Class 8 Mathematics · Chapter 1

1. A Square and A Cube

Solutions to all Figure it Out and in-text questions. CBSE Class 8 Mathematics, NCERT Ganita Prakash Part 1.

Locker Puzzle (Introduction)

1Does every number have an even number of factors?

No. Some numbers have an odd number of factors.

For example, 1 has only 1 factor. 4 has 3 factors (1, 2, 4). 9 has 3 factors (1, 3, 9).

2Can you use this insight to find more numbers with an odd number of factors?

Yes. When a factor pairs with itself (like 2 × 2 or 3 × 3), it is counted only once.

So numbers like 16 (4 × 4), 25 (5 × 5), 36 (6 × 6), 49 (7 × 7) have an odd number of factors.

Check for 36: factors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Pairs are 1 × 36, 2 × 18, 3 × 12, 4 × 9 and 6 × 6. So 36 has 9 factors, an odd number.

3Write the locker numbers that remain open.

Only lockers with square numbers remain open.

Open lockers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

4Which are the first five lockers that were touched exactly twice?

A locker is touched exactly twice when its number has only two factors. Such numbers are prime numbers.

The lockers are 2, 3, 5, 7, 11. So the passcode is 2-3-5-7-11.

1.1 Square Numbers

5Find the squares of the first 30 natural numbers and fill in the table.

1 to 10	11 to 20	21 to 30
1² = 1	11² = 121	21² = 441
2² = 4	12² = 144	22² = 484
3² = 9	13² = 169	23² = 529
4² = 16	14² = 196	24² = 576
5² = 25	15² = 225	25² = 625
6² = 36	16² = 256	26² = 676
7² = 49	17² = 289	27² = 729
8² = 64	18² = 324	28² = 784
9² = 81	19² = 361	29² = 841
10² = 100	20² = 400	30² = 900

Some patterns:

Squares end only with 0, 1, 4, 5, 6 or 9.
Squares of even numbers are even. Squares of odd numbers are odd.
Numbers ending in 5 have squares ending in 25.
Numbers ending in 0 have squares ending in 00.

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

No. For example, 26 ends in 6, but 26 is not a square. 10, 11, 14 and 15 are also not squares.

7Write 5 numbers such that you can determine by looking at their units digit that they are not squares.

Numbers ending in 2, 3, 7 or 8 are never squares.

Examples: 72, 133, 547, 1008, 2002

8The squares 1², 9², 11², 19², 21² and 29² all have 1 in their units place. Write the next two squares.

The next two are 31² = 961 and 39² = 1521.

9Which 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.

So the answer is (ii) 34², (iii) 46², (iv) 56² and (v) 74².

38² and 82² end in 4.

10If a number contains 3 zeros at the end, how many zeros will its square have at the end?

Its square will have 6 zeros at the end. For example, 1000² = 1000000.

11What do you notice about the number of zeros at the end of a number and of its square? Can squares have only an even number of zeros at the end?

The square has double the number of zeros of the number.

Yes, this always happens. So a perfect square can only have an even number of zeros at the end.

12What can you say about the parity of a number and its square?

A number and its square have the same parity.

The square of an even number is even. The square of an odd number is odd.

13See if the pattern of differences between consecutive squares continues.

Yes. 36 − 25 = 11, 49 − 36 = 13, 64 − 49 = 15, 81 − 64 = 17.

The differences are consecutive odd numbers.

14Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?

Between n² and (n + 1)² there are 2n numbers.

For example, between 1 and 4 there are 2 numbers. Between 4 and 9 there are 4 numbers. Between 9 and 16 there are 6 numbers.

15How many square numbers are there in each block of 100 up to 1000? What is the largest square less than 1000?

Block	No. of squares
1 – 100	10
101 – 200	4
201 – 300	3
301 – 400	3
401 – 500	2
501 – 600	2
601 – 700	2
701 – 800	2
801 – 900	2
901 – 1000	1

The largest square less than 1000 is 961 = 31².

16Can you see any relation between triangular numbers and square numbers?

The sum of two consecutive triangular numbers is a square number.

1 + 3 = 4, 3 + 6 = 9, 6 + 10 = 16. The next term is 10 + 15 = 25 = 5².

17Find whether 1156 and 2800 are perfect squares using prime factorisation.

1156 = 2 × 2 × 17 × 17 = (2 × 17) × (2 × 17) = 34².

So 1156 is a perfect square and √1156 = 34.

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7. The factor 7 has no pair.

So 2800 is not a perfect square.

Figure it Out (Page 10)

1Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089

2032 and 2048 end in 2 and 8. 1027 ends in 7. Squares never end in 2, 3, 7 or 8.

1089 = 33² is a perfect square.

So (i) 2032, (ii) 2048 and (iii) 1027 are not perfect squares.

2Which one among 64², 108², 292², 36² has last digit 4?

A square ends in 4 when the number ends in 2 or 8.

So 108² and 292² have last digit 4.

64² and 36² end in 6.

3Given 125² = 15625, what is the value of 126²?

126² = 125² + 126th odd number.

The 126th odd number = 2 × 126 − 1 = 251.

So 126² = 15625 + 251 = 15876. The answer is (iv) 15625 + 251.

4Find the length of the side of a square whose area is 441 m².

441 = 3 × 3 × 7 × 7 = (3 × 7)² = 21².

Side of the square = 21 m

5Find the smallest square number that is divisible by each of the following numbers: 4, 9 and 10.

LCM of 4, 9 and 10 = 180.

180 = 2 × 2 × 3 × 3 × 5. The factor 5 has no pair.

So multiply by 5: 180 × 5 = 900 = 30².

The smallest square number is 900

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.

9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7.

The factor 3 has no pair. So we multiply by 3.

9408 × 3 = 28224 = (2 × 2 × 2 × 3 × 7)².

Square root of 28224 = 2 × 2 × 2 × 3 × 7 = 168

7How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100

Between n² and (n + 1)² there are 2n numbers.

(i) 2 × 16 = 32 numbers (between 256 and 289).

(ii) 2 × 99 = 198 numbers (between 9801 and 10000).

8In the following pattern, fill in the missing numbers.

1² + 2² + 2² = 3² 2² + 3² + 6² = 7² 3² + 4² + 12² = 13² 4² + 5² + 20² = 21² 9² + 10² + 90² = 91²

The third number is the product of the first two (4 × 5 = 20, 9 × 10 = 90).

The number on the right is one more than the third number (21, 91).

9How many tiny squares are there in the picture? Write the prime factorisation of the number of tiny squares.

The picture has 7 rows and 9 columns of big squares. So there are 7 × 9 = 63 big squares.

Each big square has 5 × 5 = 25 tiny squares.

Total tiny squares = 63 × 25 = 1575

Prime factorisation: 1575 = 3 × 3 × 5 × 5 × 7 = 3² × 5² × 7

1.2 Cubic Numbers

18How many cubes of side 1 cm make a cube of side 2 cm? Of side 3 cm?

Side 2 cm: 2 × 2 × 2 = 8 cubes.

Side 3 cm: 3 × 3 × 3 = 27 cubes.

19Complete the table of cubes.

1 to 10	11 to 20
1³ = 1	11³ = 1331
2³ = 8	12³ = 1728
3³ = 27	13³ = 2197
4³ = 64	14³ = 2744
5³ = 125	15³ = 3375
6³ = 216	16³ = 4096
7³ = 343	17³ = 4913
8³ = 512	18³ = 5832
9³ = 729	19³ = 6859
10³ = 1000	20³ = 8000

Patterns: Cubes of even numbers are even and cubes of odd numbers are odd. A number ending in 1, 4, 5, 6, 9 or 0 has a cube ending in the same digit. Numbers ending in 2 give cubes ending in 8, and numbers ending in 8 give cubes ending in 2. Similarly, 3 gives 7 and 7 gives 3.

20What are the possible last digits of cubes?

A cube can end with any digit 0 to 9.

For example, 1000, 1, 512, 27, 64, 125, 216, 343, 8, 729.

21Find the number of cubes with 1 digit, 2 digits and 3 digits. What do you observe?

1 digit: 1, 8 → 2 cubes.

2 digits: 27, 64 → 2 cubes.

3 digits: 125, 216, 343, 512, 729 → 5 cubes.

Cubes are spread out more than squares. There are very few cubes among small numbers.

22Can a cube end with exactly two zeroes (00)? Explain.

No. If a number ends with one zero, its cube ends with three zeros.

A cube always has 0, 3, 6, … zeros at the end. So it cannot end with exactly two zeros.

23Find the two ways in which 4104 and 13832 can be expressed as the sum of two positive cubes.

4104 = 2³ + 16³ = 9³ + 15³

13832 = 2³ + 24³ = 18³ + 20³

2491 + 93 + 95 + … + 109 (10 consecutive odd numbers). Can you tell what this sum is without doing the calculation?

In this pattern, the sum of n consecutive odd numbers in a row gives n³.

Here there are 10 numbers. So the sum = 10³ = 1000

25Find the cube roots of these numbers: (i) ∛64 (ii) ∛512 (iii) ∛729

(i) 64 = 4 × 4 × 4, so ∛64 = 4

(ii) 512 = 8 × 8 × 8, so ∛512 = 8

(iii) 729 = 9 × 9 × 9, so ∛729 = 9

26Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?

Cubes : 1 8 27 64 125 216 Level 1 : 7 19 37 61 91 Level 2 : 12 18 24 30 Level 3 : 6 6 6

For cubes, all differences become equal (6) at Level 3. For squares, this happens at Level 2.

Figure it Out (Page 16)

1Find the cube roots of 27000 and 10648.

27000 = 27 × 1000 = (3 × 10)³. So ∛27000 = 30

10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2 × 11)³. So ∛10648 = 22

2What number will you multiply by 1323 to make it a cube number?

1323 = 3 × 3 × 3 × 7 × 7.

The factor 7 appears only two times. We need one more 7.

So multiply by 7. 1323 × 7 = 9261 = 21³.

3State true or false. Explain your reasoning.

(i) The cube of any odd number is even.

False. Odd × odd × odd is always odd. Example: 3³ = 27.

(ii) There is no perfect cube that ends with 8.

False. 2³ = 8 and 12³ = 1728 end with 8.

(iii) The cube of a 2-digit number may be a 3-digit number.

False. The smallest 2-digit number is 10 and 10³ = 1000 has 4 digits.

(iv) The cube of a 2-digit number may have seven or more digits.

False. The largest 2-digit number is 99 and 99³ = 970299 has only 6 digits.

(v) Cube numbers have an odd number of factors.

False. 8 has factors 1, 2, 4, 8, that is 4 factors. Only squares have an odd number of factors.

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.

1331: It ends in 1, so the cube root ends in 1. 10³ = 1000 and 20³ = 8000. So ∛1331 = 11

4913: It ends in 3, so the cube root ends in 7. It lies between 1000 and 8000. So ∛4913 = 17

12167: It ends in 7, so the cube root ends in 3. It lies between 8000 (20³) and 27000 (30³). So ∛12167 = 23

32768: It ends in 8, so the cube root ends in 2. It lies between 27000 (30³) and 64000 (40³). So ∛32768 = 32

5Which of the following is the greatest? Explain your reasoning. (i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²

The difference between consecutive cubes grows faster than that between consecutive squares. It is also bigger for bigger numbers.

(i) 67³ − 66³ = 13267

(ii) 43³ − 42³ = 5419

(iii) 67² − 66² = 133

(iv) 43² − 42² = 85

So (i) 67³ − 66³ is the greatest.

Puzzle: Square Pairs!

1Arrange the numbers 1 to 17 in a row so that the sum of every adjacent pair is a square. Can you arrange them in more than one way?

16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17

Check: 16 + 9 = 25, 9 + 7 = 16, 7 + 2 = 9, 2 + 14 = 16, and so on.

There is only one way (and its reverse).

Reason: 16 can pair only with 9 (16 + 9 = 25). 17 can pair only with 8 (17 + 8 = 25). So 16 and 17 must be at the two ends, and the rest of the row is then fixed.

2Can you do the same with numbers from 1 to 32 arranged in a circle?

Yes. One arrangement in a circle is:

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

The last number 15 joins back to 1, and 15 + 1 = 16.

2. Power Play →