A Square and A Cube: Chapter Solutions

Page 1

Question: ? Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer? Hint: Find out how many times each locker is toggled.

Solution: Khoisnam realized that a locker will remain open only if it is toggled an odd number of times. The number of times a locker is toggled equals its number of factors. Since only perfect square numbers have an odd number of factors, he knew that only the lockers with perfect square numbers (1, 4, 9, 16, 25, 36, 49, 64, 81, 100) would remain open.

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Page 2

Question: ? Does every number have an even number of factors?

Solution: No. While most numbers have an even number of factors because factors usually appear in pairs, perfect square numbers have an odd number of factors.

Question: ? Can you use this insight to find more numbers with an odd number of factors? For instance, 36 has a factor pair \[6\times6\] where both numbers are 6. Does this number have an odd number of factors?

Solution: Yes, 36 has an odd number of factors. Its factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Since it is a perfect square (\[6\times6\]), the factor 6 does not have a distinct “partner factor,” resulting in a total of 9 factors, which is an odd number.

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Page 3

Question: ? Write the locker numbers that remain open.

Solution: The lockers that remain open are the perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

Question: ? Which are these five lockers? (referring to the passcode consisting of the first five locker numbers that were touched exactly twice).

Solution: The lockers touched exactly twice are the prime numbers because a prime number has exactly two factors: 1 and the number itself. The first five prime numbers are 2, 3, 5, 7, and 11. The code is 2-3-5-7-11.

Question: Can we have a square of sidelength \[\frac{3}{5}\] or 2.5 units?

Solution: Yes. The area in square units would be \[
\left(\frac{3}{5}\right)^2 = \frac{9}{25}
\]
\[
(2.5)^2 = 6.25
\]

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Page 4

Question: Find the squares of the first 30 natural numbers and fill in the table below.

Answer:- \[
6^2 = 36,\; 7^2 = 49,\; 8^2 = 64,\; 9^2 = 81
\]
\[
10^2 = 100,\; 12^2 = 144,\; 13^2 = 169
\]
\[
14^2 = 196,\; 15^2 = 225,\; 16^2 = 256
\]
\[
17^2 = 289,\; 18^2 = 324,\; 19^2 = 361
\]
\[
20^2 = 400,\; 22^2 = 484,\; 23^2 = 529
\]
\[
24^2 = 576,\; 25^2 = 625,\; 26^2 = 676
\]
\[
27^2 = 729,\; 28^2 = 784,\; 29^2 = 841
\]
\[
30^2 = 900
\]

\[
29^2 = 841,\; 31^2 = 961,\; 39^2 = 1521
\]

Question: ? What patterns do you notice? Share your observations and make conjectures.

Solution: One key observation is that all perfect squares end in the digits 0, 1, 4, 5, 6, or 9. None of them end in 2, 3, 7, or 8.

Question: Study the squares in the table above. What are the digits in the units places of these numbers?

Solution: The digits in the units place are always 0, 1, 4, 5, 6, or 9.

Question: ? If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?

Solution: No. For example, 26 ends in 6 but is not a perfect square.

Question: ? Write 5 numbers such that you can determine by looking at their units digit that they are not squares.

Solution: 12, 23, 37, 48, and 52. Any number ending in 2, 3, 7, or 8 cannot be a perfect square.

Question: Write the next two squares. Notice that if a number has 1 or 9 in the units place, then its square ends in 1.

Solution: Following \[29^{2}=841\] , the next two squares of numbers ending in 1 or 9 are \[31^{2}=961\] and \[39^{2}=1521\].

Question: ? Let us consider square numbers ending in 6: … Which of the following numbers have the digit 6 in the units place? (i) \[38^{2}\] (ii) \[34^{2}\] (iii) \[46^{2}\] (iv) \[56^{2}\]] (v) \[74^{2}\] (vi) \[82^{2}\]

Solution: The numbers that will have a 6 in the units place when squared are those that end in 4 or 6. Therefore: (ii) \[34^{2}\], (iii) \[46^{2}\], (iv) \[56^{2}\], and (v) \[74^{2}\].

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Page 5

Question: ? Find more such patterns by observing the numbers and their squares from the table you filled earlier.

Solution: Numbers ending in 2 or 8 will always have a square ending in 4. Numbers ending in 3 or 7 will always have a square ending in 9. Numbers ending in 5 will always have a square ending in 5.

Question: ? If a number contains 3 zeros at the end, how many zeros will its square have at the end?

Solution: Its square will have 6 zeros at the end.

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?

Solution: The square of a number will always have twice as many zeros at the end as the original number. Yes, this will always happen, meaning a perfect square can only ever have an even number of trailing zeros.

Question: ? What can you say about the parity of a number and its square?

Solution: The parity remains the same. The square of an even number is always even, and the square of an odd number is always odd.

Question: Let us explore the differences between consecutive squares. What do you notice?

Solution: The difference between consecutive perfect squares yields the sequence of consecutive odd numbers (\[3, 5, 7, 9…\]).

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Page 6

Question: ? Using the pattern above, find \[36^{2}\], given that \[35^{2}=1225\]

Solution: Because \[35^{2}\] is the sum of the first 35 odd numbers, \[36^{2}\] is found by adding the 36th odd number to 1225. The 36th odd number is \[2(36) – 1 = 71\]. Therefore, \[1225 + 71 = 1296\], which is \[36^{2}\].

Question: ? How do we find the 36th odd number?

Solution: Use the formula for the $n^{th}$ odd number, which is \[2n – 1\]. For the 36th odd number: \[2(36) – 1 = 71\].

Question: ? What is the \[n^{th}\] odd number?

Solution: The \[n^{th}\] odd number is \[2n – 1\]].

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Page 7

Question: ? Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

Solution: Between consecutive squares \[n^{2}\] and \[(n+1)^{2}\], there are strictly \[2n\] non-square numbers. For example, between \[1^{2}\] (1) and \[2^{2}\] (4), there are \[2(1) = 2\] numbers (2 and 3).

” and cannot be part of an unbroken loop.

Question: How many square numbers are there between 1 and 100? How many are between 101 and 200?

Solution:

There are 8 square numbers strictly between 1 and 100 (4, 9, 16, 25, 36, 49, 64, 81). If the bounds are inclusive, there are 10 ($1^{2}$12 through $10^{2}$102).

There are 4 square numbers between 101 and 200: 121 ($11^{2}$112), 144 ($12^{2}$122), 169 ($13^{2}$132), and 196 ($14^{2}$142).

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Question:

Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100.

Solution:

1–100: 10

101–200: 4

201–300: 3 ($15^{2}=225, 16^{2}=256, 17^{2}=289$152=225,162=256,172=289)

301–400: 3 ($18^{2}=324, 19^{2}=361, 20^{2}=400$)

401–500: 2 ($21^{2}=441, 22^{2}=484$212=441,222=484)

501–600: 2 ($23^{2}=529, 24^{2}=576$232=529,242=576)

601–700: 2 ($25^{2}=625, 26^{2}=676$252=625,262=676)

701–800: 2 ($27^{2}=729, 28^{2}=784$272=729,282=784)

801–900: 2 ($29^{2}=841, 30^{2}=900$292=841,302=900)

901–1000: 1 ($31^{2}=961$312=961)

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Question:

What is the largest square less than 1000?

Solution:

The largest square less than 1000 is 961 ($31^{2}$312).

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Question:

Can you see any relation between triangular numbers and square numbers?

Solution:

The sum of any two consecutive triangular numbers results in a perfect square number.

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Question:

Extend the pattern shown and draw the next term.

Solution:

Following $1+3=4$1+3=4, $3+6=9$3+6=9, and $6+10=16$6+10=16, the next term combines the triangular numbers 10 and 15 to form
$10+15=25=5^{2}$10+15=25=52.

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Question:

The area of a square is 49 sq. cm. What is the length of its side?

Solution:

Because $7\times7=49$7×7=49 (or $7^{2}=49$72=49), the length of the side is 7 cm.

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Page 8

Question:

What is the square root of 64? What about −8×−8 = 64?

Solution:

The square root of 64 is 8, since $8\times8=64$8×8=64. Also, $(-8)^{2}=64$(−8)2=64, which means the square roots of 64 are both +8 and −8.

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Question:

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?

Solution:

You can first verify the units digit (e.g., 327 ends in 7, so it cannot be a perfect square). For numbers like 576, you can use successive subtraction of odd numbers, compare with nearby squares (between $24^{2}$242 and $25^{2}$252), or use prime factorization.

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Question:

Can we find the square root of 729 using this method?

Solution:

Yes, but it will take 27 steps using successive subtraction, so it is time-consuming.

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Page 9

Question:

Will looking at a number’s prime factorisation help in determining whether it is a perfect square?

Solution:

Yes. If the prime factors can be grouped into equal pairs, the number is a perfect square.

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Question:

Is 324 a perfect square?

Solution:

Yes.
$2\times2\times3\times3\times3\times3 = (2\times3\times3)\times(2\times3\times3) = 18^{2}$2×2×3×3×3×3=(2×3×3)×(2×3×3)=182
Thus, $\sqrt{324}=18$324​=18.

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Question:

Is 156 a perfect square?

Solution:

No. 156 =
$2\times2\times3\times13$
Since 3 and 13 are unpaired, it is not a perfect square.

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Question:

Find whether 1156 and 2800 are perfect squares using prime factorisation.

Solution:

$1156 = 2\times2\times17\times17 = (2\times17)^{2} = 34^{2}$1156=2×2×17×17=(2×17)2=342
So, it is a perfect square.

$2800 = 2\times2\times2\times2\times5\times5\times7$2800=2×2×2×2×5×5×7
Since 7 is unpaired, it is not a perfect square.

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Page 10

Question 1:

Which of the following numbers are not perfect squares?

Solution:

2032, 2048, and 1027 are not perfect squares.
1089 is $33^{2}$332.

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Question 2:

Which one among $64^{2}, 108^{2}, 292^{2}, 36^{2}$642,1082,2922,362 has last digit 4?

Solution:

$108^{2}$ and $292^{2}$

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Question 3:

Given $125^{2}=15625$1252=15625, what is $126^{2}$1262?

Solution:

(iv) $15625 + 251$15625+251
Since difference = $125 + 126 = 251$125+126=251.

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Question 4:

Find the side of a square with area 441 m².

Solution:

$\sqrt{441} = 21$441​=21
Side = 21 m.

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Question 5:

Find the smallest square divisible by 4, 9, and 10.

Solution:

LCM = 180
$180 = 2^{2}\times3^{2}\times5$180=22×32×5
Multiply by 5:
$180\times5 = 900$180×5=900

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Question 6:

Find smallest number to multiply 9408 to make it a square.

Solution:

$9408 = 2^{6}\times3\times7^{2}$
Multiply by 3:
$9408\times3 = 28224$

Square root:
$2^{3}\times3\times7 = 168$23×3×7=168

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Question 7:

Numbers between squares

Solution:

(i) Between $16^{2}$ and $17^{2}$: $2\times16 = 32$2×16=32

(ii) Between $99^{2}$992 and $100^{2}$1002: $2\times99 = 198$2×99=198

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Question 8:

Fill pattern

Solution:

$4^{2}+5^{2}+20^{2} = 21^{2}$42+52+202=212

$9^{2}+10^{2}+90^{2} = 91^{2}$92+102+902=912