digits can be repeated?

(b) How many different license plates are possible if letters and digits cannot be repeated?

Solution

(a) How many different license plates are possible if letters and digits can be repeated?

There are 26 choices for letters (A to Z) and 10 choices for digits (0, 1, 2, 3 …9).

The license plates have 3 letters followed by 3 digits.

LETTER LETTER LETTER DIGIT DIGIT DIGIT

Using the Counting Principle:
26 × 26 × 26 × 10 × 10 × 10

= 17,576,000

So, 17,576,000 different license plates are possible if letters and digits can be repeated.

(b) How many different license plates are possible if letters and digits cannot be repeated?

There are 26 choices for the first letter but only 25 choices for the second, and 24 for the third.

For the digits, there are 10 choices for the first, but only 9 for the second, and 8 for the third.

SO:

Using the Counting Principle:

LETTER LETTER LETTER DIGIT DIGIT DIGIT

26 × 25 × 24 × 10 × 9 × 8
= 11,232,000

So, 11,232,000 different license plates are possible if letters and digits cannot be repeated.

Here is a question for you…

How could the state increase the total number of possible license plates if letters and digits cannot be repeated?

Indirect Proof

By Chandrajeet K

The concept of proof is an important part of mathematics. There are three basic types of proofs: direct proofs, indirect proofs, and proofs by contradiction.

In this article, let’s learn about Indirect Proof. Please take time and read it carefully till the end.

Indirect proof is a type of proof that begins by ASSUMING what is to be proved is FALSE. Then we try to prove that our ASSUMPTION is true. If our ASSUMPTION leads to a contradiction then the original statement which was assumed false must be true.

Let me explain more in detail.

Suppose you wish to prove ‘statement A’ is true using an indirect proof.

The first thing you do is:

You assume statement A is false…and assume statement A’ which is a contrary of statement A to be true.

Then using valid arguments, you arrive at a contradiction (denial or disagreement) to statement A’.

Thus demonstrating that statement A is true.

This concept will be clearer when you look at some examples.

Example 1

Sarah left her house at 9:30 AM and arrived at her aunt’s house 80 miles away at 10:30 AM. Use an indirect proof to show that Sarah exceeded the 55 mph speed limit.

Solution

Suppose that the given statement is false. That is: ‘Sarah did NOT exceed the 55 mph speed limit.

She drove 80 miles at 55 mph.

At this speed, Sarah would need 80/55 (approximately) = 1 hour 27 minutes to reach her aunt’s place.

But as per the problem she drove from 9:30 AM to 10:30 AM … exactly an hour.

SO, she must have driven faster than 55 mph….a contradiction to our assumption that Sarah did NOT exceed the speed limit.

Therefore, Sarah exceeded the speed limit.


Example 2

Prove the following using an indirect proof.
For all integers ‘n’, if 3n + 1 is even, then ‘n’ is odd.

Solution

Suppose that the conclusion is false. That is: ‘n’ is NOT odd.
Assume the contrary is true. That is: ‘n’ is even.
Then the statement contrary of the given statement is:

“For all integers ‘n’, if 3n + 1 is even, then ‘n’ is EVEN”

Let’s try to prove it.

‘n’ is even means ‘n’ is a multiple of 2…that is: n = 2m for some integer ‘m’.

Then:

3n + 1 = 3(2m) + 1 = 6m + 1 --- Call it Equation (1)
Well…6m is even. So, 6m + 1 is odd.
Therefore, 3n + 1 is ODD…because 3n + 1 = 6m + 1 from Equation (1).

By assuming ‘n’ is even, we’ve shown that 3n + 1 is ODD which is a contradiction to our assumption.

Therefore:

If ‘n’ is odd then 3n + 1 is even. This is the contrapositive of the statement to be proved.

Since the contrapositive is true, it follows that the original statement “if 3n + 1 is even, then ‘n’ is odd” is true.

The next example is a classic problem where an Indirect Proof is used.

Example 3
Prove that square root of 2 or SQRT (2) is irrational using an indirect proof.

Solution
ASSUME that the given statement is false.

That is:

SQRT(2) is NOT irrational.

Assume the contrary to be true…that is…SQRT(2) is RATIONAL.

Let’s try to prove it.

A rational number is a real number that can be expressed as a quotient of two integers a/b, where b does not equal 0.

We’ve assumed SQRT(2) to be a rational number.

So:

SQRT (2) = a/b. This fraction a/b is in lowest terms - that is, a and b have no common factors.

Multiply each side by ‘b’ to get rid of the fraction.

b × SQRT(2) = a
Square both sides.

SQR (b) × 2 = SQR (a) which is the same as:
2 SQR (b) = SQR (a) --- call it Equation (2)

SQR (a) is even…because from Equation (2) above, we have, SQR(a) = 2 SQR(b)…a multiple of 2.

SQR(a) is even…implies…’a’ is even. Then, a = 2k for some integer ‘k’.

Substitute a = 2k in Equation (2). We get:

2 SQR (b) = SQR (a) --- Equation (2)
2 SQR (b) = SQR (2k)
2 SQR (b) = 4 SQR (k)

Cancel ‘2’ on either side. We have:

SQR (b) = 2 SQR (k)

The above equation shows that ‘SQR(b)’ is even…because SQR(b) = 2 SQR(k).

Again, SQR (b) is even implies ‘b’ is even.
If ‘a’ and ‘b’ is both even, then they will have a common factor…

Then…how can the fraction a/b be in lowest terms?
A contradiction…

SO, SQRT (2) is IRRATIONAL.

Example 4

Prove that “For all integers ‘n’, if ‘n’ is odd then SQR(n) is odd” using an indirect proof.

Solution

Suppose the conclusion is false.

That is:

SQR(n) is NOT odd.

ASSUME the contrary… SQR (n) is even.
Then the statement contrary of the given statement is:

“For all integers ‘n’, if ‘n’ is odd then SQR(n) is even”

Let’s try to prove it.

If SQR(n) is even, then SQR(n) can be expressed as a multiple of 4.

So:

SQR (n) = 4k for some integer ‘k’.
Take square root on either sides of the equation. We get:
n = 2 SQRT (k)

The above equation shows that ‘n’ is even, because ‘n’ is a multiple of 2…

By assuming ‘SQR(n)’ is even, we’ve shown that ‘n’ is EVEN which is a contradiction to our assumption.

So:

If ‘SQR(n)’ is odd then ‘n’ is odd. This is the contrapositive of the statement to be proved.

Since the contrapositive is true, it follows that the original statement “If ‘n’ is odd then ‘SQR (n)’ is odd” is true.

Mathematics in Physics and Chemistry

By Chandrajeet K

Mathematics is the queen of all sciences’ – those are the words of Carl Friedrich Gauss the greatest mathematician of all time.

Mathematics is an important tool for science. Math is most widely used in other sciences. Physics, Chemistry, astronomy, engineering rely most heavily upon mathematical ideas.

Students who consider studying Physics or Chemistry will need a relatively strong Math background.

Mathematics in Physics

Physics is the natural science which explores concepts like mass, energy, matter and its motions. Strong foundation in Algebra, Trigonometry, Geometry, and calculus is essential for physics. Mathematical methods are absolutely necessary to deal with important concepts in physics.

The following are some examples.

(1) Electromagnetic theory is the branch of physics that studies the group of forces associated with electric charges. Vector Analysis is very important for the understanding and developing of Electromagnetic theory.
(2) Group theory is useful in Spectroscopy, Quantum mechanics, Solid state physics, and Nuclear physics.
(3) Fourier techniques are important for the analysis of all linear systems in physics.
(4) Matrix Analysis is necessary for understanding Quantum Mechanics.
(5) Complex numbers are used extensively in physics to describe Electromagnetic Waves and Quantum Mechanics.

Mathematics in Chemistry

Chemistry is the natural science which explores the composition and properties of substances. Math is essential for chemistry. The necessary mathematical background for the study of chemistry includes basic algebra, some trigonometry, and calculus.

The following are some examples.

(1) Being able to balance chemical equations is a very important skill for chemistry students. It’s a simple mathematical exercise. Balancing a chemical equation refers to establishing the mathematical relationship between the amounts of reactants and products involved in the chemical reaction.

Let’s go more in detail.
A chemical equation is a statement that describes what happens in a chemical reaction.

In a chemical equation, we place the reactants (substances undergoing chemical reaction) on the left side of the equation and the products (substances produced in a chemical reaction) on the right side of the equation. We have reactants and products separated by an arrow and the arrow always points in the direction of the products.

Consider the reaction of carbon with oxygen gas to produce carbon-dioxide.

C + O2 ---> CO2 (2 is subscript)

The above equation is already balanced, because, it has an equal number of atoms of each element in the reactants and the product. One carbon atom (C) and two oxygen atoms (O) on the left side of the equation and it’s the same on the right side too.

Let’s look at one more example.

Sodium chloride is the common salt. Sodium and chlorine form sodium chloride.
Na + Cl2 ---> NaCl (2 is subscript)

The above equation is NOT balanced. It has two chlorine atoms on the left side, but, only one on the right side of the equation.

Let’s balance this chemical equation.
2Na + Cl2 ---> 2NaCl (2 is subscript only in Cl2)

It works! Notice that now there are equal number of atoms of each element in the reactants and the product.

Chemical equations can be balanced conveniently using matrices or simultaneous equations.

A number of fields of chemistry use a significant amount of Math.

(2) Electrochemistry is a branch of chemistry that studies the chemical action of electricity and the production of electricity by chemical reactions. Diffusion in electrochemistry is completely based on differential equations.

(3) Biochemistry is the study of the chemical processes in living organisms. Even biochemistry has important topics which depend heavily on binding theory and kinetics.