The Locker Puzzle
?In-text Question — Does every number have an even number of factors?
Write the factors of a few numbers in pairs.
\(6\): \(1\times6\), \(2\times3\).
The factors are 1, 2, 3, 6, which is 4 factors (even).
\(4\): \(1\times4\), \(2\times2\).
The factors are 1, 2, 4, which is 3 factors (odd).
\(9\): \(1\times9\), \(3\times3\).
The factors are 1, 3, 9, which is 3 factors (odd).
\(1\): \(1\times1\).
The only factor is 1, which is 1 factor (odd).
In \(2\times2\) and \(3\times3\) the two numbers in the pair are the same, so the pair gives only one factor.
?In-text Question — Can you use this insight to find more numbers with an odd number of factors?
A number has an odd number of factors when one of its factor pairs has both numbers equal.
Take 36.
It has the factor pair \(6\times6\).
The other factor pairs are \(1\times36\), \(2\times18\), \(3\times12\), \(4\times9\).
In each of these the two numbers are different.
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36, which is 9 factors (odd).
In the same way, every number of the form \(n\times n\) has an odd number of factors.
?In-text Question — Write the locker numbers that remain open.
A locker is toggled once for each factor of its number.
A locker stays open only if it is toggled an odd number of times, that is, if its number has an odd number of factors.
Only square numbers have an odd number of factors.
The square numbers from 1 to 100 are \(1^2, 2^2, 3^2, \dots, 10^2\).
Square Numbers
Patterns and Properties of Perfect Squares
AActivity — Find the squares of the first 30 natural numbers and fill in the table below.
\(n^2=n\times n\).
Multiplying each number by itself gives the table.
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |
| n | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|
| n² | 121 | 144 | 169 | 196 | 225 | 256 | 289 | 324 | 361 | 400 |
| n | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
|---|---|---|---|---|---|---|---|---|---|---|
| n² | 441 | 484 | 529 | 576 | 625 | 676 | 729 | 784 | 841 | 900 |
MTMath Talk — What patterns do you notice? Share your observations and make conjectures.
Units digit
Every square ends in 0, 1, 4, 5, 6 or 9.
None ends in 2, 3, 7 or 8.
If \(n\) ends in 1 or 9, \(n^2\) ends in 1.
If \(n\) ends in 2 or 8, \(n^2\) ends in 4.
If \(n\) ends in 3 or 7, \(n^2\) ends in 9.
If \(n\) ends in 4 or 6, \(n^2\) ends in 6.
If \(n\) ends in 5, \(n^2\) ends in 5.
If \(n\) ends in 0, \(n^2\) ends in 0.
Even and odd
The square of an even number is even.
The square of an odd number is odd.
Differences
The difference between consecutive squares is an odd number: \(4-1=3\), \(9-4=5\), \(16-9=7\), \(25-16=9\).
In general, \((n+1)^2-n^2=2n+1\).
MTMath Talk — If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
No.
Take numbers that end in 6.
16 and 36 are squares (\(4^2\) and \(6^2\)).
26 also ends in 6, but it is not a square.
It lies between \(5^2=25\) and \(6^2=36\).
So the units digit alone cannot tell us that a number is a square.
But if a number ends in 2, 3, 7 or 8, it is definitely not a square.
?In-text Question — Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
A square never ends in 2, 3, 7 or 8.
So any number that ends in 2, 3, 7 or 8 is not a square.
For example: 32, 43, 57, 78, 1002.
?In-text Question — The squares, 1², 9², 11², 19², 21², and 29², all have 1 in their units place. Write the next two squares.
The numbers 1, 9, 11, 19, 21, 29 end in 1 or 9.
The next two numbers that end in 1 or 9 are 31 and 39.
\(31^2=961\)
\(39^2=1521\)
?In-text Question — Which of the following numbers have the digit 6 in the units place?
(i) 38²
(ii) 34²
(iii) 46²
(iv) 56²
(v) 74²
(vi) 82²
A square ends in 6 only when the number ends in 4 or 6.
(i)
38 ends in 8.
\(8^2=64\), so \(38^2\) ends in 4.
(ii)
34 ends in 4.
\(4^2=16\), so \(34^2\) ends in 6.
(\(34^2=1156\))
(iii)
46 ends in 6.
\(6^2=36\), so \(46^2\) ends in 6.
(\(46^2=2116\))
(iv)
56 ends in 6, so \(56^2\) ends in 6.
(\(56^2=3136\))
(v)
74 ends in 4, so \(74^2\) ends in 6.
(\(74^2=5476\))
(vi)
82 ends in 2.
\(2^2=4\), so \(82^2\) ends in 4.
(iii) 46²,
(iv) 56² and
(v) 74² have 6 in the units place.
?In-text Question — Find more such patterns by observing the numbers and their squares from the table you filled earlier.
Numbers ending in 5
\(15^2=225\), \(25^2=625\), \(35^2=1225\).
Every such square ends in 25.
The part before 25 is \(\text{tens digit}\times(\text{tens digit}+1)\).
\(15^2\): \(1\times2=2\), so 225.
\(25^2\): \(2\times3=6\), so 625.
\(35^2\): \(3\times4=12\), so 1225.
Numbers that add up to 10
The numbers 3 and 7 add up to 10 and their squares 9 and 49 both end in 9.
The same holds for 2 and 8 (4 and 64), 1 and 9 (1 and 81), 4 and 6 (16 and 36).
Digits
\(13^2=169\) and \(31^2=961\) are made of the same digits written in reverse.
The same is true of \(12^2=144\) and \(21^2=441\).
?In-text Question — If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Take a number with 3 zeros at the end, such as 1000 or 2000.
\(1000^2=1{,}000{,}000\), which has 6 zeros.
\(2000^2=4{,}000{,}000\), which has 6 zeros.
?In-text Question — What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Observation
\(10^2=100\): 1 zero becomes 2 zeros.
\(40^2=1600\): 1 zero becomes 2 zeros.
\(100^2=10000\): 2 zeros become 4 zeros.
\(700^2=490000\): 2 zeros become 4 zeros.
The number of zeros at the end of the square is double the number of zeros at the end of the number.
Will this always happen?
Write the number as \(m\times10^k\), where \(m\) does not end in 0.
Then \(\text{square}=m^2\times10^{2k}\).
Since \(m\) does not end in 0, \(m^2\) does not end in 0 either.
So the square has exactly \(2k\) zeros at the end.
?In-text Question — What can you say about the parity of a number and its square?
Even numbers
\(2^2=4\), \(4^2=16\), \(6^2=36\).
The squares are even.
Odd numbers
\(3^2=9\), \(5^2=25\), \(7^2=49\).
The squares are odd.
Reason
An even number is \(2k\), and \((2k)^2=4k^2\), which is even.
An odd number is \(2k+1\), and \((2k+1)^2=4k^2+4k+1\), which is odd.
Perfect Squares and Odd Numbers
?In-text Question — Let us explore the differences between consecutive squares. What do you notice? 4 – 1 = 3, 9 – 4 = 5, 16 – 9 = 7, 25 – 16 = 9. See if this pattern continues for the next few square numbers.
\(36-25=11\)
\(49-36=13\)
\(64-49=15\)
\(81-64=17\)
The differences 3, 5, 7, 9, 11, 13, 15, 17, … are consecutive odd numbers.
?In-text Question — Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Between \(1^2=1\) and \(2^2=4\): the numbers 2, 3.
That is 2 numbers.
Between \(2^2=4\) and \(3^2=9\): the numbers 5, 6, 7, 8.
That is 4 numbers.
Between \(3^2=9\) and \(4^2=16\): the numbers 10 to 15.
That is 6 numbers.
Between \(4^2=16\) and \(5^2=25\): the numbers 17 to 24.
That is 8 numbers.
General rule
Let \(p\) and \(q\) be two consecutive squares.
The numbers between them are \(q-p-1\) in number.
For \(p=n^2\) and \(q=(n+1)^2\): \((n+1)^2-n^2-1=2n+1-1=2n\).
AActivity — How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?
Use the table of squares from 1 to 30.
The block 1 – 100 includes both 1 and 100.
| Block | Squares in the block | Number of squares |
|---|---|---|
| 1 – 100 | 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 | 10 |
| 101 – 200 | 121, 144, 169, 196 | 4 |
| 201 – 300 | 225, 256, 289 | 3 |
| 301 – 400 | 324, 361, 400 | 3 |
| 401 – 500 | 441, 484 | 2 |
| 501 – 600 | 529, 576 | 2 |
| 601 – 700 | 625, 676 | 2 |
| 701 – 800 | 729, 784 | 2 |
| 801 – 900 | 841, 900 | 2 |
| 901 – 1000 | 961 | 1 |
\(31^2=961\) and \(32^2=1024\), which is more than 1000.
Perfect Squares and Triangular Numbers
?In-text Question — Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term. 1 + 3 = 4 = 2², 3 + 6 = 9 = 3², 6 + 10 = 16 = 4²
The triangular numbers are 1, 3, 6, 10, 15, 21, …
The next triangular number is obtained by adding one more row of dots.
After 15 we add 6 dots, which gives 21.
Relation
The sum of two consecutive triangular numbers is a square number.
\(1+3=4=2^2\)
\(3+6=9=3^2\)
\(6+10=16=4^2\)
Next term
\(10+15=25=5^2\)
So the next figure is a \(5\times5\) square of dots, made of the triangles 10 and 15.
One more term
\(15+21=36=6^2\)
Square Roots
MTMath Talk — Given a number, such as 576 or 327, how do we find out if it is a perfect square? If it is a perfect square, how can we find its square root?
Step 1: look at the units digit
327 ends in 7, and a square never ends in 7.
So 327 is not a perfect square.
576 ends in 6, which is possible for a square.
But this does not prove that 576 is a square.
Step 2: check 576
Method 1: list the squares
\(20^2=400\), \(21^2=441\), \(22^2=484\), \(23^2=529\), \(24^2=576\)
Method 2: prime factorisation
\(576=2\times2\times2\times2\times2\times2\times3\times3\)
\(=(2\times2\times2\times3)\times(2\times2\times2\times3)\)
\(=24\times24=24^2\)
The prime factors split into two equal groups, so 576 is a perfect square.
To find the square root, split the prime factors into two equal groups, or subtract consecutive odd numbers, or list squares.
?In-text Question — Find whether 1156 and 2800 are perfect squares using prime factorisation.
1156
\(1156=2\times578\)
\(=2\times2\times289\)
\(=2\times2\times17\times17\)
\(=(2\times17)\times(2\times17)=34^2\)
The prime factors pair up, so 1156 is a perfect square, and \(\sqrt{1156}=34\).
2800
\(2800=2\times2\times2\times2\times5\times5\times7\)
The 2s pair up and the 5s pair up, but 7 is left without a partner.
So the factors cannot be split into two equal groups.
Figure it Out — Set 1
Nine questions from page 10–11 of the textbook.
1Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
A perfect square cannot end in 2, 3, 7 or 8.
(i)
2032 ends in 2.
So it is not a perfect square.
(ii)
2048 ends in 8.
So it is not a perfect square.
(iii)
1027 ends in 7.
So it is not a perfect square.
(iv)
1089 ends in 9, so it may be a square.
Check: \(30^2=900\) and \(40^2=1600\), and the last digit 9 means the root ends in 3 or 7.
Try 33, because \(35^2=1225\) is already more than 1089.
\(33^2=1089\)
So 1089 is a perfect square.
(ii) 2048 and
(iii) 1027 are not perfect squares.
2Which one among 64², 108², 292², 36² has last digit 4?
The square of a number ends in 4 only when the number ends in 2 or 8.
64 ends in 4, so \(64^2\) ends in 6.
108 ends in 8, so \(108^2\) ends in 4.
292 ends in 2, so \(292^2\) ends in 4.
36 ends in 6, so \(36^2\) ends in 6.
3Given 125² = 15625, what is the value of 126²?
(i) 15625 + 126
(ii) 15625 + 26²
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 51²
The difference between consecutive squares is an odd number.
\(126^2-125^2=(126+125)(126-125)=251\)
\(126^2-125^2=\) the 126th odd number.
The nth odd number is \(2n-1\), so the 126th odd number is \(2\times126-1=251\).
\(126^2=125^2+251=15625+251=15876\)
Check
\(126^2=(125+1)^2=125^2+2\times125+1=15625+251\) ✓
4Find the length of the side of a square whose area is 441 m².
Side \(=\sqrt{\text{area}}=\sqrt{441}\)
\(441=3\times3\times7\times7\)
\(=(3\times7)\times(3\times7)=21^2\)
\(\sqrt{441}=21\)
5Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
A number divisible by 4, 9 and 10 is divisible by their LCM.
\(4=2^2\), \(9=3^2\), \(10=2\times5\)
LCM \(=2^2\times3^2\times5=180\)
\(180=2\times2\times3\times3\times5\)
The prime 5 appears only once, so 180 is not a square.
Multiply by 5 to pair it up.
\(180\times5=900=2^2\times3^2\times5^2=(2\times3\times5)^2=30^2\)
6Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Prime factorisation of 9408:
\(9408=2\times2\times2\times2\times2\times2\times3\times7\times7\)
\(=2^6\times3\times7^2\)
The 2s form 3 pairs, and the 7s form 1 pair.
The prime 3 is left without a partner.
Multiply by 3 to complete the pair.
\(9408\times3=28224=2^6\times3^2\times7^2\)
\(=(2^3\times3\times7)^2\)
\(=(8\times3\times7)^2=168^2\)
\(\sqrt{28224}=168\)
7How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
Between \(n^2\) and \((n+1)^2\) there are \(2n\) numbers.
(i)
\(16^2=256\) and \(17^2=289\).
Numbers between them \(=289-256-1=32\)
Check: \(2\times16=32\) ✓
(ii)
\(99^2=9801\) and \(100^2=10000\).
Numbers between them \(=10000-9801-1=198\)
Check: \(2\times99=198\) ✓
(ii) 198 numbers
8In the following pattern, fill in the missing numbers:
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + 20² = (___)²
9² + 10² + (___)² = (___)²
Find the pattern
In each line the numbers are \(n\), \(n+1\) and \(n(n+1)\).
Line 1: \(n=1\): \(1\), \(2\), \(1\times2=2\)
Line 2: \(n=2\): \(2\), \(3\), \(2\times3=6\)
Line 3: \(n=3\): \(3\), \(4\), \(3\times4=12\)
The number on the right is one more than the last number on the left: \(2+1=3\), \(6+1=7\), \(12+1=13\).
Line 4
\(n=4\): \(4^2+5^2+20^2\), and \(20+1=21\).
Check: \(16+25+400=441=21^2\) ✓
Line 5
\(n=9\): \(9^2+10^2+(9\times10)^2=9^2+10^2+90^2\), and \(90+1=91\).
Check: \(81+100+8100=8281=91^2\) ✓
\(9^2+10^2+(90)^2=(91)^2\)
9How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
The picture is made of 9 rows and 9 columns of tiles.
Upright tiles and tilted tiles alternate, like a chessboard.
Number of tiles \(=9\times9=81\).
Count along one side of any tile: there are 5 tiny squares.
So each tile, whether upright or tilted, is a \(5\times5\) grid of tiny squares.
Tiny squares in one tile \(=5\times5=25\).
Total tiny squares \(=81\times25=2025\).
Prime factorisation
\(2025=5\times405\)
\(=5\times5\times81\)
\(=5\times5\times3\times3\times3\times3\)
\(=3^4\times5^2\)
Check
\(2025=45^2\), a perfect square, and \(3^4\times5^2=(3^2\times5)^2=45^2\) ✓
\(2025=3\times3\times3\times3\times5\times5=3^4\times5^2\)
Cubic Numbers
?In-text Question — How many cubes of side 1 cm make a cube of side 2 cm? How many cubes of side 1 cm will make a cube of side 3 cm?
Cube of side 2 cm
Each layer has \(2\times2=4\) unit cubes.
There are 2 layers.
Number of unit cubes \(=2\times2\times2=8\)
Cube of side 3 cm
Each layer has \(3\times3=9\) unit cubes.
There are 3 layers.
Number of unit cubes \(=3\times3\times3=27\)
AActivity — Complete the table below.
\(n^3=n\times n\times n\).
For example, \(6^3=6\times6\times6=216\).
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| n³ | 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 |
| n | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|
| n³ | 1331 | 1728 | 2197 | 2744 | 3375 | 4096 | 4913 | 5832 | 6859 | 8000 |
MTMath Talk — What patterns do you notice in the table above?
Units digit
The cube of a number ending in 1, 4, 5, 6, 9 or 0 ends in the same digit.
A cube ends in 8 if the number ends in 2, and ends in 2 if the number ends in 8.
A cube ends in 7 if the number ends in 3, and ends in 3 if the number ends in 7.
Even and odd
The cube of an even number is even.
The cube of an odd number is odd.
Zeros
\(10^3=1000\), \(20^3=8000\).
A number with one zero at the end has a cube with three zeros at the end.
?In-text Question — We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?
Take the cube of each last digit.
| Last digit of n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Last digit of n³ | 0 | 1 | 8 | 7 | 4 | 5 | 6 | 3 | 2 | 9 |
Every digit from 0 to 9 appears.
?In-text Question — Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
1-digit cubes
\(1^3=1\), \(2^3=8\).
That is 2 cubes.
2-digit cubes
\(3^3=27\), \(4^3=64\).
That is 2 cubes.
3-digit cubes
\(5^3=125\), \(6^3=216\), \(7^3=343\), \(8^3=512\), \(9^3=729\).
That is 5 cubes.
Observation
The smallest 4-digit cube is \(10^3=1000\), and \(21^3=9261\) is the largest, so there are 12 four-digit cubes.
The counts 2, 2, 5, 12 grow as the number of digits grows.
?In-text Question — Can a cube end with exactly two zeroes (00)? Explain.
Write the number as \(m\times10^k\), where \(m\) does not end in 0.
Then \(\text{cube}=m^3\times10^{3k}\).
Since \(m\) does not end in 0, \(m^3\) does not end in 0 either.
So the cube has exactly \(3k\) zeros at the end: 0, 3, 6, 9, …
\(2\) is not a multiple of 3.
TTTry This — The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.
Look for cubes that add up to the number.
4104
\(2^3+16^3=8+4096=4104\)
\(9^3+15^3=729+3375=4104\)
13832
\(2^3+24^3=8+13824=13832\)
\(18^3+20^3=5832+8000=13832\)
13832 = 2³ + 24³ = 18³ + 20³
Perfect Cubes and Consecutive Odd Numbers
?In-text Question — Later in this series, we get the following set of consecutive numbers: 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the calculation?
In the pattern, the line for \(n^3\) has \(n\) consecutive odd numbers.
\(1\), then \(3+5\), then \(7+9+11\), and so on.
The given set has 10 numbers: 91, 93, 95, 97, 99, 101, 103, 105, 107, 109.
So its sum is \(10^3\).
Check
The average of the numbers is \(\dfrac{91+109}{2}=100\).
Sum \(=10\times100=1000=10^3\) ✓
Cube Roots
?In-text Question — Find the cube roots of these numbers:
(i) ∛64
(ii) ∛512
(iii) ∛729
(i)
\(64=2\times2\times2\times2\times2\times2\)
\(=(2\times2)\times(2\times2)\times(2\times2)=4^3\)
\(\sqrt[3]{64}=4\)
(ii)
\(512=2\times2\times2\times2\times2\times2\times2\times2\times2\)
\(=(2\times2\times2)^3=8^3\)
\(\sqrt[3]{512}=8\)
(iii)
\(729=3\times3\times3\times3\times3\times3\)
\(=(3\times3)^3=9^3\)
\(\sqrt[3]{729}=9\)
(ii) 8
(iii) 9
Successive Differences
?In-text Question — Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice? Perfect Cubes: 1, 8, 27, 64, 125, 216, …
Take the cubes 1, 8, 27, 64, 125, 216, 343.
Level 1
\(8-1=7\), \(27-8=19\), \(64-27=37\), \(125-64=61\), \(216-125=91\), \(343-216=127\)
Level 2
\(19-7=12\), \(37-19=18\), \(61-37=24\), \(91-61=30\), \(127-91=36\)
Level 3
\(18-12=6\), \(24-18=6\), \(30-24=6\), \(36-30=6\)
| Perfect cubes | 1 | 8 | 27 | 64 | 125 | 216 | 343 |
|---|---|---|---|---|---|---|---|
| Level 1 | 7 | 19 | 37 | 61 | 91 | 127 | |
| Level 2 | 12 | 18 | 24 | 30 | 36 | ||
| Level 3 | 6 | 6 | 6 | 6 |
Figure it Out — Set 2
Five questions from pages 16–17 of the textbook.
1Find the cube roots of 27000 and 10648.
27000
\(27000=27\times1000=3^3\times10^3=(3\times10)^3=30^3\)
\(\sqrt[3]{27000}=30\)
10648
\(10648=2\times2\times2\times11\times11\times11\)
\(=2^3\times11^3=(2\times11)^3=22^3\)
\(\sqrt[3]{10648}=22\)
2What number will you multiply by 1323 to make it a cube number?
\(1323=3\times441=3\times3\times147\)
\(=3\times3\times3\times49\)
\(=3\times3\times3\times7\times7\)
\(=3^3\times7^2\)
The 3s form a complete group of three.
The 7 appears only twice.
Multiply by one more 7 to make a group of three.
\(1323\times7=9261=3^3\times7^3=(3\times7)^3=21^3\)
3State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
(i) False
An odd number multiplied by itself any number of times stays odd.
For example, \(3^3=27\) is odd.
(ii) False
\(2^3=8\) ends in 8, and \(12^3=1728\) ends in 8.
Any number ending in 2 has a cube ending in 8.
(iii) False
The smallest 2-digit number is 10, and \(10^3=1000\), which has 4 digits.
So the cube of a 2-digit number has at least 4 digits.
(iv) False
The largest 2-digit number is 99, and \(99^3=970299\), which has 6 digits.
So the cube of a 2-digit number has at most 6 digits.
(v) False
\(8=2^3\) has the factors 1, 2, 4, 8, which is 4 factors (even).
\(64=4^3\) has 7 factors, but 8 has 4, so not every cube has an odd number of factors.
(ii) False
(iii) False
(iv) False
(v) False
4You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Strategy
The units digit of a cube tells the units digit of the cube root.
| Cube ends in | 1 | 8 | 7 | 4 | 5 | 6 | 3 | 2 | 9 | 0 |
|---|---|---|---|---|---|---|---|---|---|---|
| Root ends in | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 0 |
Then use \(10^3=1000\), \(20^3=8000\), \(30^3=27000\), \(40^3=64000\) to find the tens digit.
1331
Ends in 1, so the root ends in 1.
\(1000<1331<8000\), so the root is between 10 and 20.
The root is 11.
4913
Ends in 3, so the root ends in 7.
\(1000<4913<8000\), so the root is between 10 and 20.
The root is 17.
12167
Ends in 7, so the root ends in 3.
\(8000<12167<27000\), so the root is between 20 and 30.
The root is 23.
32768
Ends in 8, so the root ends in 2.
\(27000<32768<64000\), so the root is between 30 and 40.
The root is 32.
Check
\(11^3=1331\), \(17^3=4913\), \(23^3=12167\), \(32^3=32768\) ✓
5Which of the following is the greatest? Explain your reasoning.
(i) 67³ – 66³
(ii) 43³ – 42³
(iii) 67² – 66²
(iv) 43² – 42²
Differences of squares
\((n+1)^2-n^2=2n+1\)
(iii) \(67^2-66^2=2\times66+1=133\)
(iv) \(43^2-42^2=2\times42+1=85\)
Differences of cubes
\((n+1)^3-n^3=3n^2+3n+1\)
(i) \(67^3-66^3=3\times66^2+3\times66+1=3\times4356+198+1=13267\)
(ii) \(43^3-42^3=3\times42^2+3\times42+1=3\times1764+126+1=5419\)
Compare
\(13267>5419>133>85\)
Square Pairs
ACTActivity — Look at the following numbers: 3 6 10 15 1. They are arranged such that each pair of adjacent numbers adds up to a square. 3 + 6 = 9, 6 + 10 = 16, 10 + 15 = 25, 15 + 1 = 16. Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square. Can you arrange them in more than one way? If not, can you explain why?
Squares we can use
The sum of two different numbers from 1 to 17 is between 3 and 33, so it can be 4, 9, 16 or 25.
The two ends
17 can be added only to 8 (\(17+8=25\)).
So 17 has just one neighbour.
16 can be added only to 9 (\(16+9=25\)).
So 16 has just one neighbour.
A number with only one possible neighbour must be at an end of the row.
So 16 and 17 are the two ends of the row.
Building the row from 16
16 is followed by 9.
After 9 the only choice is 7, then 2, then 14, then 11, then 5, then 4, then 12, then 13, then 3.
After 3 we can use 6 or 1.
If 3 is followed by 1, the row goes 1, 8, 17 and ends with 6, 10 and 15 unused.
So that choice fails.
If 3 is followed by 6, the row continues 6, 10, 15, 1, 8, 17 and uses every number.
So 3 must be followed by 6.
One arrangement
16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17
Check the sums
\(16+9=25\), \(9+7=16\), \(7+2=9\), \(2+14=16\), \(14+11=25\), \(11+5=16\), \(5+4=9\), \(4+12=16\)
\(12+13=25\), \(13+3=16\), \(3+6=9\), \(6+10=16\), \(10+15=25\), \(15+1=16\), \(1+8=9\), \(8+17=25\) ✓
There is no other arrangement, apart from writing this one in reverse order, because 16 and 17 must be at the ends and every other choice is forced.
ACTActivity — Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?
In a circle every number has two neighbours, and the last number is also next to the first one.
The sums can now be 4, 9, 16, 25, 36 or 49 (the largest sum is \(32+31=63\)).
32 can be added only to 4, 17 (\(32+4=36\), \(32+17=49\)).
31 can be added only to 5, 18 (\(31+5=36\), \(31+18=49\)).
30 can be added only to 6, 19 (\(30+6=36\), \(30+19=49\)).
Start from these forced neighbours and fix the other numbers one by one, so that every adjacent pair gives a square.
A circle of 1 to 32
1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15, and then back to 1.
Check the sums
\(1+8=9\), \(8+28=36\), \(28+21=49\), \(21+4=25\), \(4+32=36\), \(32+17=49\), \(17+19=36\), \(19+30=49\)
\(30+6=36\), \(6+3=9\), \(3+13=16\), \(13+12=25\), \(12+24=36\), \(24+25=49\), \(25+11=36\), \(11+5=16\)
\(5+31=36\), \(31+18=49\), \(18+7=25\), \(7+29=36\), \(29+20=49\), \(20+16=36\), \(16+9=25\), \(9+27=36\)
\(27+22=49\), \(22+14=36\), \(14+2=16\), \(2+23=25\), \(23+26=49\), \(26+10=36\), \(10+15=25\), \(15+1=16\) ✓
Continue with Chapter 2
Move on to Power Play — exponents, laws of powers, and scientific notation.