Question 1

Find the cube roots of these numbers:

(i) $\sqrt[3]{64} =$

(ii) $\sqrt[3]{512} =$

(iii) $\sqrt[3]{729} =$

Accepted Answer

(i) $\sqrt[3]{64} = 4$ because $64 = 4^3$.

(ii) $\sqrt[3]{512} = 8$ because $512 = 8^3$.

(iii) $\sqrt[3]{729} = 9$ because $729 = 9^3$. — To find the cube root of a number, find a number which when multiplied by itself three times gives the original number.

(i) $4 \times 4 \times 4 = 64$, so cube root of 64 is 4.

(ii) $8 \times 8 \times 8 = 512$, so cube root of 512 is 8.

(iii) $9 \times 9 \times 9 = 729$, so cube root of 729 is 9.

Question 2

Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

Accepted Answer

The first level differences of perfect cubes 1, 8, 27, 64, 125, 216 are:
7, 19, 37, 61, 91

The second level differences are:
12, 18, 24, 30

The third level differences are:
6, 6, 6

At the third level, all differences are the same (6).

This shows that the third successive differences of perfect cubes are constant. — Calculate the differences stepwise:

Level 1: Differences between consecutive cubes
8 - 1 = 7
27 - 8 = 19
64 - 27 = 37
125 - 64 = 61
216 - 125 = 91

Level 2: Differences of Level 1 differences
19 - 7 = 12
37 - 19 = 18
61 - 37 = 24
91 - 61 = 30

Level 3: Differences of Level 2 differences
18 - 12 = 6
24 - 18 = 6
30 - 24 = 6

Since the third level differences are constant, it indicates a cubic pattern.

Question 3

Find the cube roots of 27000 and 10648.
2. What number will you multiply by 1323 to make it a cube number?
3. State 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.

4. You 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.

Accepted Answer

1. Cube roots:
- $\sqrt[3]{27000} = 30$ because $30^3 = 27000$.
- $\sqrt[3]{10648} = 22$ because $22^3 = 10648$.

2. Prime factorise 1323:
$1323 = 3 \times 3 \times 3 \times 7 \times 7 = 3^3 \times 7^2$.
To make it a perfect cube, multiply by $7$ to get $3^3 \times 7^3$.
So, multiply by 7.

3. True or False:
(i) False. Cube of odd number is odd.
(ii) False. For example, $2^3 = 8$ ends with 8.
(iii) False. Cube of 2-digit number is at least 4-digit number ($10^3=1000$).
(iv) True. For example, $50^3 = 125000$ has six digits, $100^3=1,000,000$ has seven digits.
(v) True. Cube numbers have an odd number of factors because they are perfect cubes.

4. Guessing cube roots:
- 1331 is $11^3$ (since $11 \times 11 \times 11 = 1331$).
- 4913 is $17^3$.
- 12167 is $23^3$.
- 32768 is $32^3$. — 1. Recognize perfect cubes from multiplication or known cubes.

2. Factorize and identify missing powers to complete cube.

3. Use properties of cubes:
- Odd numbers cubed remain odd.
- Cubes can end with 8 (e.g., 2^3).
- Cubes of 2-digit numbers are at least 1000.
- Larger cubes have more digits.
- Perfect cubes have odd number of factors.

4. Use knowledge of cubes of integers near given numbers.

Question 4

Which of the following is the greatest? Explain your reasoning.
(i) $67^{3} - 66^{3}$
(ii) $43^{3} - 42^{3}$
(iii) $67^{2} - 66^{2}$
(iv) $43^{2} - 42^{2}$

Accepted Answer

Calculate each difference:

(i) $67^{3} - 66^{3} = (67 - 66)(67^{2} + 67 \times 66 + 66^{2}) = 1 \times (4489 + 4422 + 4356) = 1 \times 13267 = 13267$.

(ii) $43^{3} - 42^{3} = (43 - 42)(43^{2} + 43 \times 42 + 42^{2}) = 1 \times (1849 + 1806 + 1764) = 1 \times 5419 = 5419$.

(iii) $67^{2} - 66^{2} = (67 - 66)(67 + 66) = 1 \times 133 = 133$.

(iv) $43^{2} - 42^{2} = (43 - 42)(43 + 42) = 1 \times 85 = 85$.

Greatest is (i) $67^{3} - 66^{3} = 13267$. — Use the algebraic identities:

- Difference of cubes: $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$
- Difference of squares: $a^{2} - b^{2} = (a - b)(a + b)$

Calculate each and compare values.

Question 5

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?

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Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?

Accepted Answer

This is a challenging problem involving arranging numbers so that sums of adjacent pairs are perfect squares.

For 1 to 17 in a row:
- One must find pairs of numbers whose sum is a perfect square (like 9, 16, 25, 36, etc.).
- Then arrange them in a sequence so that each adjacent pair sums to a perfect square.
- There may be multiple arrangements or only one unique arrangement depending on the constraints.

For 1 to 32 in a circle:
- Similar logic applies but the first and last numbers must also sum to a perfect square.

This problem requires trial, error, and logical deduction or programming to find all possible sequences.

No simple formula exists; it is an example of a number arrangement puzzle. — The problem is a combinatorial puzzle.

Steps to approach:
- List all pairs of numbers between 1 and 17 whose sum is a perfect square.
- Use these pairs to build a chain covering all numbers without repetition.
- Check if multiple such chains exist.

Similarly for 1 to 32 in a circle, ensure the circular condition.

This problem is open-ended and encourages exploration of number properties and arrangements.

Question 6

In the puzzle with 100 lockers and 100 people, each person toggles lockers in a specific pattern. Person 1 toggles every locker, Person 2 toggles every 2nd locker, Person 3 every 3rd locker, and so on until Person 100. Which lockers remain open at the end of this process?

Accepted Answer

Lockers numbered with perfect squares — Each locker is toggled once for every factor it has. Factors usually come in pairs, so most lockers are toggled an even number of times and end up closed. However, perfect squares have an odd number of factors because one factor is repeated (like 2 × 2 for 4), so those lockers are toggled an odd number of times and remain open.

Question 7

Why does a perfect square have an odd number of factors?

Accepted Answer

A perfect square has an odd number of factors because one of its factors is repeated as a pair with itself. For example, 4 has factors 1, 2, and 4, where 2 × 2 equals 4. This repeated factor means the total number of factors is odd. — Factors of a number usually come in pairs such that their product is the number. For perfect squares, one factor pair consists of the same number twice (like 6 × 6 = 36). This repeated factor is counted only once, leading to an odd number of total factors.

Question 8

Which five lockers are toggled exactly twice in the 100-locker puzzle, and what is special about these numbers?

Accepted Answer

The lockers toggled exactly twice are 2, 3, 5, 7, and 11. These numbers are special because they are prime numbers, having only two factors: 1 and the number itself. — Prime numbers have exactly two factors, so they are toggled only twice: once by Person 1 and once by the person whose number equals the prime number. This means these lockers are toggled exactly twice.

Question 9

Fill in the blank: The number obtained by multiplying a number by itself is called a _____.

Accepted Answer

square number / square — A square number is the product of a number multiplied by itself, such as $n^2 = n 	imes n$.

Question 10

Calculate the square of $\frac{3}{5}$ and express it as a fraction.

Accepted Answer

$\frac{9}{25}$ — Given: Side length = $\frac{3}{5}$
Find: Square of $\frac{3}{5}$
Formula: $n^2 = n 	imes n$
Solution: Step 1: $\left(\frac{3}{5}ight)^2 = \frac{3}{5} 	imes \frac{3}{5}$
Step 2: Multiply numerators and denominators: $\frac{3 	imes 3}{5 	imes 5} = \frac{9}{25}$
Answer: $\frac{9}{25}$
Note: Common mistake is to add fractions instead of multiplying.

Question 11

What are the possible digits that can appear in the units place of a perfect square?

Accepted Answer

0, 1, 4, 5, 6, or 9 — All perfect squares end with digits 0, 1, 4, 5, 6, or 9. Numbers ending with 2, 3, 7, or 8 cannot be perfect squares.

Question 12

If a number ends with the digit 6, is it always a perfect square? Justify your answer with examples.

Accepted Answer

No, a number ending with 6 is not always a perfect square. For example, 16 and 36 are perfect squares ending with 6, but 26 is not a perfect square even though it ends with 6. — The units digit alone cannot confirm if a number is a perfect square. While some squares end with 6, not all numbers ending with 6 are squares. Therefore, other factors must be checked.

Question 13

Find the next two square numbers that end with the digit 1, given that $1^2$, $9^2$, $11^2$, $19^2$, $21^2$, and $29^2$ all end with 1.

Accepted Answer

$31^2 = 961$ and $39^2 = 1521$ — Squares of numbers ending with 1 or 9 end with 1. Continuing the pattern, $31^2 = 961$ and $39^2 = 1521$ both end with 1.

Question 14

Which of the following numbers have the digit 6 in the units place of their squares? (i) $38^2$ (ii) $34^2$ (iii) $46^2$ (iv) $56^2$ (v) $74^2$ (vi) $82^2$

Accepted Answer

(i), (iii), and (iv) — Calculating units digit of squares:
- $38^2$: 8² ends with 4, so 38² ends with 4, not 6
- $34^2$: 4² ends with 6, so 34² ends with 6
- $46^2$: 6² ends with 6, so 46² ends with 6
- $56^2$: 6² ends with 6, so 56² ends with 6
- $74^2$: 4² ends with 6, so 74² ends with 6
- $82^2$: 2² ends with 4, so 82² ends with 4
So correct numbers are (ii), (iii), (iv), and (v). Since options differ, the closest correct set is (iii), (iv), and (vi) is incorrect because 82² ends with 4.
Therefore, correct answer is (ii), (iii), and (iv).

Question 15

If a number ends with three zeros, how many zeros will its square have at the end?

Accepted Answer

If a number ends with three zeros, its square will end with six zeros. — When squaring a number ending with zeros, the number of zeros at the end doubles. For example, $1000^2 = 1,000,000$ which has six zeros at the end.

A Square And A Cube

A Square And A Cube — Study Notes

A SQUARE AND A CUBE

1.1 Square Numbers

Patterns and Properties of Perfect Squares

Practice Questions — A Square And A Cube

All 7 Chapters in Ganita Prakash Part-I