Maths Mysteries
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The Missing 8
Without 812 345 679 x 9 = 111 111 11112 345 679 x 18 = 222 222 22212 345 679 x 27 = 333 333 33312 345 679 x 36 = 444 444 444and notice…12 345 679 x 999 999 999 =12 345 678 987 654 321 
With the 8123 456 789 x 9 = 1 111 111 101123 456 789 x 18 = 2 222 222 202123 456 789 x 27 = 3 333 333 303123 456 789 x 36 = 4 444 444 404and 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: 131=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 TwoDigit 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 5^{2} = 25.
Example 1: How much is 25^{2 }?
2 x 3 = 6 Write down 6 and affix 25… = 625
Example 2: How much is 75^{2} ?
7 x 8 = 56 Write down 56 and affix 25… = 5 625
Challenge:
1. How much is 85^{2 }– 35^{2} ? 7225 – 1225 =
2. How much is 45^{2} + 95^{2} ? 2025 + 9025 =
3.How much is 115^{2} ? (use same method) 11 x 12 = 132; so 115^{2} =
4.How much is 195^{2} ? 19 x 20 = 380; so 195^{2} =
5. How much is 2 995^{2} ? 299 x 300 = 89 700; so 2 995^{2} =
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 rightangled 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.
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 rightangled triangles. This triangle (345) is known as a Pythagorean triplet.
Another rightangled triangle is the one with sides 5, 12 and 13 (because 13^{2 }= 12^{2 }+ 5^{2}); 51213 is another Pythagorean triplet.
Two other Pythagorean triplets are 72425 (because 25^{2 }= 24^{2 }+ 7^{2}) and 94041 (because 41^{2 }= 40^{2 }+ 9^{2}).
The first six Pythagorean triplets (NB: there are other sidelength combinations that are found in rightangled 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 rightangles 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 13^{2 }+ 60^{2 }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

If: 
Example: 
2 
The last digit is even (0,2,4,6,8) 
128 is

3 
The sum of the digits is divisible by 3 
381 (3+8+1=12, and 12÷3 = 4) Yes217 (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)

5 
The last digit is 0 or 5 
175 is

6 
The number is divisible by both 2 and 3 
114 (it is even, and 1+1+4=6 and 6÷3 = 2) Yes308 (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, 674=63, and 63÷7=9) Yes905 (Double 5 is 10, 9010=80, and 80÷7=11 ^{3}/_{7}) No 
8 
The last three digits are divisible by 8 
109816 (816÷8=102) Yes216302 (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) Yes2013 (2+0+1+3=6) No 
10 
The number ends in 0 
220 is

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) Yes25176 ((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) Yes916 (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 
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.
Thank you Peter.
Glad you’re getting something out of the website and thanks so much for taking the time to write.
Best wishes,
Ron and Jacqueline
Great looking resources and even better being able to download/ share these with other teachers,
Thanks, keep up the good work,
Sue
Thanks Sue.
Glad you’re getting good use out of them.
Have a great year!
Best,
Ron & Jacqueline
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 crosscurriculum 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.
Thanks for your kind words and your very interesting contribution Steve.
Wiki tells us, “Since if every even number greater than 4 is the sum of two odd primes, merely adding 3 to each even number greater than 4 will produce the odd numbers greater than 7.”
You’re right…this would indeed be a great exercise for nottooyoung maths students.
Many thanks again.
Ron and Jacqueline
Hi again Steve,
You may have seen this already…
http://au.news.yahoo.com/thewest/a//world/14831400/mathematicianproofprimenumbers/
All the best,
Ron
PS
I know it’s not the same conjecture you discussed in your comment above.
The study mat for children is great!! I am from India and my sixyearold 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
These are great! Haven’t used them yet but I can definitely see them in my maths groups. Thanks a lot.
Many thanks for your positive comment Jaki.
We hope to put more maths activities up on the website soon.
All the best,
Ron and Jackie
Thanks so much for these amazing resources! We are homeschoolers and appreciate the time you have put into these activities. We especially appreciate the care and consideration you have when explaining each lesson.
Thank you for your kind comments Lori.
It’s nice to know you find our ideas useful.
Best wishes,
Ron and Jacqueline