Maths

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Maths Mysteries
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The Missing 8

                    Without 8
12 345 679 x   9 = 111 111 111
12 345 679 x 18 = 222 222 222
12 345 679 x 27 = 333 333 333
12 345 679 x 36 = 444 444 444
     and notice…
12 345 679 x 999 999 999 =
12 345 678 987 654 321
                     With the 8
123 456 789 x   9 = 1 111 111 101
123 456 789 x 18 = 2 222 222 202
123 456 789 x 27 = 3 333 333 303
123 456 789 x 36 = 4 444 444 404
     and notice…
123 456 789 x 999 999 999 =
123 456 788 876 543 211

Challenge:             

Complete these:

                  Without the 8…..

1) 12 345 679 x 45 =           2) 12 345 679 x 54 =         3) 12 345 679 x 63 =

4) 12 345 679 x 72 =           5) 12 345 679 x 81 =

                         With the 8…. 

6)   123 456 789 x 45 =       7)   123 456 789 x 54 =       8)   123 456 789 x 63 =

9)   123 456 789 x 72 =       10) 123 456 789 x 81 =

The ‘without 8’ pattern continues below:

12 345 679 x 90   =   1 111 111 110

12 345 679 x 99   =   1 222 222 221

12 345 679 x 108 =   1 333 333 332

   Fill in the answers to these:

11) 12 345 679 x 117 =       12) 12 345 679 x 126 =       13) 12 345 679 x 135 =

14) 12 345 679 x 144 =       15) 12 345 679 x 153 =       16) 12 345 679 x 162 =

The ‘with the  8’ pattern continues below:

123 456 789 x 90 =  11 111 111 010

123 456 789 x 99 =  12 222 222 111

123 456 789 x 108 =13 333 333 212

   Now fill in the answers to these:

17) 123 456 789 x 117 =       18) 123 456 789 x 126 =       19) 123 456 789 x 135 =

20) 123 456 789 x 144 =       21) 123 456 789 x 153 =       22) 123 456 789 x 162 =

 Solutions to The Missing 8

1) 555 555 555                  2) 666 666 666              3) 777 777 777                    

4) 888 888 888                 5) 999 999 999                6) 5 555 555 505

7) 6 666 666 606             8) 7 777 777 707             9) 8 888 888 808            

10) 9 999 999 909           11) 1 444 444 443               12) 1 555 555 554

13) 1 666 666 665             14) 1 777 777 776               15) 1 888 888 887                

16) 1 999 999 998              17) 14 444 444 313             18) 15 555 555 414

19) 16 666 666 515            20) 17 777 777 616              21) 18 888 888 717            

22) 19 999 999 818

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Multiples of 9

987 654 321 x   9 =   8 888 888 889

987 654 321 x 18 = 17 777 777 778

987 654 321 x 27 = 26 666 666 667

987 654 321 x 36 = 35 555 555 556

987 654 321 x 45 = 44 444 444 445

Challenge:             

1) What do you notice about the first and last digits of the product?

Complete these:

2) 987 654 321 x 54 =

3) 987 654 321 x 63 =

4) 987 654 321 x 72 =

5) 987 654 321 x 81 =

Solutions to Multiples of 9

1) Same as the multiplier        2) 53 333 333 334
3) 62 222 222 223                    4) 71 111 111 112
5) 80 000 000 001

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Amazing Number 2 520

The number 2 520 can be divided by 1 and 2 and 3 and 4 and 5 and 6 and 7 and 8 and 9 and 10.

Challenge:

1) Write down the first ten factors of 2 520.

2) 7 and 8 are factors of 2 520; does that mean that 56 is a factor of 2 520?

3) 28, 35, 42. Which of these are not factors of 2 520?

4) 11, 12, 14, 15, 16, 18, 20, 21, 22, 24.
Seven of these ten numbers are also factors of 2 520. Which ones are they?

5) Can 5 040 be divided by 1 and 2 and 3 and 4 and 5 and 6 and 7 and 8 and 9 and 10?

6) 2 520 divided by 2 equals 1 260. Which of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 are not factors of 1 260?

Solutions to Amazing Number 2 520

1) 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10        2) yes
3) None of them. They all divide evenly into 2 520.
4) 12, 14, 15, 18, 20, 21, 24        5) yes        6) 8 

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Mysterious Primes

Every prime number except for 2 and 3 is evenly divisible by 6 if you either subtract 1 from it or add 1 to it.

Example 1:    13-1=12 and 12 is divisible by 6

Example 2:    17+1=18 and 18 is divisible by 6.

Challenge:

1. Does the rule work for the prime 37? Test it.

2. Does the rule work for the prime 41? Test it.

3. Does the rule work for the prime 43? Test it.

4. Does the rule work for the prime 47? Test it.

5. Does the rule work for the prime 59? Test it.

6. What prime number gives a quotient of 4 after 1 is added to it and the result is divided by 6?

7. What prime number gives a quotient of 5 after 1 is subtracted from it and the result is divided by 6?

8. What prime number gives a quotient of 2 after 1 is added to it and the result is divided by 6?

9. What prime number gives a quotient of 3 after 1 is subtracted from it and the result is divided by 6?

10. What prime number gives a quotient of 9 after 1 is added to it and the result is divided by 6?

Solutions to Mysterious Primes

1. yes      2. yes      3. yes      4. yes      5. yes

6. 23      7. 31      8. 11      9. 19      10. 53 

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Curiosities & Perplexities
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Positive Even Integers Greater Than 2

Every positive even integer can be written as the sum of two primes.

Examples:     6=3+3       8=3+5       10=3+7       18=7+11       20=7+13

Challenge: Write two primes that add to the following even numbers.

(1) 14         (2) 22        (3) 30        (4) 32        (5) 34        

(6) 36        (7) 38         (8) 40        (9) 50        (10) 60      

(11) 64       (12) 68      (13) 70       (14) 76       (15) 78       

(16) 80     (17) 84      (18) 92      (19) 96      (20) 100    

 Solutions to Positive Even Integers Greater Than 2
(in cases where there are multiple solutions up to three solutions are given)

(1)         11+3,   7+7           (2)        17+5,   3+19,   11+11

(3)        19+11,   7+23,   13+17      (4)        19+13,   3+29

(5)        23+11,   5+29,   17+17      (6)        17+19,   5+31,   7+29

(7)         7+31,   19+19        (8)        23+17,   29+11,   3+37

(9)        31+19,   37+13,   47+3      (10)      31+29,   23+37,   19+41

(11)       23+41,   5+59,   11+53        (12)      31+37,   7+61

(13)       3+67,   11+59,   17+53        (14)       47+29,   5+71,   17+59

(15)       41+37,   5+73,   7+71        (16)      19+61,   7+73,   13+67

(17)       79+5,   11+73,   13+71        (18)      89+3,   13+79,   19+73

(19)      79+17,   13+83,   23+73        (20)     53+47,   7+93,   11+89

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Odd Numbers Greater Than 7

Odd numbers greater than 7 can be expressed as the sum of three odd primes.

Examples:      9=3+3+3        17=3+3+11       19=3+5+11       37=3+11+23       

Challenge:
Write three odd primes that add to the following odd numbers.

1. 41

2. 29

3. 23

4. 53

5. 43

6. 61

7. 67

8. 47

9. 59

10. 31

11. 71

12. 79

13. 83

14. 89

15. 97

16. 101

17. 103

18. 113

19. 127

20. 131

 Solutions to Odd Numbers Greater Than 7
(in cases where there are multiple solutions up to three solutions are given)
1. 3+7+31, 5+13+23, 7+11+23
2. 3+7+19, 5+5+19, 3+3+23
3. 3+7+13, 5+5+13, 5+7+11
4. 3+7+43, 3+13+37, 5+5+43
5. 5+19+19, 11+13+19, 11+15+17
6. 7+13+41, 3+17+41, 3+5+53
7. 11+19+37, 13+17+37, 7+23+37
8. 5+13+29, 7+11+29, 7+17+23
9. 5+7+47, 31+23+5, 17+19+23
10. 7+7+17, 3+11+17, 5+7+19
11. 3+31+37, 29+11+31, 29+29+13
12. 7+31+41, 19+19+41, 19+23+27
13. 11+31+41, 13+29+41, 33+37+13
14. 11+37+41, 7+41+41, 5+41+43
15. 19+37+41, 23+37+37, 29+31+37
16. 19+41+41, 23+37+41, 29+31+41
17. 23+37+43, 19+41+43, 19+31+53
18. 23+37+53, 19+41+53, 5+7+101
19. 37+37+53, 3+5+119, 7+11+109
20. 37+41+53, 5+7+119, 5+19+107

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Speedy Maths
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Multiplying Two-Digit Numbers by 11

The first step is to write the number, leaving some space between the two digits.

Then insert the sum of the number’s two digits in between the two digits themselves; you will have to carry when the sum of the digits exceeds 9.

Example 1:  36 x 11

1) 3 6

2) 3+6=9

3) 396

Example 2:  78 x 11.

1) 7 8

2) 7+8=15

3) 858

Challenge:

How fast can you do these?

1)  25 x 11      2)  42 x 11     3)  72 x 11     4)  34 x 11   

5)  26 x 11     6)  84 x 11     7)  76 x 11     8)  97 x 11   

9)  67 x 11     10)  46 x 11   

Use the same method  but work backwards:

11)  ? x 11 = 396      12)  ? x 11 = 781      13)  ? x 11 = 187    

14)  ? x 11 = 385     15)  ? x 11 = 484     16)  ? x 11 = 748   

17)  ? x 11 = 209     18)  ? x 11 = 638     19)  ? x 11 = 858   

20)  ? x 11 = 1 078 

Solutions to Multiplying Two Digit Numbers by 11

 1)  275    2)  462    3)  792    4)  374    5)  286

6)  924    7)  836    8)  1 067    9)  737    10)  506

11)  36    12)  71    13)  17    14)  35    15)  44

16)  68    17)  19    18)  58    19)  78    20)  98

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Squaring Two Digit Numbers ending in 5

To square a number ending in 5, first multiply the number formed by the digit(s) in front of the 5 by the next whole number. To that product, affix the number 25. The number to affix (25) is easy to remember, because 52 = 25.

Example 1: How much is 252 ?

      2 x 3 = 6         Write down 6 and affix 25…  = 625

Example 2: How much is 752 ?

      7 x 8 = 56       Write down 56 and affix 25…  = 5 625

Challenge:

1. How much is 852 – 352 ?     7225 – 1225 =

2. How much is 452 + 952 ?     2025 + 9025 =

3.How much is 1152 ?     (use same method) 11 x 12 = 132; so 1152 =

4.How much is 1952 ?     19 x 20 = 380; so 1952 =

5. How much is 2 9952 ?     299 x 300 = 89 700; so 2 9952 =

Solutions to Squaring Two Digit Numbers ending in 5

1. 6 000    2. 11 050    3. 13 225    

4. 38 025    5. 8 970 025

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Mr Pythagoras
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Pythagoras

You might know that Pythagoras is the famous Greek mathematician who told everyone that if you make squares on each side of a right-angled triangle then the area of the square on the triangle’s hypotenuse (the longest side) will always equal the sum of the two squares on the triangle’s other two sides.

Look at the diagram below.

Image result for pythagorean triangle

The triangle’s hypotenuse is 5 units in length and the other two sides measure 3 and 4 units.
The square drawn on the hypotenuse is 25 square units (5×5) in area; this equals the sum of the other two sides’ areas… 16 (4×4) plus 9 (3×3). 

The triangle above (sides 3, 4 and 5) is just one of an infinite number of right-angled triangles. This triangle (3-4-5) is known as a Pythagorean triplet.  

Another right-angled triangle is the one with sides 5, 12 and 13 (because 132 = 122 + 52); 5-12-13 is another Pythagorean triplet.   

Two other Pythagorean triplets are 7-24-25 (because 252 = 242 + 72) and 9-40-41 (because 412 = 402 + 92).

The first six Pythagorean triplets   (NB: there are other side-length combinations that are found in right-angled triangles but Pythagorean triplets are the only ones with three whole numbers).

side 1

side 2

side 3

3

4

5

5

12

13

7

24

25

9

40

41

11

60

61

13

84

85

Class or Group Activity:    Let’s say your friend tells you they know of a Pythagorean triplet.   They then say they will tell you the length of just one of the sides but you have to work out the lengths of the other two sides.  Can you do it?  Yes, it’s easy! Read on.

In the table above, can you see any patterns? (there are several). Notice that the middle side is always an even number and that the shortest and longest sides are always odd. Note too that the longest side (the hypotenuse) is always just one unit more than the middle side).

Now, back to your friend, the one who gave you the length of just one side of a right-angles triangle and asked you to come up with the other two sides. By noticing patterns in the Pythagorean triplets table you might just have figured out a way to do it.       

Solutions:

Say your friend gives you a side with an even number and asks you to complete the triplet with the other two sides. Simply add 1 to the even number and that will give you the hypotenuse. To work out the 3rd side (in this case, the smallest side) all you have to do is get the square root of the sum of the two sides you already know.                            

Example 1:
Your friend tells you that the side he/she is thinking of is 12 units in length.   

12 is an even number so add 1 to 12 and you get 13.                                                             
Now you have two sides of the triplet…12 and 13.                                                                  
Add 12 and 13 together and you get 25.                                                                                  
Now get the square root of 25, which is 5.                                                                               
Your Pythagorean triplet is now complete and the sides are 5, 12 and 13. 

Example 2:
Your friend tells you that the side he/she is thinking of is 11 units in length.   
11 is midway between 10 and 12 so multiply 10 and 12 together to get 120.                       
Half of 120 is 60 which is the 2nd number of the triplet.                                                       
The square root of 132 + 602 is 61.                                                                                                  
You have now completed the triplet… 11, 60, 61.

Example 3:
Your friend tells you that the side he/she is thinking of is 41 units in length.   
41 is midway between 40 and 42 so multiply 40 and 42 together to get 1 680.                       
Half of 1 680 is 840 which is the 2nd number of the triplet.                                                                                                                                           As 840 is an even number the other side must be 840+1 units in length.                         
Your Pythagorean triplet is now complete and the sides are 41, 840 and 841.

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Divisibility Rules 

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The Divisibility Rules

These rules let you test if one number can be evenly divided by another, without having to do too much calculation.

A number is
divisible by:

If:

Example:

2

The last digit is even (0,2,4,6,8)

128 is
129 is not

3

The sum of the digits is divisible by 3

381 (3+8+1=12, and 12÷3 = 4) Yes

217 (2+1+7=10, and 10÷3 = 3 1/3) No

4

The last 2 digits are divisible by 4

1312 is (12÷4=3)
7019 is not

5

The last digit is 0 or 5

175 is
809 is not

6

The number is divisible by both 2 and 3

114 (it is even, and 1+1+4=6 and 6÷3 = 2) Yes

308 (it is even, but 3+0+8=11 and 11÷3 = 3 2/3) No

7

If you double the last digit and subtract it from the rest of the number and the answer is divisible by 7 or 0.

(Note: you can apply this rule to that answer again if you want)

672 (Double 2 is 4, 67-4=63, and 63÷7=9) Yes

905 (Double 5 is 10, 90-10=80, and 80÷7=11 3/7) No

8

The last three digits are divisible by 8

109816 (816÷8=102) Yes

216302 (302÷8=37 3/4) No

9

The sum of the digits is divisible by 9

(Note: you can apply this rule to that answer again if you want)

1629 (1+6+2+9=18, and again, 1+8=9) Yes

2013 (2+0+1+3=6) No

10

The number ends in 0

220 is
221 is not

11

If you sum every second digit and then subtract the other digits and the answer is divisible by 11 or 0

7392 ((7+9) – (3+2) = 11) Yes

25176 ((5+7) – (2+1+6) = 3) No

12

The number is divisible by both 3 and 4

648 (6+4+8=18 and 18÷3=6, also 48÷4=12) Yes

916 (9+1+6=16, 16÷3= 5 1/3) No

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Famous Mathematicians
**click Pascal’s pic for print version** 

Blaise Pascal

Blaise Pascal was born in France in 1623.

Blaise, who had 3 sisters, was saddened by the death of his mother when he was just 3 years old. Blaise’s father had his own views on education and he decided to teach Blaise himself.

His father did not want Blaise to study mathematics until he was 15 so he removed all mathematics books from their house. However this made Blaise curious about maths and, at age 12, he started work on geometry himself. He discovered that the sum of the angles of a triangle equals two right angles.

Blaise Pascal spent his lifetime studying mathematics and, though he did not discover it, is best known for his work on Pascal’s Triangle (see below).

He died in 1662.

        1          
        1   1        
      1   2   1      
    1   3   3   1    
  1   4   6   4   1  
1   5   10   10   5   1

Notice that each of the numbers is obtained by finding the sum of the two numbers above it (the first a little to the left, the second a little to the right).

Challenge: 

Six rows are shown in Pascal’s Triangle above.

1. Write out the 7th row in the triangle.      

2. Sum the numbers in the 7th row.       

3. Write out the 8th row in the triangle.     

4. What is the total of the numbers in the 8th row?     

5. What pattern is formed by the row totals in Pascal’s Triangle?

Solutions to Challenge

1.    1   6   15   20   15   6   1        2.   6

3.   1   7   21   35   35   21   7   1        4.   128

5.   Each row is double the one before.

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Fractions (Fundamentals)
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Fractions (Method)
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Fractions (Conversions)
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Fractions (Problem Solving)
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Mathematics Crosswords (1)
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Mathematics Crosswords (2)
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Addition, Subtraction, Multiplication & Division (without calculators): Vol 1
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Addition, Subtraction, Multiplication & Division (without calculators): Vol 2
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Multiplication Tables (1)
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Multiplication Tables (2)
**click on graphic for printable lessons**

  Image result for multiplication tables cartoon

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Request
So that we can make this site as classroom-friendly as possible we would be pleased to hear from any teachers who have tried any of our ideas, suggestions or lessons with their classes.
Just a very short note mentioning year level, idea/suggestion used, whether it was a written exercise, class discussion or debate, and any other useful feedback would be appreciated.
(school name optional but State would be of interest)
                               Send feedback to info[at]australianteacher.org 

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If using material from this website please attribute source…

Australian Teacher:  http://australianteacher.org

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11 Responses to Maths

  1. Peter Fedder says:

    These resources are great. Some of the most practical and useful I have had sent to me for quite some time. All were useful not just ‘one out of the bunch’. Thanks again.

  2. Sue MacGibbon says:

    Great looking resources and even better being able to download/ share these with other teachers,

    Thanks, keep up the good work,

    Sue

  3. Steve Thornton says:

    There are some nice activities here. Thanks.
    The activity about even numbers being the sum of two odd primes is called Goldbach’s conjecture, and is a famous unproved conjecture in mathematics. This is a really nice example that shows mathematics as a living subject, with problems for which we don’t yet have an answer. Researching some of the historical aspects of primes and Goldbach’s conjecture is a great cross-curriculum activity.
    Also, if Goldbach’s conjecture for even numbers happens to be true, the result about odd numbers greater than 7 being the sum of 3 odd primes follows logically. Why? This is a good exercise in logical reasoning for children.

  4. dev says:

    The study mat for children is great!! I am from India and my six-year-old is bright in maths. (usually Indians are great at mathematics (although not much in research part). I hope this material helps my kid to understand the nuances of maths in order to delve deep into it.
    Thanks and all the best

    Dev

    • Thank you Dev.

      It’s nice to know that some of our maths material is being used in India, a nation that has produced some of the world’s great mathematicians.

      Best wishes to you and your little six year old.

      Warm regards,

      Ron and Jacqueline

  5. Jaki says:

    These are great! Haven’t used them yet but I can definitely see them in my maths groups. Thanks a lot.

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