MATH PUZZLE & IDEAS

checkerboard puzzle

Take a look at the checkerboard below. A typical checkerboard puzzle could ask you the following question:

How many squares, of all sizes, are there on this 8 × 8 checkerboard?

However, isn't important to know how many different sizes are there? 

The different sizes are 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, 6 × 6, 7 × 7, and 8 × 8

In this checkerboard puzzle, it is easy to know how many 1 × 1 there are. Since there are 8 such square on each size, there are a total of 8 × 8 = 64

It is also easy to see that there is only 1 square that has a size of 8 × 8

To find out how many there are for any other size is a big headache. However, I will illustrate the technique or trick for 2 sizes, the 2 × 2 and the 6 × 6.

I leave it up to you to find it for the remaining sides! And if you do find it, I am happy.

First, let us find out how many 2 × 2 there are. You will need to carefully examine the following illustration:

I carefully numbered all the 2 × 2 squares we can get on one side starting from the very top and going down 1 unit each time

Since there are 7 such square on one side, we know there will be 7 such square on any other side.

since 7 × 7 = 49, 49 squares have a size of 2 × 2

Next, let us find out how many 6 × 6 there are. Again, you will need to carefully examine the following illustration:

I carefully numbered all the 6 × 6 squares we can get starting from the very top and going down 1 unit each time

Since there are 3 such square on one side, we know there will be 3 such square on any other side.

since 3 × 3 = 9, 9 squares have a size of 6 × 6

Following a similar course, there are 

36 squares with a size of 3 × 3

25 squares with a size of 4 × 4

16 squares with a size of 5 × 5

4 squares with a size of 7 × 7

Adding the values of all sizes, we get 64 + 1 + 49 + 9 + 36 + 25 + 16 + 4 = 204

Therefore, there are 204 squares of sizes 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, 6 × 6, 7 × 7, and 8 × 8

Let's get started with some fun math puzzles!

Puzzle #1:

Use four 5s and some of the symbols +, ×, −, and ÷ to give expressions for 0, 1, 2, and 5

Puzzle #2:

Use the two isosceles triangles below to make a star?

Puzzle #3:

It takes 1629 digits to number the pages of a book. How many pages does the book have?

Puzzle #4:

Use only 4 straight lines to cut a pizza into 11 pieces?

Puzzle #5:

would you rather work for 30 days and get paid 5 millions dollars, or be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, and so on...?

Puzzle #6:

Place 10 dishes along the 4 edges of a table so that each edge has the same amount of dishes

Puzzle #7:

In a family with 4 siblings, John is older than Mary, Peter is younger than John, Mary is older than Peter, and Sarah is older than John. Who is the second oldest in the family? Who is the youngest?

Puzzle #1 solution:

5 − 5 + 5 − 5 = 0

5 − 5 + 5 ÷5 = 1

5 ÷5 + 5 ÷5 = 1 + 1 = 2

5 × (5 − 5) + 5 = 5

Puzzle #2 solution:

Puzzle #3:

From page #1 to page # 9, there are 9 pages

From page #10 to page # 99, there are 90 pages. However, since each page number has 2 digits, there are 180 digits from page # 10 to page # 99 

From page #100 to page # 999, there are 900 pages. However, since each page number has 3 digits, there are 2700 digits from page # 100 to page # 999

Since there are 1629 digits, we know so far that number of pages is less than 900

Now, we need to find out how many digits there are for those pages greater than 99 or those pages with 3 digits

Since there are 189 digits for pages with less than 3 digits, we can subtract 189 from 1629 digits to find the number of pages with 3 digits

1629 − 189 = 1440

1440 represents the pages with 3 digits, so number of pages with 3 digits = 1440/3 = 480

Thus, total number of pages = 480 + 90 + 9 = 579 pages

Puzzle #4 solution:

Puzzle #5:

The trick is to find out how much money you make in 30 days if you get 1 cent the first day, 2 cents the second day, 4 cents the third day, and so on... Day 1: 1 cents

Day 2: 2 cents

Day 3: 4 cents

Day 4: 8 cents 

Day 5: 16 cents

Day 6: 32 cents

Day 7: 64 cents

Day 8: 128 cents

Day 9: 256 cents

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Keep mutliplying by 2

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Day 30: 536,870,912 cents

536,870,912 cents/100 = 5,368,709.12 dollars and 5,368,709.12 is bigger than 5,000,000

Therefore, you are better off taking 1 cent the first day, 2 cents the second day, 4 cents the third day, and so on... until the 30th day

Puzzle #6:

If the blue rectangle represents the table and the red circles represent dishes, here is the solution

Are you not tired yet with these fun math puzzles? Read on...

Puzzle #7:

In a family with 4 siblings, John is older than Mary, Peter is younger than John, Mary is older than Peter, and Sarah is older than John. Who is the second oldest in the family? Who is the youngest?

Metric measurement

The gram is used to measure mass or weight.

For instance, a person weighing 170 pounds in customary measurement, weighs 77110 grams or 77.11 kilograms in the metric system.

the liter is used to measure capacity. 

for instance, 1 liter of coke measures about 33.81 ounces in customary measurement.

The meter is used to measure length.

For instance, a person whose height is 1 meter is equivalent 3.2808 feet in customary measurement.

In the metric system, other units are named by using prefixes such as kilo, hecto, deka, deci, centi, and milli etc...

1 kilo = one thousand and the symbol we use is k

For example, 1 kilometer = 1000 meters

1 hecto = one hundred and the symbol we use is h

1 hectogram = 100 grams

1 deka = ten and the symbol we use is da

1 dekameter = 10 meters

1 deci = one-tenth and the unit is d

1 deciliter = 0.1 liter

1 centi = one-hundredth and the unit is c

1 centimeter = 0.01 meter

1 milli = one-thousandth and the unit is m

1 millimeter = 0.001 meter 

Commutative property

Certainly, in commutative property, we see the word commute which means exchange from the latin word commutare

The word exchange in turn may mean switch. For examples, washing my face and combing my hair is a good example of this property.

Another good example is doing my math homework and then finishing my science reading. 

The important thing to notice in the two examples above is that the order we do things can be switched, so does not matter or will never cause any problems or conflicts.

However, reading a math lesson and then answering the review questions is not commutative.

Here the order does matter because I have to read the lesson before knowing how to answer the review questions

In mathematics, we know that

2 + 5 = 5 + 2

12 + 4 = 4 + 12

-1 + 8 = 8 + -1

All the above illustrates the commutative property of addition. This means that when adding two numbers, the order in which the two numbers are added does not change the sum

All three examples given above will yield the same answer when the left and right side of the equation are added

For example, 2 + 5 = 7 and 5 + 2 is also equal to 7

The property is still valid if we are doing multiplication

Again, we know that

3 × 4 = 4 × 3

12 × 0 = 0 × 12

9 × 6 = 6 × 9

Again, 3 × 4 = 12 and 4 × 3 = 12

More examples: Take a close look at them and study them carefuly

(3 + 2) × 4 = 4 × (3 + 2)

x + y = y + x 

x × y = y × x

2 × x = x × 2

(x + z) × (m + n) = (m + n) × (x + z)

4 + y = y + 4

Warning! Although addition is commutative, subtraction is not commutative

Notice that 3 − 2 is not equal to 2 − 3

3 − 2 = 1 , but 2 − 3 = -1

Therefore, switching the order yield different results

Associative property

The word associate in associative property, may mean to join or to combine

For examples, suppose I go to the supermarket and buy ice cream for 12 dollars, bread for 8 dollars, and milk for 15 dollars.

How much money do I owe the cashier? The situation above is associative

When I do my total in my head, I can combine or add the price of the ice cream and the bread first and add the result to the price of milk.

Otherwise, I can combine or add the price of bread and milk first and add the result to the price of ice cream

Both ways of approaching the problem gives the same answer

Mathematically, you are trying to do the following:

12 + 8 + 15

You can add these three numbers in the order they appear

12 + 8 = 20 ( This is adding price of ice cream and bread first) 

20 + 15 = 35

You can use parentheses to show the order in which you are adding

(12 + 8) + 15

Another way to add is to add not according the order in which they appear

You may decide you will add first 8 and 15

8 + 15 = 23 ( This is adding price of bread and milk first)

12 + 23 = 35

Again, using parentheses to show the order in which you are adding, you get:

12 + (8 + 15)

We conclude that (12 + 8) + 15 = 12 + ( 8 + 15)

The above example illustrates the associative property of addition

Terms added in different combinations or grouping yield the same answer

Associative property of multiplication

Again, we know that

(3 × 4) × 5 = 3 × (4 × 5)

(2 × 6) × 7 = 2 × (6 × 7)

(1 × 9) × 8 = 1 × (9 × 8)

All three examples given above will yield the same answer when the left and right side of the equation are multiplied

For example, 3 × 4 = 12 and 12 × 5 = 60 

Also, 4 × 5 = 20 and 3 × 20 = 60

Warning! Although mutiplication is associative, division is not associative

Notice that ( 24 ÷ 6) ÷ 2 is not equal to 24 ÷( 6 ÷ 2)

( 24 ÷ 6) ÷ 2 = 4 ÷ 2 = 2

However, 24 ÷( 6 ÷ 2) = 24 ÷ 3 = 8

Therefore, different combination may yield different results.

Notice that it may happen that a different grouping gives the same result.

( 24 ÷ 6) ÷ 1 = 24 ÷( 6 ÷ 1)

( 24 ÷ 6) ÷ 1 = 4 ÷ 1 = 4 and 24 ÷( 6 ÷ 1) = 24 ÷ 6 = 4 

Distributive property

Example #1:

Look at the following illustration. How would you get the area?

Area = width × length

Since width = 6 and length = 4 + 10, area = 6 × (4 + 10)

You can do the math two ways

You can add 4 and 10 and multiply what you get by 6. 

Otherwise, you can use the distributive property illustrated above by multiplying 6 by 4 and 6 by 10 and adding the results

Example #2:

You go to the supermarket. 1 bag of apples costs 4 dollars. 1 gallon of olive oil costs 10 dollars. You get 6 bags of apples and 6 gallons of olive oil.

How much money do you pay the cashier?

Total cost = # of items you get × (cost for apples + cost for olive oil)

Total cost = 6 × (4 + 10) = 6 × 4 + 6 × 10 = 24 + 60 = 84 dollars. 

Same answer you would get for example #1! 

Example #3:

Robert has 8 notebooks and his brother has 6. If we double both amount, how many do they now have altogether?

We get 2 × ( 8 + 6) = 2 × 8 + 2 × 6 = 16 + 12 = 28

Notice that we get the same asnwer if we add 8 and 6 and multiply the result by 2

Properties of zero

The two properties of zero are the addition property and the multiplication property.

Addition property:

The addition property says that a number does not change when adding or subtracting zero from that number

Examples

2 + 0 = 2

12 + 0 = 12

5 − 0 = 5

48 − 0 = 48

0 + 1 = 1

0 − 9 = - 9

x + 0 = x

(a + b) + 0 = a + b

Additive inverse property

If you add two numbers and the sum is zero, we call the two numbers additive inverses or opposites of each other

For example, 2 is the additive inverse of -2 because 2 + -2 = 0

-2 is also the additive inverse of 2 because -2 + 2 = 0

Multiplication property

The multiplication property says that zero times any number is equal to zero

Examples

2 × 0 = 0

0 × 12 = 0

-5 × 0 = 0

23344555677888882 × 0 = 0

x × 0 = 0

(x + y + z + r )× 0 = 0

Identity property of multiplication

The identity property of multiplication, also called the multiplication property of one says that a number does not change when that number is multiplied by 1.

Examples

3 × 1 = 3

10 × 1 = 10

6 × 1 = 6

68 × 1 = 68

1 × 4 = 4

1 × -9 = - 9

x × 1 = x

(a + b) × 1 = a + b

Multiplicative inverse property

If you multiply two numbers and the product is 1, we call the two numbers multiplicative inverses or reciprocals of each other

For example, 4 is the multiplicative inverse of 1/4 because 4 × 1/4 = 1

1/4 is also the multiplicative inverse of 4 because 1/4 × 4 = 1

Notice that the multiplicative inverse of 1 is 1. In fact, 1 and -1 are the only two numbers that can be their own multiplicative inverse

Notice also that any number divided by 1 return the same number

We call this the identity property of division

Examples

2 ÷ 1 = 2

50 ÷ 1 = 50

-5 ÷ 1 = -5

Properties of equality

We will show 8 properties of equality. When appropriate, we will illustrate with real life examples of properties of equality.

Let x, y, and z represent real numbers

Reflexive property: x = x

Example: 2 = 2 or I am equal to myself

Symetric property: If x = y, then y = x

Example: Suppose fish = tuna, then tuna = fish

transitive property: If x = y and y = z, then x = z

Example: Suppose John's height = Mary's height and Mary's height = Peter's height, then John's height = Peter's height

Addition property: If x = y, then x + z = y + z

Example: Suppose John's height = Mary's height, then John's height + 2 = Mary's height + 2

Or suppose 5 = 5, then 5 + 3 = 5 + 3

Subtraction property: If x = y, then x − z = y − z

Example: Suppose John's height = Mary's height, then John's height − 5 = Mary's height − 5

Or suppose 8 = 8, then 8 − 3 = 8 − 3

Multiplication property: If x = y, then x × z = y × z

Example: Suppose Jetser's weight = Darline's weight, then Jetser's weight × 4 = Darline's weight × 4 

Or suppose 10 = 10, then 10 × 10 = 10 × 10

Division property: If x = y, then x ÷ z = y ÷ z

Example: Suppose Jetser's weight = Darline's weight, then Jetser's weight ÷ 4 = Darline's weight ÷ 4 

Or suppose 20 = 20, then 20 ÷ 10 = 20 ÷ 10

Substitution property: If x = y, then y can be substituted for x in any expression

Example: x = 2 and x + 5 = 7, then 2 can be substituted in x + 5 = 7 to obtain 2 + 5 = 7

Properties of inequality

We will show 6 properties of inequality. When appropriate, we will illustrate with real life examples of properties of inequality.

Let x, y, and z represent real numbers

Addition property:

If x < y, then x + z < y + z

Example: Suppose Sylvia's weight < Jennifer's weight, then Sylvia's weight + 4 < Jennifer's weight + 4 

Or suppose 1 < 4, then 1 + 6 < 4 + 6

If x > y, then x + z > y + z

Example: Suppose Sylvia's weight > Jennifer's weight, then Sylvia's weight + 9 > Jennifer's weight + 9 

Or suppose 4 > 2, then 4 + 5 > 2 + 5

Subtraction property:

If x < y, then x − z < y − z

Example: Suppose Sylvia's weight < Jennifer's weight, then Sylvia's weight − 4 < Jennifer's weight − 4 

Or suppose 4 < 8, then 4 − 3 < 8 − 3

If x > y, then x − z > y − z

Example: Suppose Sylvia's weight > Jennifer's weight, then Sylvia's weight − 9 > Jennifer's weight − 9 

Or suppose 8 > 3, then 8 − 2 > 3 − 2

Multiplication property:

If x < y, and z > 0 then x × z < y × z

Example: Suppose 2 < 5, then 2 × 10 < 5 × 10 ( Notice that z = 10 and 10 > 0)

If x > y, and z > 0 then x × z > y × z

Example: Suppose 20 > 10, then 20 × 2 > 10 × 2

If x < y, and z < 0 then x × z > y × z

Example: Suppose 2 < 5, then 2 × -4 > 5 × -4 ( -8 > -20. z = -4 and -4 < 0 ) 

If x > y, and z < 0 then x × z < y × z

Example: Suppose 5 > 1, then 5 × -2 < 1 × -2 ( -10 < -2 ) 

Division property:

It works exactly the same way as multiplication

If x < y, and z > 0 then x ÷ z < y ÷ z

Example: Suppose 2 < 4, then 2 ÷ 2 < 4 ÷ 2

If x > y, and z > 0 then x ÷ z > y ÷ z

Example: Suppose 20 > 10, then 20 ÷ 5 > 10 ÷ 5

If x < y, and z < 0 then x ÷ z > y ÷ z

Example: Suppose 4 < 8, then 4 ÷ -2 > 8 ÷ -2 ( -2 > -4 ) 

If x > y, and z < 0 then x ÷ z < y ÷ z

Example: Suppose 5 > 1, then 5 ÷ -1 < 1 ÷ -1 ( -5 < -1 ) 

Transitive property:

If x > y and y > z, then x > z

Example: Suppose 10 > 5 and 5 > 2, then 10 > 2

x < y and y < z, then x < z

5 < 10 and 10 < 20, then 5 < 20

Comparison property:

If x = y + z and z > 0 then x > y

Example: 6 = 4 + 2, then 6 > 4

The properties of inequality are more complicated to understand than the property of equality. 

Properties of exponents

We will show 8 properties of exponents. Let x and y be a number not equal to zero and let n and m be any integers 

Property #1

x⁰ = 1 

Example: 4⁰ = 1 and (2500000000000000000000)⁰ = 1 

Property #2

xⁿ × x^m = x^{n + m}

Example: 4⁶ × 4⁵ = 4^{6 + 5} = 4¹¹ 

Property #3

xⁿ ÷ x^m = x^{n − m}

Example: 4⁶ ÷ 4⁵ = 4^{6 − 5} = 4¹ 

Property #4

(xⁿ)^m = x^{n × m}

(5²)⁴ = 5^{2 × 4} = 5⁸

Property #5

(x × y)ⁿ = xⁿ × yⁿ 

(6 × 7)⁵ = 6⁵ × 7⁵ 

Property #6

x^-n = 1 ÷(xⁿ) = 1/(xⁿ)

8^-4 = 1 ÷ (8⁴) = 1 / (8⁴)

Property #7

(x/y)ⁿ = xⁿ / yⁿ

(8/5)⁴ = 8⁴ / 5⁴

Property #8

Multiplying by powers of ten

Follow the following shortcut when multiplying by powers of ten

Whole numbers multiplied by powers of 10

When multiplying a whole number by a power of ten, just count how many zero you have and attached that to the whole number

Examples:

1) 56 × 10

There is only one zero, so 56 × 10 = 560

2) 45 × 10,000

There are 4 zeros, so 45 × 10000 = 450000

3) 18 × 10,000,000

There are 7 zeros, so 18 × 10,000,000 = 180,000,000

Decimals multiplied by powers of 10

When multipying a decimal by a positive power of ten (positive exponent), move the decimal point one place to the right for each zero you see after the 1

Examples:

1) 0.56 × 10

There is only one zero, so move the decimal point one place to the right.

0.56 × 10 = 5.6

2) 0.56 × 100

There are 2 zeros, so move the decimal point two places to the right

0.56 × 100 = 56

3) 0.056 × 1000

There are three zeros, so move the decimal point 3 places to the right.

0.056 × 1000 = 56

4) 0.056 × 100,000

0.056 × 100,000 = 0.056 × 1000 × 100 = 56 × 100 = 5600

When multipying a decimal by a negative power of ten (negative exponent), move the decimal point one place to the left for each zero you see before the 1

Note that 0.1 = 10^-1, 0.01 = 10^-2, 0.001 = 10^-3, and so forth....

We call 10^-1, 10^-2, and 10^-3 negative powers of 10 because the exponents are negative

Examples:

1) 56 × 0.1

There is only one zero, so move the decimal point one place to the left.

56 × 0.1 = 5.6

2) 560 × 0.01

There are 2 zeros, so move the decimal point two places to the left

560 × 0.01 = 5.6

2) 560 × 0.001

There are 3 zeros, so move the decimal point two places to the left

560 × 0.001 = 0.560

3) 0.56 × 0.1

There is only one zero, so move the decimal point one place to the left.

0.56 × 0.1 = 0.056

4) 0.56 × 0.01

There are 2 zeros, so move the decimal point two places to the left

0.56 × 0.01 = 0.0056

Any questions about multiplying by powers of ten? Let me know...

Multiplication by 11

To understand the multiplication by 11 trick, take a look at the following multiplication below:

Did you make the following two important observations?

First, notice that the digits of the number that is multiplied by 11 appear again in the answer

Second, the number in the middle of the answer is always found by adding the digits of the number that is multiplied by 11

This is the basic facts that you have to remember to quickly multiply numbers by 11

Therefore, if you want to use this trick, all you have to do is add the digits of the number and put the answer between the number

Other examples

1) 35 × 11

Just add 3 and 5 to get 8

Put 8 between 3 and 5

The answer is 385

2) 18 × 11

Just add 1 and 8 to get 9

Put 9 between 1 and 8

The answer is 198

3) 43 × 11

Just add 4 and 3 to get 7

Put 7 between 4 and 3

The answer is 473


Now, look at this example 48 × 11
Add 4 and 8 to get 12

Just put 12 between 4 and 8???

No, it does not work like this!

You can still put 2 in between. However, since the number is bigger than 9, you have to carry the 10 represented with a 1 over the 4

1 + 4 = 5. Thus the answer is 528

4) 85 × 11

Just add 8 and 5 to get 13

Add 1 to 8 to get 9

Put 3 between 9 and 5

The answer is 935


5) 78 × 11

Just add 7 and 8 to get 15

Add 1 to 7 to get 8

Put 5 between 8 and 8

The answer is 858

Squaring any two digit number ending in five

In this lesson, we will show you how straightforward squaring any two digit number ending in five can be 


First look at the following multiplication

Did you make the following important observation?

The answer always ends in 25. This will be the case when you square any two digit number ending with 5

Now, how did we get the number(s) before 25? Easy!

Look for the number in the tens place.For the multiplications above, these are 2, 4, and 1

Multiply each number by its next higher digit

So,

2 × 3 = 6

4 × 5 = 20

1 × 2 = 2

Put these numbers on the left of 25

Other examples

1) 35 × 35

The digit in the tens place is 3

Multiply 3 by its next higher digit, which is 4

3 × 4 = 12

Write down 12 and put 25 next to it

The answer is 1225

2) 55 × 55

The digit in the tens place is 5

Multiply 5 by its next higher digit, which is 6

5 × 6 = 30

Write down 30 and put 25 next to it

The answer is 3025

3) 95 × 95

The digit in the tens place is 9

Multiply 9 by its next higher digit, which is 10

9 × 10 = 90

Write down 90 and put 25 next to it

The answer is 9025

Important note:

Before you can really apply these tricks, you must know your multiplication table by heart

Finger multiplication

The use of finger multiplication has been widespread through the years. It is not the traditional way of doing multiplication in school today. 

However, if your kid has trouble remembering the whole multiplication table, multiplication with fingers is a good alternative

To multiply with fingers, you are only required to remember the multiplication table up to 5 × 5.

After that, all multiplication can be performed with your fingers

Here is the technique:

The two numbers to be multiplied are each represented on a different hand

Each hand may have some raised fingers and some closed fingers at the same time

Number of fingers to raise = Factor − 5. Remember that a factor is a number in a multiplication problem

The sum of the raised fingers is the number of tens

The product of the closed fingers is the number of ones

Example #1:

7 × 8 

For 7, use your left hand and raise 2 fingers (Factor − 5 = 7 − 5 = 2) . This means there are 3 closed fingers

For 8, use your right hand and raise 3 fingers. This means that there are 2 closed fingers

Sum of raised fingers = 2 + 3 = 5. This means we have 5 tens or 50

Product of closed fingers = 3 × 2 = 6. This means that we have 6 ones

50 + 6 = 56 

Example #2:

9 × 6 

For 9, use your right hand and raise 4 fingers. This means there is 1 closed finger

For 6, use your left hand and raise 1 finger. This means that there are 4 closed fingers

Sum of raised fingers = 4 + 1 = 5. This means we have 5 tens or 50

Product of closed fingers = 1 × 4 = 4. This means that we have 4 ones

50 + 4 = 54

Example #3:

8 × 5 

For 8, use your left hand and raise 3 fingers. This means there are 2 closed fingers

For 5, use your left hand and raise no finger. This means that there are 5 closed fingers

Sum of raised fingers = 3 + 0 = 3. This means we have 3 tens or 30

Product of closed fingers = 2 × 5 = 10. This means that we have 10 ones or 1 ten or 10

30 + 10 = 40

Feel free to use the following multiplication calculator as you practice your finger multiplication to check your answer

Number trick with 1089

The number trick with 1089 has been around for centuries. To impress someone with this trick, he or she will need paper and pencils:


Here is how it goes:

Step #1:

Have the person write down any three digits number with decreasing digits (432 or 875).

Step #2:

Reverse the number you wrote in step #1. 

Step #3:

Subtract the number obtained in step #2 from the number you wrote in step #1

Step #4:

Reverse the number obtained in step #3

Step #5:

Add the numbers found in step #3 and step #4


Example #1:


Step #1:

Have the person write down any three digits number with decreasing digits: 752

Step #2:

Reverse the number you wrote in step #1: 257

Step #3:

Subtract the number obtained in step #2 from the number you wrote in step #1: 752 - 257 = 495

Step #4:

Reverse the number obtained in step #3: 594

Step #5:

Add the numbers found in step #3 and step #4: 495 + 594 = 1089


Example #2:


Step #1:

Have the person write down any three digits number with decreasing digits: 983

Step #2:

Reverse the number you wrote in step #1: 389

Step #3:

Subtract the number obtained in step #2 from the number you wrote in step #1: 983 - 389 = 594

Step #4:

Reverse the number obtained in step #3: 495

Step #5:

Add the numbers found in step #3 and step #4: 594 + 495 = 1089


Example #3:


Step #1:

Have the person write down any three digits number with decreasing digits: 210

Step #2:

Reverse the number you wrote in step #1: 012

Step #3:

Subtract the number obtained in step #2 from the number you wrote in step #1: 210 - 012 = 198

Step #4:

Reverse the number obtained in step #3: 891

Step #5:

Add the numbers found in step #3 and step #4: 198 + 891 = 1089

This number trick with 1089 works with any 3 digits number as long as you choose a number with decreasing digits in step #1

Basic geometry

Point: A point is a location in space. It is represented by a dot. Point are usually named with a upper case letter. For example, we refer to the following as "point A"

Line: A line is a collection of points that extend forever. The following is a line. The two arrows are used to show that it extends forever.

We put two points in order to name the line as line AF. However, there are an infinite amount of points. You can also name it line FA

Line segment: A line segment is part of a line. The following is a segment. A segment has two endpoints. The endpoints in the following segments are A and F. Notice also that the line above has no endpoints.

Ray: A ray is a collection of points that begin at one point (an endpoint) and extend forever on one direction. The following is a ray.

Angle: Two rays with the same endpoint is an angle. The following is an angle.

Plane: A plane is a flat surface like a piece of paper. It extends in all directions. We can use arrows to show that it extends in all directions forever. The following is a plane

Parallel lines When two lines never meet in space or on a plane no matter how long we extend them, we say that they are parallel lines The following lines are parallel.

Intersecting lines: When lines meet in space or on a plane, we say that they areintersecting lines The following are intersecting lines.



Vertex: The point where two rays meet is called a vertex. In the angle above, point A is a vertex.

Common geometry formulas

Here, we provide you with common geometry formulas for some basic shapes

---

Rectangle:

Perimeter = l + l + w + w = 2 × l + 2 × w 

Area = l × w

---

Square:

Perimeter = s + s + s + s = 4 × s

Area = s²

---

Parallelogram:

Perimeter = a + a + b + b = 2 × a + 2 × b 

Area = b × h

---

Rhombus:

Perimeter = b + b + b + b = 4 × b 

Area = b × h

---

Triangle:

Perimeter = a + b + c

Area = (b × h)/2 

---

Trapezoid:

Perimeter = a + b + c + d

---

Circle:

Perimeter = 2 × pi × r or Perimeter = pi × d

Area = pi × r² or Area = (pi × d²)/4 

The circle

An easier way to define such geometric figure is to say that it is a closed curved line with all points equally distant from the center.

The following shows an example:



Real life examples bicycle wheels, coins, such as dimes and pennies, CDs, and MP3 players.

A line segment joining two points on the figure is a chord.The following are examples of two chords



When a chord passes through the center,we call it a diameter. A diameter usually divides such figure into two equal halves. Each half is called a semi-circle



Half a diameter is called a radius.

In other words, 2 radii= diameter

Definition of a polynomial

Before giving you the definition of a polynomial, it is important to provide the definition of a monomial

Definition of a monomial:

A monomial is a variable, a real number, or a multiplication of one or more variables and a real number with whole-number exponents

Examples of monomials and non-monomials

Monomials	9	x	9x	6xy	0.60x4y
Not monomials	y - 6	x-1 or 1/x	√(x) or x1/2	6 + x	a/x

Polynomial definition:

A polynomial is a monomial or the sum or difference of monomials. Each monomial is called a term of the poynomial

Important!:Terms are seperated by addition signs and subtraction signs, but never by multiplication signs

A polynomial with one term is called a monomial

A polynomial with two terms is called a binomial

A polynomial with three terms is called a trinomial


Examples of polynomials:

Polynomial	Number of terms	Some examples
Monomial	1	2, x, 5x3
Binomial	2	2x + 5, x2 - x, x - 5
Trinomial	3	x2 + 5x + 6, x5 - 3x + 8

Difference between a monomial and a polynomial:

A polynomial may have more than one variable.

For example, x + y and x² + 5y + 6 are still polynomials although they have two different variables x and y

By the same token, a monomial can have more than one variable. For example, 2 × x × y × z is a monomial



Exercices

For all expressions below, look for all expressions that are polynomials. 

For those that are polynomials, state whether the polynomial is a monomial, a binomial, or a trinomial

1) 3.4 + 3.4x

2) z² + 5z^-1 + 6

3) -8

4) 2c² + 5b + 6

5) 14 + x

6) 5x - 2^-1

7) 4 b² - 2 b^-2

8) f² + 5f + 6

Answer: 1), 3), 4), 5), 6), and 8) are polynomials. 1), 5), and 6) are binomials. 3) is a monomial. 4) and 8) are trinomials

2) and 7) are not because they have negative exponents

Notice that 6) is still a polynomial although it has a negative exponent. It is because it is the exponent of a real number, not a variable

In fact, 5x - 2^-1 = 5x + 1/2 = 5x + 0.5

Graphing

Graphing equalities is probably is the easiest thing to do. If you are familiar with a number line, you should have no problems.

An equation is a mathematical sentence with an equal sign. Equations such as 2 + 6 = 8 and x = 2 are examples of equalities

equalities such as x = 2 are what we will be dealing with.

Simply locate on the number line the value of x or any other variables used for the equality and make a shaded circle at the value of x

Example #1

x = 3

Just locate 3 on the number line and draw a shaded circle at 3



Example #2

x = -5

Just locate -5 on the number line and draw a shaded circle at -5



Example #3

x = 6

Just locate 6 on the number line and draw a shaded circle at 6



Example #4

x = -2

Simplifying algebraic expressions

Simplifying algebraic expressions with one or more variables by combining like terms is what this lesson will show you. 

We will show you first how to simplify numerical expressions. 

Let's say you are adding 5 + 5 + 5 + 5. What is a quick way to get the answer?

You can do 4 × 5 and 4 × 5 is a simplifed version of 5 + 5 + 5 + 5

If you had the following problem to add 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5, you can quickly see why it would be useful to just count how many fives there are and then times the amount by 5.

Since there are 11 fives, the simplified expression is 11 × 5

What does this have to do with simplifying algebraic expressions?

Try to simplify v + v + v + v + v + v

Well, in the same manner, just count how many v's there are and multiply the amount by v

Since there are 6 of them, v + v + v + v + v + v = 6 × v

To simplify 6 × v even further, you can write 6v. However, remember that there is a multiplication between 6 and v

Notice that putting a 1 next to each v will not change the answer.

v + v + v + v + v + v = 1v + 1v + 1v + 1v + 1v + 1v = ( 1 + 1 + 1 + 1 + 1 + 1)v = 6v

This shows you a couple of things. First, 1v = v. Then, when simplifying algebraic expressions, you can just do your math with the number on the left on the variable

Now, try to add v + v + v + v + x + x + x

I hope it makes sense to you that you cannot add v and x. What is v + x? You cannot say 2v because it is v + v that is equal to 2v

You cannot say 2x because it is x + x that is equal to 2x. Basically v and x are not like terms, so v + x is just equal to v + x

By the same token, there is another thing you need to watch for. What is v + 2? It is not 2v as already shown

Again, v and 2 are not like terms either, so v + 2 = v + 2 

However, since you can add all the v's, we call the v's like terms. Same for the x's, they are called like terms

v + v + v + v + x + x + x = ( v + v + v + v) + ( x + x + x) = 4 × v + 3 × x = 4v + 3x

What about 4v + 3x + 5v + 4x? 

You cannot add 4v to 3x, but you can add 4v to 5v. And you can add 3x to 4x

4v + 3x + 5v + 4x = 4v + 5v + 3x + 4x = 9v + 7x

When simplifying algebraic expressions, subtraction and/or putting negative numbers work pretty much the same. You can only subtract like terms. The only difficulty you will encounter is the subtraction or addition of integers.

Do 4v - 3x - 2v + 4x

I can do 4v with 2v. However, be very careful! The operation on the left of 2v is subtraction, so you have to do 4v - 2v = 2v

In the same way, I can do 3x with 4x, but the operation next to 3x is subtraction. You have to change the subtraction to + -

Then, you will do -3x + 4x = 1x. I hope you have mastered how to add integers

4v - 3x - 2v + 4x = 2v + 1x

Useful guidelines when simplifying algebraic expressions:

Step 1: Look for like terms

Step 2: Watch out for minus or negative sign next to variables (on the left of the variables) 

Step 3: When you move this term around, you got to move it with the minus or negative sign as well 

Step 4: Add or subtract only the integers on the left of the like terms.

You are ready to do some challenging problems

Example #1

9x + 4 + 15v - 7x + 8 - 10v 

Put like terms together

9x + 4 + 15v - 7x + 8 - 10v = 9x - 7x + 15v - 10v + 4 + 8 = 2x + 5v + 12

Example #2

-5 - 6x - 5x + 10y - 4x + 1 + 10x + y 

Put like terms together

-5 - 6x - 5x + 10y - 4x + 1 + 10x + y = -6x - 5x - 4x + 10x + 10y + y - 5 + 1 

Just add all the numbers next to x, called coefficients.

-6 - 5 - 4 + 10 = -11 - 4 + 10 = -15 + 10 = -5

Then, add all the numbers next to y.

10 + 1 = 11

-5 - 6x - 5x + 10y - 4x + 1 + 10x + y = -5x + 11y + -4

More on like terms when simplifying algebraic expressions

When simplifying algebraic expressions, as long as you are adding the same thing, you have like terms even if you see more than one variable in each term

For example what is xy + xy + xy + xy + xy + xy?

How many xy's do you see? Since there are 6 of them, xy + xy + xy + xy + xy + xy = 6xy

We do have a multiplication between x and y, but it does ot change anything

In the same way, 8xyzf + 12xyzf + xyzf + xyzf + xyzf = 20xyzf + 3xyzf = 23xyzf

Algebra proofs

Many algebra proofs are done using proof by mathematical induction. To demonstrate the power of mathematical induction, we shall prove an algebraic equation and a geometric formula with induction. 

If you are not familiar with with proofs using induction, carefully study proof by mathematical induction given as a reference above. Otherwise, you could struggle with these algebra proofs below

Algebra equation:

Prove by mathematical induction that 1 + 2 + 4 + 8 + ... + 2^n-1 = 2ⁿ - 1 

Step # 1: 

Show that the equation is true for n = 2. n = 2 means adding the first two terms

1 + 2 = 3 and 2² - 1 = 4 - 1 = 3. So, it is true for n =2

Just for fun, let's show it is true also for n = 4. n =4 means adding the first 4 terms

1 + 2 + 4 + 8 = 15 and 2⁴ - 1 = 16 - 1 = 15. So, it is true also for n =4

Step # 2: 

Suppose it is true for n = k

Just replace n by k

1 + 2 + 4 + 8 + ... + 2^k-1 = 2^k - 1

Step # 3: 

Prove it is true for n = k + 1

You need to write down what it means for the equation to be true for n = k + 1

Caution: Writing down what it means is not the same as proving the equation is true. In fact, it just shows you what you need to prove

Here is what it means for n = k + 1: 

After you replace k by k+1, you get :

1 + 2 + 4 + 8 + ... + 2^{k + 1 -1} = 2^{k + 1} - 1

1 + 2 + 4 + 8 + ... + 2^k = 2^{k + 1} - 1

You can now complete the proof by using the hypothesis in step # 2 and then show that 

1 + 2 + 4 + 8 + ... + 2^k = 2^{k + 1} - 1

starting with the hypothesis, 1 + 2 + 4 + 8 + ... + 2^k-1 = 2^k - 1

ask yourself, " What does the next term look like? "

Since the last term now is 2^k-1, the next term should be 2^{k + 1 -1} = 2^k after replacing k by k + 1

Add 2^k to both sides of the hypothesis

1 + 2 + 4 + 8 + ... + 2^k-1 + 2^k = 2^k - 1 + 2^k 

The trick here is to see that 2^k + 2^k = 2 × 2^k = 2¹× 2^k = 2^{k + 1}

1 + 2 + 4 + 8 + ... + 2^k-1 + 2^k = 2^k - 1 + 2^k 

                                             = 2 × 2^k -1

                                             = 2¹× 2^k -1 

                                             = 2^{k + 1} -1 

Geometric formula:

Show by mathematical induction that 

the sum of the angles in an n-gon = ( n - 2 ) × 180^°

A couple of good observations before we prove it:

Observation #1:

An n-gon is a closed figure with n sides. For example, 

an n-gon with 4 sides is called a quadrilateral

an n-gon with 3 sides is called a triangle

Observation #2:

After a close examination of the figure above, can you see that every time you add a side, you are also adding one more triangle

I am now ready to show the proof.

Step # 1: 

Show that the equation is true for n = 3. Notice that n cannot be smaller than 3 since we cannot make a closed figure with just two sides or one side.

When n = 3, we get a triangle and the sum of the angles in a triangle is equal to 180^°

When n = 3, ( 3 - 2 ) × 180^° = 1 × 180^° = 180^°

When n = 4, you are adding one more triangle to get two triangles and the sum of the angles of the two triangles is equal to 360^°

When n = 4, ( 4 - 2 ) × 180^° = 2 × 180^° = 360^°

Thus, the formula is true for n = 3 and n = 4

Step # 2: 

Suppose it is true for n = k

Just replace n by k

The sum of the angles in a k-gon = ( k - 2 ) × 180^°

Step # 3: 

Prove it is true for n = k + 1

You need to write down what it means for the equation to be true for n = k + 1

Here is what it means for n = k + 1: 

After you replace k by k+1, you get : 

The sum of the angles in a (k+1)-gon = ( k +1 - 2 ) × 180^°

The sum of the angles in a (k+1)-gon = ( k - 1 ) × 180^°

You can now complete the proof by using the hypothesis in step # 2 and then show that 

The sum of the angles in a (k+1)-gon = ( k - 1 ) × 180^°

starting with the hypothesis, the sum of the angles in a k-gon = ( k - 2 ) × 180^°

ask yourself, " What does the next term look like? "

Since the last term now is k-gon or a figure with k sides, the next term should be a figure with k + 1 sides after replacing k by k + 1

Now, what are we adding to both sides?

First, recall the meaning of adding one side. It means that you will be adding also one triangle

( k + 1) gon = k-gon + ( 1 side or one more triangle)

( k + 1) gon = k-gon + one more triangle

( k + 1) gon = k-gon + 180^°

So add 180^° to both sides of the hypothesis

The sum of the angles in a k-gon + 180^° = ( k - 2 ) × 180^° + 180^°

The sum of the angles in a (k+1)-gon = ( k - 2 ) × 180^° + 180^°

                                             = 180^°k + -2 × 180^° + 180^°

                                             = 180^°k + -1 × 180^° 

                                             = 180^° ( k + -1 ) 

Proof of the quadratic formula

The following is a proof of the quadratic formula. It will show you how the quadratic formula that is widely used was developed. 

The proof is done using the standard form of a quadratic equation and solving the standard form by completing the square 

Start with the the standard form of a quadratic equation: 

ax² + bx + c = 0

Divide both sides of the equation by a so you can complete the square






Subtract c/a from both sides






Complete the square:

The coefficient of the second term is b/a

Divide this coefficient by 2 and square the result to get (b/2a)² 

Add (b/2a)² to both sides:




Since the left side of the equation right above is a perfect square, you can factor the left side by using the coefficient of the first term (x) and the base of the last term(b/2a) 

Add these two and raise everything to the second.

Then, square the right side to get (b²)/(4a²)




Get the same denominator on the right side:






Now, take the square root of each side:





Simplify the left side:








Rewrite the right side:





Subtract b/2a from both sides:








Adding the numerator and keeping the same denominator, we get the quadratic formula:





The + - between the b and the square root sign means plus or negative. In other words, most of the time, you will get two answers when using the quadratic formula.

Rational numbers

They have the form a/b with a and b are integers and b not equal to zero.

Examples of numbers that are rational are:

2/3     5/2    1/4    2     -8/2    0

Now, why are 2 and 0 examples of such numbers?

It is because 2 and 0 can be written as 2/1 and 0/1

We can also write rational numbers as a decimals.

We do this by performing a quick division that is dividing the numerator by the denominator

For instance,

2/5 = 0.4    or   

2/5 = 0.4000000 because you can continue division to keep getting zeros for the decimal places after 4 



The bar on top of 0 means that if we continue to perform long division, we will keep getting an infinite number of zeros. Another way to convert 2/5 into a decimal is to notice that we can multiply 5 by 20 and 2 by 20 to get 40/100 and 40/100 = 0.40.

Diving by 100 or any other power of 10 is a straightforward process.

If you are dividing by 10, just move the decimal point one place to the left.

If you are dividing by 100, just move the decimal point 2 places to the left.

and so forth...

For 40/100, the decimal point is after 0 for 40.

Moving that two places to the left bring the decimal point right before the 4.

Exponents

Exponents can make your math problems a lot easier to handle. Simply put, it is a shortcut for multiplying numbers over and over again. 

Look at the following multiplication problem:

8 × 8 × 8 × 8 × 8 × 8 

Instead of multiplying 8 six times,we can just write 8⁶ and it will mean the same thing.

When reading 8⁶, we say eight to the sixth power or eight to the power of six.

Other examples:

5³ = 5 × 5 × 5

9⁴ = 9 × 9 × 9 × 9

7² = 7 × 7

6⁶ = 6 × 6 × 6 × 6 × 6 × 6 × 6

2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

In general,

yⁿ = y × y × y × y ×...× y × y  (n times) 

Sequences and patterns

Say for instance you go to the bank to deposit money and the bank gives you the following two options to choose from:

Option A Deposit 1000 dollars.

The second day you receive 1100

The third day you receive 1200

The fourth day your receive 1300

And so forth....

Option B Deposit 1 dollar.

The second day you receive 3

The third day you receive 9

The fourth day your receive 27

And so forth....

Which option gives you more money in 10 days?

At first, the tendency is say that option A is the best option

However, let us take a look and see what is going on here

If you choose option A, 

The fifth day you receive 1400

The sixth day you receive 1500

The seventh day your receive 1600

The eighth day you receive 1700

The ninth day you receive 1800

The tenth day your receive 1900

On the other hand, if you choose option B, 

The fifth day you receive 81

The sixth day you receive 243

The seventh day your receive 729

The eighth day you receive 2187

The ninth day you receive 6561

The tenth day your receive 19683

No doubt now you can see clearly that option B is the best option.

Notice that in option A, to get to the next number, just add 100 every time

We call this pattern an arithmetic sequence. 

In option B, to get to the next number, just multiply by 3 every time

We call this pattern a geometric sequence. 

The reason the money grew so fast in option B is because the pattern is an exponential growth, which usually grows fast.

Thw exponential growth above can be modeled with an exponential function

The exponential function is 3ⁿ

when n = 1, 3¹= 3

when n = 2, 3²= 9

Fibonacci sequence

The Fibonacci sequence is a naturally occuring phenomena in nature. It was discovered by Leonardo Fibonacci. 

Leonardo was an Italian mathematician who lived from about 1180 to about 1250 CE. Mathematicians today are still finding interesting way this series of numbers describes nature

To see how this sequence decribes nature, take a close look at the figure below:

This spiral shape is found in many flowers, pinecones, and snails' shell to mention just a few

What exactly is happening here as far as math is concerned?


You can see that we begin with two squares with a side length that is equal to 1

Then, to get the side length of the third square, we add the side lengths of the two previous squares that is 1 and 1 ( 1 + 1 = 2)

To get the side length of a fourth square, we add 1 and 2 ( 1 + 2 = 3)

To get the side length of a fifth square, we add 2 and 3 ( 2 + 3 = 5)

If we continue this pattern we get:

3 + 5 = 8

5 + 8 = 13

8 + 13 = 21

13 + 21 = 34

21 + 34 = 55

34 + 55 = 89 

55 + 89 = 144

89 + 144 = 233


Here is a short list of the fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233

Each number in the sequence is the sum of the two numbers before it

We can try to derive a Fibonacci sequence formula by making some observations

F₁ = 1

F₂ = 1

F₃ = F₂ + F₁ = 1 + 1 = 2

F₄ = F₃ + F₂ = 2 + 1 = 3

F₅ = F₄ + F₃ = 3 + 2 = 5

F₆ = F_6-1 + F_6-2 = F₅ + F₄ = 5 + 3 = 8 

F₇ = F_7-1 + F_7-2 = 8 + 5 = 13 

......

......

......

F_n = F_n-1 + F_n-2 

Try this

Find the sum of the first ten terms of the fibonacci sequence

1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143

Now, we will choose numbers other than 1 and 1 to create other Fibonacci-like sequences

2, 2 , 4, 6, 10, 16, 26, 42, 68, 110

The sum is 2 + 2 + 4 + 6 + 10 + 16 + 26 + 42 + 68 + 110 = 286

what if we start with 3 and 3?

3, 3, 6, 9, 15, 24, 39, 63, 102, 165

3 + 3 + 6 + 9 + 15 + 24 + 39 + 63 + 102 + 165 = 429

Now, we shall make a nice observation?

143/11 = 13

286/11 = 26

429/11 = 39

143 = 11 × 13 = 11 × 13 × 1 

286 = 11 × 26 = 11 × 13 × 2

429 = 11 × 39 = 11 × 13 × 3 

You can thus see that the sum of the first 10 terms follow this pattern

11× 13 × 1 

11× 13 × 2 

11× 13 × 3 

11× 13 × 4 

......

......

......

11× 13 × 4 

11 × 13 × n 

Just remember that n = 1 is the Fibonacci sequence starting with 1 and 1

n = 2 is the one starting with 2 and 2

Understanding probability

Permutations

Permutations are ordered arrangement of objects

Objects stand for anything you are trying to arrange or put in a certain order.

Examples of arrangements are:

1)You have 5 CDs. How many different ways can you listen to them?

2)You are seating 3 people on 3 chairs. How many different ways can people sit?

3)You have 6 books to read.How many different ways can you read your books?

You can use the fundamental counting principle to find out how many different permutations or arrangements you can have.

Before solving the situations above, let us start with this example.

and 4 boxes to arrange those books.

You want to know how many different arrangements are possible.

Since you have 4 boxes, you have 4 choices to put the first book. You can put the first book in any of those boxes. The important thing is to see that you have 4 choices to do this.

Say you put algebra in box #1 

After you put the algebra book in box #1, you can no longer use box #1. You can use box #2, box #3, or box #4 to place your next book.

This means you have 3 choices to place your next book. Say that you put basic math in box #2

After you put the basic math book in box #2, you can no longer use box #2. You can use box #3 or box #4 to place your next book. 

Say that you put the algebra 1 book in box #3

After you put the algebra 1 book in box #3, you have only 1 choice ( box #4) to place your last book, which is geometry

Now to find the total number of arrangement, you need to multiply all your choices as seen in fundamental counting principle

Therefore,

Number of permutations = 4 × 3 × 2 × 1 = 24

Now,let us solve the situations above.

First, how many ways can 3 people sit on 3 chairs?

The first person has 3 choices

once the first person sits, there are only 2 seats left.

Thus, the second person who sits has 2 choices

The last person has 1 choice

The number of ways people can sit = 3 × 2 × 1 = 6 ways


Second, How many different ways can you listen to five CDs?

Assuming you do not listen twice to a CD, you have 5 choices to listen to the first CD.

4 choices to listen to the second CD

3 choices to listen to the third CD

2 choices to listen to the fifth CD

1 choice for the last CD

So you have 5 × 4 × 3 × 2 × 1 = 120 different ways to listen to the 5 CDs

Theoretical probability

The theoretical probability is found whenever you make use of a formula to find the probability of an event.

To find the probability of an event, also called likelihood of an event, use the formula below:

probability of an event = 

number of favorable outcomesnumber of possible outcomes

The number of favorable outcomes is the likelihood to get a specific outcome

For example, suppose you throw a die numbered from 1 to 6.

Count all the possible numbers you can get. This is called number of possible outcomes. 

All the possible numbers are 1, 2, 3, 4, 5, and 6. Thus the number of possible outcomes is 6 

You could make up different types of favorable outcomes

You could say...

a. Likelihood to get an even number

b. Likelihood to get a prime number

c. Likelihood to get an odd number

d. Likelihood to get a 4.

e. Likelihood to get a 1.

f. Likelihood to get a number bigger than 4

g. Likelihood to geta number less than 6

All the above are favorable outcomes


Exercise #1


Throw a die once. What is the probability of getting a number less than 6?

Ask yourself, "How many number are less than 6?" 

Since there are 5 numbers less than 6, the number of favorable outcomes is 5

Since the die had a total of 6 numbers, the number of possible outcomes is 6

probability of getting a number less than 6 = 

number of favorable outcomesnumber of possible outcomes

probability of getting a number less than 6 = 

56

probability of getting a number less than 6 = 0.8333 

probability of getting a number less than 6 = 83.33% 

probability of getting a yellow ball = 0.3333 

This means that it is very likely you will get a number less than 6 


Exercise #2


A bag contains 6 blue balls, 4 yellow balls, and 2 red balls

What is the theoretical probability of getting a yellow ball?

Since you have 4 yellow balls playing on your favor, the number of favorable outcomes is 4

To get the number of possible outcomes, just count all the balls. Number of possible outcomes is 12 

probability of getting a yellow ball = 

412

probability of getting a yellow ball = 0.3333

probability of getting a yellow ball = 33.33%

Probability of compound events

The probability of compound events combines at least two simple events. The probability that a coin will show head when you toss only one coin is a simple event

However, if you toss two coins, the probability of getting 2 heads is a compound events because once again it combines two simple events

Suppose you say to a friend, " I will give you 10 dollars if both coins land on head."

Let's see what happens when your friend toss two coins:

The different outcomes are HH, HT, TH, or TT. 

As you can see, out of 4 possibilities, only 1 will give you HH. 

Therefore, the probability of getting 2 heads is 

14

Your friend has 25% chance of getting 10 dollars since one-fourth = 25%.

The example above is a good example of independent events. What are independent events?

When the outcome of one event does not affect the outcome of another event, the two events are said to be independent.

In our example above, when you toss two coins, neither coin has the power to influence the other coin.

This compound events is independent then. When two events are independent, you can use the following formula

probability(A and B) = probability(A) × probability(B)

Let's use this formula to find the probability of getting 2 heads when two coins are tossed

probability(Heads and Heads) = probability(Heads) × probability(Heads)

Coin #1: 

Probability of getting head = 

12

Coin #2: 

Probability of getting another head = 

12

probability(Heads and Heads) = probability(Heads) × probability(Heads)

probability(Heads and Heads) =

12

×

12

probability(Heads and Heads) = 

1 × 12 × 2

probability(Heads and Heads) = 

14

Sometimes, compound events can be dependent. What are dependents events?

When the outcome of one event has the power to affect the outcome of another event, the two events are said to be dependent.

When two events are dependent, you can use the following formula

probability(A and B) = probability(A) × probability(B given A)

Soppose, a bag has 4 red balls and 6 blue balls. What is the probability of choosing 2 blue balls at random?

These events are dependent since after you choose one blue ball, it changes the number of blue balls and the number of balls all together

Blue ball #1: 

Probability of getting a blue ball = 

610

Blue ball #2: Now there are 5 blue balls and 9 balls all together 

Probability of getting another blue ball = 

59

probability(Blue and Blue) = probability(Blue) × probability(Blue)

probability(Blue and Blue) =

610

×

59

Probability of getting blue and blue = 

6 × 510 × 9

Probability of getting blue and blue = 

3090

Probability of getting blue and blue = 

13

You have 33.33% chance of doing this since 1/3 is equal to 33.33%

Lastly, sometimes, as opposed to having two events happening at the same time, you may need to choose between two events.

When two events cannot both occur, they are called mutually exclusive events.

To find the probability of compound events when the events are mutually exclusive, use the formula:

probability (A or B) = probability (A) + probability (B)

Suppose you and your brother both throw a die. Whoever get a 4 wins!

These are mutually exclusive events because you cannot both win this game.

probability (you win or your brother wins) = probability (you win) + probability (your brother wins)

You: 

Probability you win = 

16

Your brother: 

Probability your brother wins = 

16

probability(you win or your brother wins) = probability(you win) + probability(your brother wins)

probability(you win or your brother wins) =

16

+

16

probability(you win or your brother win) = 

1 + 16

probability(you win or your brother wins) = 

26

Coin toss probability

Coin toss probability is explored here with simulation.

When asked the question, what is the probability of a coin toss coming up heads, most people answer without hesitation that it is 50%, 1/2, or 0.5

we get this probability by assuming that the coin is fair, or heads and tails are equally likely

The probability for equally likely outcomes is:

Number of outcomes in the event ÷ Total number of possible outcomes

For the coin, number of outcomes to get heads = 1 

Total number of possible outcomes = 2

Thus, we get 1/2

However, if you suspect that the coin may not be fair, you can toss the coin a large number of times and count the number of heads

Suppose you flip the coin 100 and get 60 heads, then you know the best estimate to get head is 60/100 = 0.6

This way of looking at probability is called the relative frequency estimate of a probability

The interesting thing with this is that the more you flip the coin, the closer you get to 0.5

If you have a computer, you can simulate coin toss probability with different numbers of coin tosses, the result might be a table like this. 

Number of tosses	Number of heads	Probability to get heads
4	1	0.25
100	56	0.56
1000	510	0.510
10000	4988	0.4988

Notice that for 10000 flip, the probability is close to 0.5

Try the same experiment to get the coin toss probability with the following coin flip simulation.

After you have flipped the coin so many times, you should get answers close to 0.5 for both heads and tails 

Fundamental counting principle

We will introduce the fundamental counting principle with an example. 

This counting principle is all about choices we might make given many possibilities.

Suppose most of your clothes are dirty and you are left with 2 pants and 3 shirts.

How many choices do you have or how many different ways can you dress?

Let's call the pants: pants #1 and pants #2 

Let's call the shirts: shirt #1, shirt #2, and shirt #3

Then, a tree diagram as the one below can be used to show all the choices you can make

As you can see on the diagram, you can wear pants #1 with shirt # 1. That's one of your choices.

Count all the branches to see how many choices you have.

Since you have six branches, you have 6 choices.

However, notice that a quick multiplication of 2 × 3 will yield the same answer.

In general, if you have n choices for a first task and m choices for a second task, you have n × m choices for both tasks

In the example above, you have 2 choices for pants and 3 choices for shirts. Thus, you have 2 × 3 choices.

Another example:

You go a restaurant to get some breakfast. The menu says pancakes, waffles, or home fries. And for drink, coffee, juice, hot chocolate, and tea. How many different choices of food and drink do you have?

There 3 choices for food and 4 choices for drink.

Thus, you have a total of 3 × 4 = 12 choices.

Finding the average

Let us illustrate with examples:

Example #1:

Get the average of the following set of numbers:

5,4,12,2,1,6

Step 1:

Find the sum:

Sum = 5+4+12+2+1+6= 30

Step 2:

Divide 30 by 6 (the total number of numbers)



30 divided by 6 is 5, so the average is 5

Example #2:

Get the average of the following set of numbers:

5,10,20,5,,10

Find the sum:

Sum = 5+10+20+5+10=50

Divide the 50 by 5 (the total number of numbers added)



50 divided by 5 is 10, so 10 is the average.

Finding the mode

When finding the mode of a set a data, first put the data in order from least to greatest although it is not necessary to do so.

Then, Look for the most occurring item(s) or number(s)

For example, 

Find the mode of the following set:

15, 14, 8, 15, 16

The mode is 15 because it shows up twice

Other examples:

8, 5, 8, 2, 1, 8, 4

The mode is 8 because it occurs 3 times.

Sometimes, sets may have more than one mode as the following sets show

2, 4, 3, 4, 5, 6, 7, 8, 10, 6

Put the set in order

2, 3, 4, 4, 5, 6, 6, 7, 8, 10

The set above has 2 modes because both 4 and 6 occurred twice.

10, 15, 13, 11, 13, 16, 13, 10, 15, 14, 15, 18, 11, 20, 9, 11, 12

After putting the set in order, we get:

9, 10, 10, 11, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 18, 20

Here, we have 3 mode because 11, 13, and 15 occurred 3 times.

Notice, and it is very important, that 10 is not a mode although it occurs twice.

It is because 10 is not among the most occurring numbers like 13, which occurred 3 times.

Finding the median

When finding the median of a set of data, first put the data in order and then find the number located right in the middle.

For example, find the median for the following set:

S₁ = {15, 14, 11}

Put the numbers in order

11, 14, 15

The median is 14 because it is in the middle

Other examples:

S₂ = {5, 3, 7, 2, 4}

Put the set in order

2, 3, 4, 5, 7

In the example above, the median is 4 because 4 is in the middle

When the number of numbers in the set is an odd number as in the two sets above, your median is right in the middle.

However, when the number of numbers in the set is an even number, you will end up with two numbers in the middle

In this case, just take the average of the numbers

Examples

S₃ = { 15, 14, 11,16}

Put S₃ in order

11, 14, 15, 16

The two values in the middle are 14 and 15

The average is (14+15)/2 = 29/2 = 14.5

So, the median is 14.5

S₄ = { 6, 2, 8, 9, 1, 10, 4, 12}

Put S₄ in order

Tips

When a set contains many numbers, cross out numbers as you put them in order to keep yourself organized

For example for S₄, put 1 in your new ordered list and then cross it out. Then, put 2 and cross it out...

1, 2, 4, 6, 8, 9, 10, 12

The two numbers in the middle are 6 and 8

(6 + 8)/2 is 7, so the median is 7

Box and whiskers plot

Before studying this lesson, you need to understand the median. Basically a box and whiskers plot looks like this:

Just like the name suggests, the rectangle you see is called a box. And the reason we call the two lines extending from the edge of the box whiskers is simply because they look like whiskers or mustache, especially mustache of a cat

The five points or dot that you see represents the followings starting from left to right

Lower extreme: the lowest or smallest value in a set of data

Lower quartile or first quartile: the median of all data below the median

Median or second quartile: the middle value of the set of data. If there are two values in the middle, the median is the average of the two values

Upper quartile or third quartile: the median of all data above the median

Upper extreme: The biggest value in the set

Example:

Construct a box and whiskers plot for the data set: {5, 2, 16, 9, 13, 7, 10}

First, you have to put the data set in order from greatest to least or from least to greatest

From least to greatest we get : 2    5    7    9    10     13    16

Since the smallest value in the set is 2, the lower extreme is 2

Since the greatest value in the set is 16, the upper extreme is 16

Now, look carefully at the set:  2    5    7    9    10     13    16

You can see that 9 is located right in the middle of the set of data

Therefore, 9 is the median

Now to get the lower quartile, you need all data before the median or 9

2    5    7    9    10     13    16

In bold right above we show all data before 9, so  2    5    7   

Since the value in the middle for the set   2    5    7    is 5, the lower quartile is 5

Finally, to get the upper quartile, you need all data after the median or 9

2    5    7    9    10     13    16

In bold right above we show all data after 9, so  10    13    16   

Since the value in the middle for the set   10    13    16    is 13, the upper quartile is 13

Now make a number line and graph above the number line 2, 5, 9, 13, and 16 with five dots: one dot will represent the median, one dot will represent each extreme, finally, one dot will represent each quartile.

Your graph should look like this after you are done

Draw a rectangle or box starting from the lower quartile to the upper quartile. Draw a vertical segment too to represent the median

Finally, draw horizontal segments or whiskers that connect all five dots together. 

The box and whiskers plot for {5, 2, 16, 9, 13, 7, 10} is :

Stem and leaf plot

A stem and leaf plot organizes data by showing the items in order using stems and leaves.

The leaf is the last digit on the right or the ones digits. The stem is the remaining digit or digits 

Examples:

For 12, 2 is the leaf and 1 is the stem. For 45.7, 7 is the leaf and 45 is the stem

Basically, a straight vertical line is used. Then, we put all stems on the left and all leaves on the right

We will now illustrate with examples

Example #1:

24, 10, 13, 2, 28, 34, 65, 67, 55, 34, 25, 59, 8, 39, 61

First, put this data in order

2, 6, 10, 13, 24, 25, 28, 34, 34, 39, 55, 59, 61, 65, 67

We will use 0, 1, 2, 3, 4, 5, and 6 as stems. The plot is displayed below:

Example #2:

This time, the data is already in order

104, 107, 112, 115, 115, 116, 123, 130, 134, 145, 147

We will use 10, 11, 12, 13, and 14 as stems. The plot is displayed below:

Sometimes, it is useful to show leaves on both sides of the stem. Say for instance you teach algebra in two different classes.

You may in this case want to compare performance for the classes to see which class performed better

Example #3:

Grade for class A: 60, 68, 70, 75, 84, 86, 90, 91, 92, 94, 94, 96, 100, 100

Grade for class B: 60, 60, 70, 71, 73, 73, 75, 76, 77, 84, 85, 86, 91, 92

The plot is displayed below:

A quick look at the graph and you will see that class A performed a lot better than class B

Class B has more scores in the 70s than class A

Class A has more scores in the 90s than class B

Types of graphs

In this unit, you will learn about pictographs, bar graphs,double bar graphs,line graphs, and circle graphs.

Let us start with pictographs:

A graph is a visual way to display information.

A pictograph is a graph that uses pictures or symbols to display information.

The pictures in a pictograph usually represent more than one item.

The following is a pictograph:

The graph above has a title and a scale. Let say that the circles were drawn by David, Peter, John, Brown, and Jennifer.

The title is "number of circles drawn"

The scale used is the following: one circle represents 5 circles.

David drew 15 circles since we have 3 circles and each circle represents 5 circles.

John drew 7.5 circles. How did we get 7.5?

Note that if a circle represents 5 circles, half a circle will represent 5/2 circles or 2.5 circles.

Add 5 circles to 2.5, give 7.5

Exercise:

Construct a pictograph with the following table:

Here is the pictographs:

Triangular numbers

Triangular numbers are numbers that represent the shapes that you see below. My goal is to help you examine the pattern and derive a formula. 

           

Looking at the pattern, you should see that the first 4 numbers are 1, 3, 6, and 10

If we can find how many dots there are in the 100th triangular number, it will be fairly easy to derive a general formula

Here is how to proceed:

First number: 1

Second number: 3 = 1 + 2

Third number: 6 = 1 + 2 + 3

Fourth number: 10 = 1 + 2 + 3 + 4

Hundredth number: ? = 1 + 2 + 3 + 4 + 5 + 6 +....+ 100

Instead of adding in the order, you can add as shown below (credited to Gauss)

(1 + 100) + (2 + 99) + (3 + 98) + (4 + 97) + ......+ (50 + 51)

Notice that each pair is equal to 101. Furthermore, since we are pairing the numbers, and there are 100 numbers, there will be 50 pairs

Therefore, instead of adding 101 fifty times, you can just multiply 101 by 50

Since 50 × 101 = 5050, the sum for 1 + 2 + 3 + 4 + 5 + 6 +....+ 100 is equal to 5050

You can play with 50 × 101 to get a general formula. 

If we can rewrite 50 × 101 and make 100 appear into the expression, we can just make a prediction and say that that 100 represents the hundredth number.

Then, we can simply replace 100 by n and n will represent the nth number.

It is not a complete proof. You just make a sound and logical conclusion based on a pattern

50 × 101 = (100/2) × 101 = (100/2) × (100 + 1)

If we substitute 100 for n, the formula we get is (n/2) × (n + 1)

Now let's test the formula for the first 4 numbers above

First number: (1/2) × (1 + 1) = (1/2) × 2 = 1 

Second number: (2/2) × (2 + 1)= 1 × (2 + 1) = 1 × 3 = 3

Third number: (3/2) × (3 + 1) = 3/2 × 4 = 12/2 = 6

Fourth number: (4/2) × (4 + 1) = 4/2 × 5 = 2 × 5 = 10 

Since the formula is working for 5 numbers, you have a pattern and it is reasonable to conclude that it will work for all triangular numbers

Palindrome

In mathematics, a palindrome is a number that reads the same forward and backward. For example, 353 and 787.

By definition, all numbers that have the same digits such as 4, 11, 55, 222, and 6666 are other examples of such number. 

Given any numbers, you can use the following simple algorithm to find other palindromes.

Step 1:

Start with any number. Call it original number. Reverse the digits of the original number

Step 2:

Call the number whose digits are reversed new number. Add the new number to your original number.

Call the number found by adding the new number to the original number test number

Step 3:

If test number is a palindrome, you are done. If not, use your test number as your original number and repeat the steps above

Sound complicated? However, it is not!Let us illustrate

Example #1:

75

Reversing 75 gives 57

Adding 75 and 57 gives 132

Reversing 132 gives 231

Adding 132 and 231 gives 363 and we are done!

Example #2:

255

Reversing 255 gives 552

Adding 255 and 552 gives 807

Reversing 807 gives 708

Adding 807 and 708 gives 1515

Reversing 1515 gives 5151

Adding 1515 and 5151 gives 6666. Now we are done!

Now, here is your puzzle. Find 3 numbers less than 100 that require at least 4 additions to obtain palindromes