1. Chapter at a Glance
A square floor tiled with 1 m × 1 m tiles needs 1, 4, 9, 16, … tiles as its side grows; a cube built from small 1 cm cubes needs 1, 8, 27, 64, … cubes. These are square numbers and cube numbers. This chapter explores their patterns and properties, shows how to find square roots and cube roots, and includes a famous puzzle about lockers and the remarkable number 1729 linked with Srinivasa Ramanujan.
- Perfect squares, their patterns and properties.
- Square roots by prime factorisation, by subtracting odd numbers, and by estimation.
- Perfect cubes and their properties.
- Cube roots by prime factorisation.
- The locker puzzle and the Hardy–Ramanujan number 1729.
2. Square Numbers
A number of the form n × n = n² (n a natural number) is a perfect square or square number: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, …
The locker puzzle
A school has 100 closed lockers and 100 students. Student 1 opens every locker; student 2 toggles every 2nd locker; student 3 toggles every 3rd; and so on. Which lockers stay open at the end? A locker is toggled once for each of its factors. Factors come in pairs (like 2 × 6 for 12), so most numbers have an even number of factors — except perfect squares, where one pair is a repeat (6 × 6 for 36). So only lockers 1, 4, 9, 16, …, 100 — the 10 perfect squares — stay open.
- A perfect square ends only in 0, 1, 4, 5, 6 or 9 — never in 2, 3, 7 or 8.
- If a square ends in 0, it ends in an even number of zeros (100, 3600).
- Squares of even numbers are even; squares of odd numbers are odd.
- Perfect squares have an odd number of factors.
- The sum of the first n odd numbers is n²: 1 + 3 + 5 + 7 = 16.
- Between n² and (n + 1)² there are exactly 2n non-square numbers.
Useful patterns
- (n + 1)² − n² = 2n + 1: e.g. 26² − 25² = 51.
- Squares ending in 5: (10a + 5)² = 100a(a + 1) + 25. So 35² = 12|25 = 1225; 75² = 56|25 = 5625.
- Pythagorean triples come from squares: 3² + 4² = 5², and for any m > 1, (2m, m² − 1, m² + 1) is a triple.
3. Square Roots
The square root of a number is the number which, multiplied by itself, gives it. √49 = 7 because 7 × 7 = 49. (−7 also gives 49, but √ means the positive root.)
Method 1: Prime factorisation
- Write the number as a product of primes.
- Pair equal primes.
- Take one prime from each pair and multiply.
Find √1764.
Solution: 1764 = 2 × 2 × 3 × 3 × 7 × 7 = (2 × 3 × 7)². So √1764 = 42.
If a prime is left without a partner, the number is not a perfect square. 72 = 2 × 2 × 2 × 3 × 3 — one 2 is unpaired. Multiplying by 2 gives 144 = 12², the smallest perfect square that is a multiple of 72.
Method 2: Repeated subtraction of odd numbers
Since n² = 1 + 3 + 5 + … (n terms), subtract 1, 3, 5, … from the number until you reach 0; the number of steps is the square root. 36 − 1 − 3 − 5 − 7 − 9 − 11 = 0 in 6 steps, so √36 = 6.
Method 3: Estimation
For non-perfect squares, find the squares on either side. √250: 15² = 225 and 16² = 256, so √250 is between 15 and 16, closer to 16 (≈ 15.8).
4. Cube Numbers
A number of the form n × n × n = n³ is a perfect cube: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, …
- Cubes of even numbers are even; of odd numbers are odd.
- Cubes can end in any digit. The last digit of n³ depends only on the last digit of n: 2 → 8, 3 → 7, 7 → 3, 8 → 2, and 0, 1, 4, 5, 6, 9 stay the same.
- If a cube ends in zeros, the number of zeros is a multiple of 3 (1000, 8000).
- Cubes as sums of consecutive odd numbers: 1 = 1, 8 = 3 + 5, 27 = 7 + 9 + 11, 64 = 13 + 15 + 17 + 19.
The Hardy–Ramanujan number
The mathematician G. H. Hardy once visited Srinivasa Ramanujan in hospital and remarked that his taxi number, 1729, seemed dull. Ramanujan replied that it is very interesting: it is the smallest number that can be written as the sum of two cubes in two different ways: 1729 = 1³ + 12³ = 9³ + 10³. Such numbers are now called taxicab numbers.
5. Cube Roots
The cube root of a number, written ∛, is the number whose cube gives it: ∛125 = 5.
- Write the number as a product of primes.
- Group equal primes in threes.
- Take one prime from each group and multiply.
Find ∛5832.
Solution: 5832 = 2³ × 3⁶ = 2³ × 3³ × 3³. ∛5832 = 2 × 3 × 3 = 18.
Find the smallest number by which 1350 must be multiplied to make a perfect cube.
Solution: 1350 = 2 × 3³ × 5². To complete groups of three: 2 needs two more 2s, 5² needs one more 5. Multiply by 2² × 5 = 20, giving 27000 = 30³.
6. Solved Problems (NCERT Style)
Without calculating, which of 1057, 23453, 7928, 222222, 1069 cannot be perfect squares?
Solution: Squares never end in 2, 3, 7 or 8. So 1057, 23453, 7928 and 222222 are not squares. 1069 ends in 9, so it might be (it isn't actually, since 32² = 1024 and 33² = 1089).
A gardener has 1000 plants to plant in rows so that rows and columns are equal. How many more plants does he need for the smallest such arrangement?
Solution: 31² = 961 < 1000 < 32² = 1024. He needs 1024 − 1000 = 24 more plants.
A cube has volume 2197 cm³. Find its side.
Solution: 2197 = 13³, so side = 13 cm.
7. Master Formula Table
| Result | Formula / fact |
|---|---|
| Square | n² = n × n |
| Sum of first n odd numbers | n² |
| Consecutive squares | (n + 1)² − n² = 2n + 1 |
| Non-squares between n² and (n+1)² | 2n |
| Pythagorean triple | (2m, m² − 1, m² + 1) |
| Square ending in 5 | (a5)² = a(a + 1) followed by 25 |
| Cube | n³ = n × n × n |
| √ by primes | pairs → one from each |
| ∛ by primes | triples → one from each |
| Hardy–Ramanujan | 1729 = 1³ + 12³ = 9³ + 10³ |
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 |
| n³ | 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 | 1331 | 1728 |
8. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Perfect squares | Odd number of factors; last digits | Identify non-squares |
| Square roots | Prime pairs, odd subtraction, estimation | Exact and approximate roots |
| Perfect cubes | Groups of three primes | Smallest multiplier/divisor |
| Cube roots | One prime per triple | Volume problems |
| Famous facts | Lockers, 1729 | Reasoning questions |
9. Mind Map
- Last digits 0,1,4,5,6,9
- Odd-number sums
- Locker puzzle
- Prime pairs
- Subtract odds
- Estimate
- Consecutive odd sums
- Last-digit rules
- Prime triples
- 1729 Ramanujan
10. Tips and Tricks
- Learn squares up to 30 and cubes up to 15 — they save time.
- For ∛ of a perfect cube up to 6 digits: the last digit of the root comes from the last digit of the cube; the first part comes from the thousands group. ∛ 103823: last digit 3 → root ends in 7; 103 lies between 4³ = 64 and 5³ = 125 → root is 47.
- Square of a number ending in 5: multiply the tens part by the next number and write 25 after it.
- Confusing n² with 2n, or n³ with 3n.
- Pairing primes when finding a cube root (you need triples).
- Thinking every number ending in 6 is a square (26 is not).
11. Important One-liners
- Perfect squares never end in 2, 3, 7 or 8.
- A perfect square has an odd number of factors.
- The sum of the first n odd numbers is n².
- There are 2n non-squares between n² and (n + 1)².
- Only the perfect-square lockers remain open in the locker puzzle.
- Cubes of odd numbers are odd.
- 1729 is the smallest number expressible as a sum of two cubes in two ways.
- √ uses prime pairs; ∛ uses prime triples.
12. Exam-pattern Questions with Detailed Answers
Which of these is a perfect square?
- 1022
- 1764
- 2353
- 1847
Show answer
1022, 2353 and 1847 end in 2, 3 and 7. 1764 = 42².
Answer: (B) 1764
Find ∛343.
Show answer
343 = 7 × 7 × 7.
Answer: 7
How many non-square numbers lie between 15² and 16²?
Show answer
2n with n = 15 gives 30.
Answer: 30
Find √2025 by prime factorisation.
Show answer
2025 = 3⁴ × 5² = (3² × 5)², so √2025 = 45.
Answer: 45
Find the smallest number by which 252 must be multiplied to get a perfect square.
Show answer
252 = 2² × 3² × 7. 7 is unpaired, so multiply by 7: 1764 = 42².
Answer: 7
What is 85²?
- 7025
- 7225
- 6425
- 7255
Show answer
8 × 9 = 72, then append 25: 7225.
Answer: (B) 7225
Find the smallest number by which 3087 must be divided to get a perfect cube.
Show answer
3087 = 3² × 7³. Divide by 3² = 9 to get 343 = 7³.
Answer: 9
Write a Pythagorean triple with 14 as one member.
Show answer
2m = 14 → m = 7: m² − 1 = 48, m² + 1 = 50. Triple (14, 48, 50).
Answer: 14, 48, 50
In the locker puzzle with 200 lockers, how many remain open?
Show answer
Open lockers are perfect squares up to 200: 1², 2², …, 14² = 196 (15² = 225 > 200). That is 14 lockers.
Answer: 14
Without full factorisation, find ∛ 74088 (a perfect cube).
Show answer
Last digit 8 → root ends in 2. Thousands part 74 lies between 4³ = 64 and 5³ = 125 → tens digit 4. Root = 42. Check: 42³ = 74088 ✔.
Answer: 42
Show that 1729 can be written as a sum of two cubes in two ways.
Show answer
1³ + 12³ = 1 + 1728 = 1729; 9³ + 10³ = 729 + 1000 = 1729.
Answer: 1³ + 12³ = 9³ + 10³ = 1729
The area of a square field is 5184 m². Find the cost of fencing it at ₹15 per metre.
Show answer
5184 = 2⁶ × 3⁴ → side = 2³ × 3² = 72 m. Perimeter = 288 m. Cost = 288 × 15 = ₹4320.
Answer: ₹4,320
13. Quick Revision Summary
- Squares end in 0, 1, 4, 5, 6, 9; have an odd number of factors; equal sums of odd numbers.
- √ by prime pairs, by subtracting odd numbers, or by estimating between squares.
- Cubes: groups of three primes; ∛ takes one from each group.
- Make a perfect square/cube by multiplying or dividing by the unpaired/ungrouped primes.
- 1729 = 1³ + 12³ = 9³ + 10³ (Ramanujan).