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?
Solution: He realised that a locker will only remain open if it is touched an odd number of times. The number of times a locker is touched equals its number of factors. Because only perfect squares have an odd number of factors, he knew that lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 would be the ones left open.
Page 2
Question: Does every number have an even number of factors?
Solution: No, perfect square numbers always 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 × 6… Does this number have an odd number of factors?
Solution: Yes, any perfect square (like 1, 4, 9, 16, 25, etc.) will have an odd number of factors. Since 36 is a perfect square (6²), it has an odd number of factors (1, 2, 3, 4, 6, 9, 12, 18, 36).
Page 3
Question: Write the locker numbers that remain open.
Solution: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Question: Which are these five lockers? (The passcode consists of the first five locker numbers that were touched exactly twice).
Solution: Lockers touched exactly twice represent prime numbers. The first five prime-numbered lockers are 2, 3, 5, 7, and 11.
Question: Can we have a square of sidelength 3/5 or 2.5 units?
Solution: Yes. The area would simply be the square of those values: (3/5)² = 9/25 square units, and 2.5² = 6.25 square units.
Page 4
Question: Find the squares of the first 30 natural numbers and fill in the table below.
Solution:
- 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100
- 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225, 16² = 256, 17² = 289, 18² = 324, 19² = 361, 20² = 400
- 21² = 441, 22² = 484, 23² = 529, 24² = 576, 25² = 625, 26² = 676, 27² = 729, 28² = 784, 29² = 841, 30² = 900
Question: What patterns do you notice? Share your observations and make conjectures.
Solution: Perfect squares always end in the digits 0, 1, 4, 5, 6, or 9. They never end in 2, 3, 7, or 8.
Question: If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
Solution: No. For example, 14, 15, and 26 end in those digits but are not perfect squares.
Question: Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Solution: 22, 33, 47, 58, and 92 (any number ending in 2, 3, 7, or 8).
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
Solution: Numbers ending in 4 or 6 will have squares ending in 6. Therefore, the answers are (ii) 34², (iii) 46², (iv) 56², and (v) 74².
Page 5
Question: If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution: 6 zeros.
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 has exactly twice the number of zeros as the original number. Yes, this will always happen, which means perfect squares can only end with an even number of trailing zeros.
Question: What can you say about the parity of a number and its square?
Solution: The parity is conserved. The square of an even number is always even, and the square of an odd number is always odd.
Page 6
Question: Using the pattern above, find 36², given that 35² = 1225.
Solution: 36² is equal to 35² plus the 36ᵗʰ odd number. The 36ᵗʰ odd number is 2(36) – 1 = 71. So, 1225 + 71 = 1296.
Question: How do we find the 36th odd number?
Solution: By substituting n = 36 into the formula for the nᵗʰ odd number (2n – 1).
Question: What is the nᵗʰ odd number?
Solution: 2n – 1.
Page 7
Question: Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Solution: Between any two consecutive squares n² and (n+1)², there are exactly 2n non-square numbers.
Question: How many square numbers are there between 1 and 100? How many are between 101 and 200?
Solution: Strictly between 1 and 100, there are 8 square numbers (4, 9, 16, 25, 36, 49, 64, 81). Between 101 and 200, there are 4 square numbers (121, 144, 169, 196).
Question: What is the largest square less than 1000?
Solution: 961 (which is 31²).
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.
Question: The area of a square is 49 sq. cm. What is the length of its side?
Solution: 7 cm.
Page 8 & 9
Question: What is the square root of 64?
Solution: 8 (and -8).
Question: Is 324 a perfect square?
Solution: Yes. Its prime factorization is 2² × 3⁴, which can be cleanly grouped into pairs: (2 × 3 × 3)², proving 324 = 18².
Question: Is 156 a perfect square?
Solution: No. Its prime factorization is 2² × 3 × 13. The factors 3 and 13 cannot be paired.
Question: Find whether 1156 and 2800 are perfect squares using prime factorisation.
Solution: * 1156 = 2² × 17². Yes, it is a perfect square (34²).
- 2800 = 2⁴ × 5² × 7. The factor 7 does not have a pair, so it is not a perfect square.
Page 10 (Figure it Out)
1. Which of the following numbers are not perfect squares?
Solution: (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares because they end in 2, 8, and 7 respectively.
2. Which one among 64², 108², 292², 36² has last digit 4?
Solution: 108² and 292². (Numbers ending in 2 or 8 have squares ending in 4).
3. Given 125²=15625 what is the value of 126²?
Solution: (iv) 15625 + 251. Using the property that (n+1)² = n² + n + (n+1), this translates to 15625 + 125 + 126.
4. Find the length of the side of a square whose area is 441 m².
Solution: 21 m.
5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Solution: The LCM of 4, 9, and 10 is 180 (2² × 3² × 5). To make it a perfect square, multiply by the unpaired factor 5 to get 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.
Solution: 9408 = 2⁶ × 3 × 7². The unpaired factor is 3, so you must multiply by 3. The product is 28224, and 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
Solution: (i) 32 numbers (calculated as 2 × 16). (ii) 198 numbers (calculated as 2 × 99).
8. In the following pattern, fill in the missing numbers:
Solution: Based on the pattern n² + (n+1)² + (n(n+1))² = (n(n+1)+1)²:
- 4² + 5² + 20² = 21²
- 9² + 10² + 90² = 91²
Page 11
Question: How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Solution: Assuming the grid represents a standard 10×10 block common in such curriculum graphics, there are 100 tiny squares. The prime factorization of 100 is 2² × 5².
Question: How many cubes of side 1 cm make a cube of side 2 cm?
Solution: 8 cubes (2³).
Question: How many cubes of side 1 cm will make a cube of side 3 cm?
Solution: 27 cubes (3³).
Page 12
Question: Is 9 a cube?
Solution: No. It falls between the perfect cubes 2³ (8) and 3³ (27).
Question: Can you estimate the number of unit cubes in a cube with an edge length of 4 units?
Solution: 64 unit cubes (4³).
Question: Complete the table below (1³ to 20³).
Solution:
- 1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000
- 11³=1331, 12³=1728, 13³=2197, 14³=2744, 15³=3375, 16³=4096, 17³=4913, 18³=5832, 19³=6859, 20³=8000
Question: What patterns do you notice in the table above?
Solution: Cubes of even numbers are even, cubes of odd numbers are odd. Furthermore, unlike squares, cubes can end in any digit from 0 to 9.
Page 13
Question: Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Solution: * 1-digit cubes: 1, 8 (2 cubes)
- 2-digit cubes: 27, 64 (2 cubes)
- 3-digit cubes: 125, 216, 343, 512, 729 (5 cubes)
Question: Can a cube end with exactly two zeroes (00)? Explain.
Solution: No. A perfect cube must end in a number of zeroes that is a multiple of 3 (e.g., 3, 6, 9 zeroes). Cubing a base number multiplies its prime factors (including the pairs of 2 and 5 which make 10) by 3.
Question: 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.
Solution: * 4104 = 16³ + 2³ = 15³ + 9³
- 13832 = 24³ + 2³ = 20³ + 18³
Page 14 & 15
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?
Solution: 1000. This is the sum of 10 consecutive odd numbers in the sequence, which corresponds directly to 10³.
Question: Let us check if 3375 is a perfect cube.
Solution: Yes. Its prime factorization is 3³ × 5³, confirming it is 15³.
Question: Is 500 a perfect cube?
Solution: No. Its prime factorization is 2² × 5³. The factor 2 cannot form a complete triplet.
Question: Find the cube roots of these numbers: (i) ∛64 (ii) ∛512 (iii) ∛729
Solution: (i) 4 (ii) 8 (iii) 9.
Question: Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?
Solution: After subtracting the sequential differences three times, the 3rd level differences of perfect cubes become a constant (6).
Page 16 & 17 (Figure it Out)
1. Find the cube roots of 27000 and 10648.
Solution: ∛27000 = 30. ∛10648 = 22.
2. What number will you multiply by 1323 to make it a cube number?
Solution: 1323 = 3³ × 7². Multiply by 7 to make it a perfect cube.
3. State true or false. Explain your reasoning.
- (i) The cube of any odd number is even. False. The product of odd numbers is always odd (e.g., 3³ = 27).
- (ii) There is no perfect cube that ends with 8. False. 2³ = 8, 12³ = 1728.
- (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 (which 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 (which has 6 digits).
- (v) Cube numbers have an odd number of factors. False. For example, 8 is a perfect cube and has factors 1, 2, 4, 8 (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.
Solution: * 1331 ends in 1 -> 11
- 4913 ends in 3 -> 17
- 12167 ends in 7 -> 23
- 32768 ends in 8 -> 32
5. Which of the following is the greatest? Explain your reasoning. (i) 67³ – 66³ (ii) 43³ – 42³ (iii) 67² – 66² (iv) 43² – 42²
Solution: (i) 67³ – 66³ is the greatest. The geometric difference between consecutive cubes grows much faster (scaling on the order of 3n²) than the difference between consecutive squares (scaling on the order of 2n).
Page 18 (Puzzle Time)
Question: 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.
Solution: 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17.
Question: Can you arrange them in more than one way? If not, can you explain why?
Solution: No. The number 16 only has one possible partner from the group 1 to 17 to make a perfect square (which is 9, forming 25). Since it only has one valid adjacent link, it must be forced to the start or end of the row, anchoring a unique sequence.
Question: Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?
Solution: Yes, a mathematically valid continuous circular sequence is: 1, 15, 10, 26, 23, 2, 14, 22, 27, 9, 16, 20, 29, 7, 18, 31, 5, 11, 25, 24, 12, 13, 3, 6, 30, 19, 17, 32, 4, 21, 28, 8, (and back looping to 1).