AP 8th Maths Textbook Solutions – 1. A Square and A Cube (2026-27)

1 A SQUARE AND A CUBE

1.0 Introduction

• 1. Does every number have an even number of factors?
  No, not every number has an even number of factors. Numbers that are perfect squares have an odd number of factors.
• 2. For instance, 36 has a factor pair 6 × 6 where both numbers are 6. Does this number have an odd number of factors? Check if this is true.
  Yes. The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Since it has exactly 9 factors, it has an odd number of factors.
• 3. Write the locker numbers that remain open.
  The lockers that remain open are those whose numbers are perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
• 4. Which are these five lockers? (First five locker numbers that were touched exactly twice)
  The lockers that are toggled exactly twice are the prime numbers. The first five prime numbers are 2, 3, 5, 7, and 11.

1.1.1 Patterns and Properties of Perfect Squares

• 1. Find the squares of the first 30 natural numbers and fill in the table below.

  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

• 2. What patterns do you notice? Share your observations and make conjectures.
  All perfect squares end with the digits 0, 1, 4, 5, 6, or 9. None of them end in 2, 3, 7, or 8.
• 3. Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
  12, 23, 37, 48, and 52. (Any number ending in 2, 3, 7, or 8 is not a perfect square).
• 4. The squares, 1², 9², 11², 19², 21², and 29², all have 1 in their units place. Write the next two squares.
  The next two squares ending in 1 are 31² (which is 961) and 39² (which is 1521).
• 5. 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
  Numbers ending in 4 or 6 will have squares ending in 6. Therefore, (ii) 34², (iii) 46², (iv) 56², and (v) 74² have the digit 6 in the units place.
• 6. Find more such patterns by observing the numbers and their squares from the table you filled earlier.
  Any number ending in 5 always has its square ending in 25. Any number ending in 0 has its square ending in an even number of zeroes.

• 1. If 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. The number of zeros gets doubled.
• 2. 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?
  The number of zeros at the end of a square is exactly twice the number of zeros at the end of the original number. Yes, this will always happen.
• 3. Can we say that squares can only have an even number of zeros at the end?
  Yes, perfect squares always end with an even number of zeros.
• 4. What can you say about the parity of a number and its square?
  The parity remains the same. The square of an even number is even, and the square of an odd number is odd.
• 5. Let us explore the differences between consecutive squares. What do you notice? 4 - 1 = 3, 16 - 9 = 7, 25 - 16 = 9. See if this pattern continues for the next few square numbers.
  Yes, the pattern continues: 36 - 25 = 11, 49 - 36 = 13, 64 - 49 = 15. The differences between consecutive perfect squares are consecutive odd numbers.

• 1. Using the pattern above, find 36², given that 35² = 1225.
  1225 is the sum of the first 35 odd numbers. To find 36², we add the 36th odd number (which is 2 × 36 - 1 = 71). 1225 + 71 = 1296.
• 2. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
  Between any two consecutive perfect squares n² and (n+1)², there are exactly 2n non-square numbers.
• 3. 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.

  1-100	101-200	201-300	301-400	401-500
  10	4	3	3	2

  
  

  501-600	601-700	701-800	801-900	901-1000
  2	2	2	2	1

• 4. What is the largest square less than 1000?
  The largest perfect square less than 1000 is 961 (which is 31²).

• 1. Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.
  The sum of two consecutive triangular numbers results in a perfect square. The next term in the pattern is 10 + 15 = 25 (which is 5²).
• 2. The area of a square is 49 sq. cm. What is the length of its side?
  The length of the side is 7 cm, since 7 × 7 = 49.
• 3. What is the square root of 64?
  The square roots of 64 are 8 and -8, because 8² = 64 and (-8)² = 64. (Typically we consider the positive square root, 8).

1.1.5 Estimating

• 1. Find whether 1156 and 2800 are perfect squares using prime factorisation.
  For 1156: Prime factorisation is 2 × 2 × 17 × 17 = (2 × 17)² = 34². Since the factors can be grouped into pairs, 1156 is a perfect square.
  
  For 2800: Prime factorisation is 2 × 2 × 2 × 2 × 5 × 5 × 7. The prime factor 7 is left without a pair. Therefore, 2800 is not a perfect square.

Figure it Out - 1.1

• 1. Which of the following numbers are not perfect squares?
  (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
  (i) 2032 (ends in 2), (ii) 2048 (ends in 8), and (iii) 1027 (ends in 7) are not perfect squares. Any number ending in 2, 3, 7, or 8 cannot be a perfect square.
• 2. Which one among 64², 108², 292², 36² has last digit 4?
  Both 108² (since 8 × 8 = 64) and 292² (since 2 × 2 = 4) have 4 as their last digit.
• 3. Given 125² = 15625, what is the value of 126²?
  (i) 15625 + 126
  (ii) 15625 + 26²
  (iii) 15625 + 253
  (iv) 15625 + 251
  (v) 15625 + 51²
  Using the property (n+1)² = n² + n + (n+1), we get 126² = 125² + 125 + 126 = 15625 + 251. Therefore, (iv) 15625 + 251 is the correct value.
• 4. Find the length of the side of a square whose area is 441 m².
  Side = √441 = 21 m.
• 5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
  The LCM of 4, 9, and 10 is 180. The prime factorisation of 180 is 2² × 3² × 5. To make it a perfect square, we must multiply by 5. 180 × 5 = 900. The smallest square number is 900.
• 6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
  The prime factorisation of 9408 is 2⁶ × 3 × 7². The factor 3 has no pair. We must multiply by 3 to make it a perfect square. The new product is 9408 × 3 = 28224. Its square root is 2³ × 3 × 7 = 168.
• 7. How many numbers lie between the squares of the following numbers?
  (i) 16 and 17
  (ii) 99 and 100
  (i) Between 16² and 17², there are 2 × 16 = 32 numbers.
  (ii) Between 99² and 100², there are 2 × 99 = 198 numbers.
• 8. In 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
  Based on the pattern n² + (n+1)² + (n(n+1))² = (n(n+1)+1)²:
  4² + 5² + 20² = (21)²
  9² + 10² + (90)² = (91)²
• 9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
  Count the total number of squares arranged in the grid pattern provided in your textbook. Once you find the total number (which is a perfect square), break it down into its prime factors and write them in pairs.

1.2 Cubic Numbers

• 1. How many cubes of side 1 cm make a cube of side 2 cm?
  2 × 2 × 2 = 8 cubes.
• 2. How many cubes of side 1 cm will make a cube of side 3 cm?
  3 × 3 × 3 = 27 cubes.
• 3. Can you estimate the number of unit cubes in a cube with an edge length of 4 units?
  It will have 4 × 4 × 4 = 64 unit cubes.

• 1. Complete the table below.

  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

• 2. What patterns do you notice in the table above?
  The cube of an even number is always even, and the cube of an odd number is always odd.
• 3. What are the possible last digits of cubes?
  Cubes can end with any digit from 0 to 9. (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).
• 4. Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
  There are two 1-digit cubes (1, 8), two 2-digit cubes (27, 64), and five 3-digit cubes (125, 216, 343, 512, 729).
• 5. Can a cube end with exactly two zeroes (00)? Explain.
  No, a perfect cube cannot end with exactly two zeros. If a number ends with a zero, its cube will end with exactly three zeros (or a multiple of three zeros).

1.2.1 Taxicab Numbers

• 1. 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.
  4104 = 2³ + 16³ and 4104 = 9³ + 15³.
  13832 = 2³ + 24³ and 13832 = 18³ + 20³.
• 2. Can you tell what this sum is without doing the calculation?
  91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109
  Based on the consecutive odd numbers pattern for cubes, this is the sum of 10 consecutive odd numbers corresponding to the 10th cube. Therefore, the sum is 10³ = 1000.

1.2.3 Cube Roots

• 1. Find the cube roots of these numbers:
  (i) ∛64
  (ii) ∛512
  (iii) ∛729
  (i) ∛64 = 4
  (ii) ∛512 = 8
  (iii) ∛729 = 9
• 2. 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
  Level 1 differences: 7, 19, 37, 61, 91
  Level 2 differences: 12, 18, 24, 30
  Level 3 differences: 6, 6, 6
  Notice that for perfect cubes, all the differences become equal (to 6) at Level 3.

Figure it Out - 1.2

• 1. Find the cube roots of 27000 and 10648.
  ∛27000 = ∛(30 × 30 × 30) = 30.
  ∛10648 = ∛(22 × 22 × 22) = 22.
• 2. What number will you multiply by 1323 to make it a cube number?
  The prime factorisation of 1323 is 3³ × 7². To make it a perfect cube, we need to multiply it by 7.
• 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.
  (i) False. The cube of an odd number is always odd.
  (ii) False. The cube of a number ending in 2 ends in 8 (e.g., 2³ = 8).
  (iii) False. The smallest 2-digit number is 10, and 10³ = 1000, which is a 4-digit number.
  (iv) False. The largest 2-digit number is 99, and 99³ = 970299, which has exactly 6 digits.
  (v) False. Perfect squares have an odd number of factors, but perfect cubes (unless they are also perfect squares) have an even 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.
  Yes, by estimation based on the last digit and nearest tens:
  ∛1331 = 11 (ends in 1, between 10³ and 20³)
  ∛4913 = 17 (ends in 3 so root ends in 7, between 10³ and 20³)
  ∛12167 = 23 (ends in 7 so root ends in 3, between 20³ and 30³)
  ∛32768 = 32 (ends in 8 so root ends in 2, between 30³ and 40³)
• 5. Which of the following is the greatest? Explain your reasoning.
  (i) 67³ - 66³
  (ii) 43³ - 42³
  (iii) 67² - 66²
  (iv) 43² - 42²
  (i) 67³ - 66³ is the greatest. The differences between consecutive cubes grow much faster than differences between consecutive squares, and 67 & 66 are the largest pair.

IT'S PUZZLE TIME! Square Pairs!

• 1. 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?
  The arrangement is: 17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.
  No, you cannot arrange them in any other fundamentally different way (other than exact reverse). This is because numbers like 16 and 17 only have one possible partner each (9 and 8 respectively) to form a perfect square within the 1-17 range, forcing them to be fixed at the endpoints.
• 2. Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?
  Yes. In a circle, every number requires exactly two partners to sum up to perfect squares. An arrangement is possible since within the range up to 32, every number can be paired with two distinct numbers that yield perfect square sums.

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