📚 Class 8 Mathematics · NCERT
Chapter 1: A Square and a Cube
Complete Study Notes · Ganita Prakash · Grade 8
✔ Square Numbers
✔ Square Roots
✔ Cube Numbers
✔ Cube Roots
✔ Exam Questions
✔ Square Roots
✔ Cube Numbers
✔ Cube Roots
✔ Exam Questions
📋 Contents
The Locker Puzzle – How the Chapter Begins
The Story
Queen Ratnamanjuri leaves a puzzle for her son Khoisnam and 99 relatives. There are 100 lockers numbered 1–100. Person 1 opens every locker. Person 2 toggles every 2nd locker. Person 3 toggles every 3rd… and so on up to Person 100. Which lockers stay open at the end?
Queen Ratnamanjuri leaves a puzzle for her son Khoisnam and 99 relatives. There are 100 lockers numbered 1–100. Person 1 opens every locker. Person 2 toggles every 2nd locker. Person 3 toggles every 3rd… and so on up to Person 100. Which lockers stay open at the end?
🔑 Key Insight
A locker stays open if it is toggled an odd number of times. The number of toggles for each locker = number of factors of that locker’s number.
👉 A locker stays OPEN ↔ its number has an ODD number of factors ↔ the number is a PERFECT SQUARE
Why? Factors always come in pairs (e.g., for 6: pairs are 1×6 and 2×3). But perfect squares have one factor that pairs with itself (e.g., for 4: 2×2), giving an odd total count of factors.
Open lockers = Perfect square numbers
Lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 remain open (10 lockers in total). These are 1², 2², 3², 4², 5², 6², 7², 8², 9², 10².
Lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 remain open (10 lockers in total). These are 1², 2², 3², 4², 5², 6², 7², 8², 9², 10².
The Passcode Bonus
Lockers toggled exactly twice are prime numbers (factors: 1 and the number itself). The first five are: 2, 3, 5, 7, 11.
Lockers toggled exactly twice are prime numbers (factors: 1 and the number itself). The first five are: 2, 3, 5, 7, 11.
1.1 Square Numbers
📌 What is a Square Number?
When a number is multiplied by itself, the result is called a square number.
n × n = n² (read as “n squared”)
Examples: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25 …
Examples: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25 …
The squares of natural numbers are called perfect squares.
Not just whole numbers!
We can square fractions and decimals too: (3/5)² = 9/25 | (2.5)² = 6.25
We can square fractions and decimals too: (3/5)² = 9/25 | (2.5)² = 6.25
📊 Table of Squares (1–20)
| n | n² | n | n² | n | n² |
|---|---|---|---|---|---|
| 1 | 1 | 8 | 64 | 15 | 225 |
| 2 | 4 | 9 | 81 | 16 | 256 |
| 3 | 9 | 10 | 100 | 17 | 289 |
| 4 | 16 | 11 | 121 | 18 | 324 |
| 5 | 25 | 12 | 144 | 19 | 361 |
| 6 | 36 | 13 | 169 | 20 | 400 |
| 7 | 49 | 14 | 196 | 30 | 900 |
Properties of Perfect Squares
1️⃣ Units Digit Rule
Perfect squares ALWAYS end in: 0, 1, 4, 5, 6, or 9
If a number ends in 2, 3, 7, or 8 → It is DEFINITELY NOT a perfect square.
If a number ends in 2, 3, 7, or 8 → It is DEFINITELY NOT a perfect square.
| Units digit of n | Units digit of n² |
|---|---|
| 0 | 0 |
| 1 or 9 | 1 |
| 2 or 8 | 4 |
| 3 or 7 | 9 |
| 4 or 6 | 6 |
| 5 | 5 |
Caution!
Ending in 0,1,4,5,6,9 does NOT guarantee a number is a perfect square. (E.g., 26 ends in 6 but is NOT a perfect square). The units digit only tells us when a number is NOT a square.
Ending in 0,1,4,5,6,9 does NOT guarantee a number is a perfect square. (E.g., 26 ends in 6 but is NOT a perfect square). The units digit only tells us when a number is NOT a square.
2️⃣ Zeros at the End
- Perfect squares can only have an even number of zeros at the end.
- 10² = 100 (two zeros), 100² = 10000 (four zeros)
- A number like 1000 (three zeros) is NOT a perfect square.
3️⃣ Parity Rule
Even number → Even square
2² = 4, 4² = 16, 6² = 36 …
2² = 4, 4² = 16, 6² = 36 …
Odd number → Odd square
1² = 1, 3² = 9, 5² = 25 …
1² = 1, 3² = 9, 5² = 25 …
4️⃣ Perfect Squares and Odd Numbers
Adding consecutive odd numbers starting from 1 always gives a perfect square!
1 = 1²
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 + 7 + 9 = 25 = 5²
…
Sum of first n odd numbers = n²
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 + 7 + 9 = 25 = 5²
…
Sum of first n odd numbers = n²
5️⃣ Differences of Consecutive Squares
(n+1)² – n² = 2n + 1 (always an odd number!)
Example: 5² – 4² = 25 – 16 = 9 = 2(4)+1
Example: 5² – 4² = 25 – 16 = 9 = 2(4)+1
6️⃣ Numbers Between Consecutive Squares
Between n² and (n+1)², there are exactly 2n natural numbers.
Example
Between 4² (=16) and 5² (=25): numbers 17,18,19,20,21,22,23,24 → that’s 8 = 2×4 numbers.
Between 4² (=16) and 5² (=25): numbers 17,18,19,20,21,22,23,24 → that’s 8 = 2×4 numbers.
7️⃣ Perfect Squares and Triangular Numbers
Any two consecutive triangular numbers add up to a perfect square!
1 + 3 = 4 = 2²
3 + 6 = 9 = 3²
6 + 10 = 16 = 4²
10 + 15 = 25 = 5²
3 + 6 = 9 = 3²
6 + 10 = 16 = 4²
10 + 15 = 25 = 5²
Square Roots
📌 Definition
If y = x², then x is the square root of y, written as √y = x.
√49 = 7 | √81 = 9 | √100 = 10
In general: √(n²) = ±n (positive and negative roots exist)
We only consider the POSITIVE square root in this chapter.
In general: √(n²) = ±n (positive and negative roots exist)
We only consider the POSITIVE square root in this chapter.
Methods to Find Square Roots
Method 1: Successive Subtraction of Odd Numbers
Subtract consecutive odd numbers (1, 3, 5, …) until you reach 0. The number of steps taken = the square root.
Subtract consecutive odd numbers (1, 3, 5, …) until you reach 0. The number of steps taken = the square root.
Find √81:
81–1=80, 80–3=77, 77–5=72, 72–7=65, 65–9=56,
56–11=45, 45–13=32, 32–15=17, 17–17=0
9 steps ∴ √81 = 9
81–1=80, 80–3=77, 77–5=72, 72–7=65, 65–9=56,
56–11=45, 45–13=32, 32–15=17, 17–17=0
9 steps ∴ √81 = 9
Method 2: Prime Factorisation
Find prime factors. If each prime appears an even number of times (can be split into 2 equal groups) → it’s a perfect square.
Find prime factors. If each prime appears an even number of times (can be split into 2 equal groups) → it’s a perfect square.
- Find prime factorisation of the number.
- Group the prime factors into pairs.
- If all factors pair up exactly → perfect square.
- Square root = product of one factor from each pair.
Is 324 a perfect square?
324 = 2×2 × 3×3 × 3×3
Pairs: (2×2), (3×3), (3×3) ✓ All paired!
√324 = 2×3×3 = 18
324 = 2×2 × 3×3 × 3×3
Pairs: (2×2), (3×3), (3×3) ✓ All paired!
√324 = 2×3×3 = 18
Is 156 a perfect square?
156 = 2×2 × 3 × 13 ← 3 and 13 have no pair!
∴ 156 is NOT a perfect square
Method 3: Estimation
Find the two perfect squares the number lies between, then narrow down.
Find the two perfect squares the number lies between, then narrow down.
- Identify the perfect squares just below and above the number.
- Use the units digit to narrow down possible answers.
- Check the midpoint square to halve the range.
- Verify your guess.
Example: Find √1936
40² = 1600 < 1936 < 2500 = 50² → answer is between 40 and 50.
Units digit is 6 → root ends in 4 or 6 → try 44 or 46.
45² = 2025 > 1936 → root is between 40 and 45 → try 44.
44² = 1936 ✓ → √1936 = 44
40² = 1600 < 1936 < 2500 = 50² → answer is between 40 and 50.
Units digit is 6 → root ends in 4 or 6 → try 44 or 46.
45² = 2025 > 1936 → root is between 40 and 45 → try 44.
44² = 1936 ✓ → √1936 = 44
1.2 Cubic Numbers
📌 What is a Cube Number?
When a number is multiplied by itself three times, the result is called a cube number (or perfect cube).
n × n × n = n³ (read as “n cubed”)
Examples: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125 …
Examples: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125 …
Geometrically: a cube with side n contains n³ unit cubes.
📊 Table of Cubes (1–20)
| n | n³ | n | n³ |
|---|---|---|---|
| 1 | 1 | 11 | 1,331 |
| 2 | 8 | 12 | 1,728 |
| 3 | 27 | 13 | 2,197 |
| 4 | 64 | 14 | 2,744 |
| 5 | 125 | 15 | 3,375 |
| 6 | 216 | 16 | 4,096 |
| 7 | 343 | 17 | 4,913 |
| 8 | 512 | 18 | 5,832 |
| 9 | 729 | 19 | 6,859 |
| 10 | 1,000 | 20 | 8,000 |
Properties of Perfect Cubes
1️⃣ Units Digit of Cubes
| Units digit of n | Units digit of n³ |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 7 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 3 |
| 8 | 2 |
| 9 | 9 |
Unlike squares, cubes can end in ANY digit (0–9)!
Also: a cube cannot end in exactly two zeros (the number of zeros at the end of n³ is always a multiple of 3).
Also: a cube cannot end in exactly two zeros (the number of zeros at the end of n³ is always a multiple of 3).
2️⃣ Perfect Cubes and Consecutive Odd Numbers
1 = 1³
3 + 5 = 8 = 2³
7 + 9 + 11 = 27 = 3³
13 + 15 + 17 + 19 = 64 = 4³
21 + 23 + 25 + 27 + 29 = 125 = 5³
n³ = sum of n consecutive odd numbers
3 + 5 = 8 = 2³
7 + 9 + 11 = 27 = 3³
13 + 15 + 17 + 19 = 64 = 4³
21 + 23 + 25 + 27 + 29 = 125 = 5³
n³ = sum of n consecutive odd numbers
3️⃣ Successive Differences for Cubes
Taking successive differences of perfect cubes, all differences become equal after three levels (the constant difference is 6).
Cubes: 1 8 27 64 125 …
Level 1: 7 19 37 61 …
Level 2: 12 18 24 …
Level 3: 6 6 6 …
Level 1: 7 19 37 61 …
Level 2: 12 18 24 …
Level 3: 6 6 6 …
🚕
The Hardy–Ramanujan Number: 1729
Mathematician G.H. Hardy visited Ramanujan in hospital, arriving by taxicab number 1729. Hardy called it “dull,” but Ramanujan instantly replied it was the smallest number expressible as a sum of two cubes in two different ways:
Mathematician G.H. Hardy visited Ramanujan in hospital, arriving by taxicab number 1729. Hardy called it “dull,” but Ramanujan instantly replied it was the smallest number expressible as a sum of two cubes in two different ways:
1729 = 1³ + 12³ = 9³ + 10³
Such numbers are now called Taxicab Numbers.
Cube Roots
📌 Definition
If y = x³, then x is the cube root of y, written as ∛y = x.
∛8 = 2 | ∛27 = 3 | ∛1000 = 10
In general: ∛(n³) = n
In general: ∛(n³) = n
Finding Cube Roots using Prime Factorisation
- Find prime factorisation of the number.
- Group prime factors into triplets (groups of 3).
- If all factors form complete triplets → it’s a perfect cube.
- Cube root = product of one factor from each triplet.
Is 3375 a perfect cube?
3375 = 3×3×3 × 5×5×5 = 3³ × 5³
Triplets: (3,3,3) and (5,5,5) ✓ All grouped!
∛3375 = 3×5 = 15
3375 = 3×3×3 × 5×5×5 = 3³ × 5³
Triplets: (3,3,3) and (5,5,5) ✓ All grouped!
∛3375 = 3×5 = 15
Is 500 a perfect cube?
500 = 2×2 × 5×5×5 ← only one 2-triplet partial!
∴ 500 is NOT a perfect cube
Key Property
Each prime factor of a number appears exactly 3 times in the prime factorisation of its cube. So, for a number to be a perfect cube, each prime in its factorisation must appear a multiple of 3 times.
Each prime factor of a number appears exactly 3 times in the prime factorisation of its cube. So, for a number to be a perfect cube, each prime in its factorisation must appear a multiple of 3 times.
Guessing Cube Roots of Large Numbers
Trick for 4–6 digit perfect cubes
Split the number: last 3 digits and the rest. The units digit of the cube gives the units digit of the cube root (see table above). The first part tells you the tens digit.
Split the number: last 3 digits and the rest. The units digit of the cube gives the units digit of the cube root (see table above). The first part tells you the tens digit.
Find ∛4913:
Split: 4 | 913
Units digit of 913 → 3, so cube root ends in 7
4913 > 1³=1 and < 2³=8 → tens digit is 1
∛4913 = 17 ✓ (17³ = 4913)
Split: 4 | 913
Units digit of 913 → 3, so cube root ends in 7
4913 > 1³=1 and < 2³=8 → tens digit is 1
∛4913 = 17 ✓ (17³ = 4913)
1.3 A Pinch of History
- The first known list of perfect squares and cubes was compiled by the Babylonians around 1700 BCE on clay tablets.
- In ancient India, the Sanskrit word varga meant both the square figure and the square power.
- The Sanskrit word ghana was used for both the cube solid and the cube power.
- The fourth power was called varga-varga.
- Aryabhata (499 CE) defined: “A square figure of four equal sides and the number representing its area are called varga.”
- The word “root” (√) comes from the Sanskrit word mula (meaning root of a plant/basis/origin), used for square root and cube root in ancient India.
- The Arabic word jidhr and the Latin word radix (both meaning root of a plant) were later borrowed for the same mathematical concept.
- Brahmagupta (628 CE): “The pada (root) of a krti (square) is that of which it is a square.”
Chapter Summary — Quick Revision
Square Number
n × n = n² · Perfect squares: 1, 4, 9, 16, 25 … · Squares of natural numbers.
n × n = n² · Perfect squares: 1, 4, 9, 16, 25 … · Squares of natural numbers.
Units Digit (Squares)
End in 0,1,4,5,6,9 only. · Never end in 2,3,7,8. · Even number of zeros at end.
End in 0,1,4,5,6,9 only. · Never end in 2,3,7,8. · Even number of zeros at end.
Sum of Odd Numbers
1+3+5+…+(2n–1) = n² · Differences of consecutive squares are odd. · (n+1)²−n² = 2n+1
1+3+5+…+(2n–1) = n² · Differences of consecutive squares are odd. · (n+1)²−n² = 2n+1
Square Root (√)
Inverse of squaring. · √(n²) = n · Methods: subtraction, prime factorisation, estimation.
Inverse of squaring. · √(n²) = n · Methods: subtraction, prime factorisation, estimation.
Cube Number
n×n×n = n³ · Perfect cubes: 1,8,27,64,125 … · Can end in any digit 0–9.
n×n×n = n³ · Perfect cubes: 1,8,27,64,125 … · Can end in any digit 0–9.
Cube Root (∛)
Inverse of cubing. · ∛(n³) = n · Use prime factorisation: group into triplets.
Inverse of cubing. · ∛(n³) = n · Use prime factorisation: group into triplets.
📌 Key Formulae at a Glance
1. n² = n × n 2. n³ = n × n × n
3. Sum of first n odd numbers = n²
4. (n+1)² − n² = 2n + 1
5. Numbers between n² and (n+1)² = 2n
6. Perfect square: prime factors split into 2 equal groups
7. Perfect cube: prime factors split into 3 equal groups
8. Hardy–Ramanujan Number: 1729 = 1³+12³ = 9³+10³
3. Sum of first n odd numbers = n²
4. (n+1)² − n² = 2n + 1
5. Numbers between n² and (n+1)² = 2n
6. Perfect square: prime factors split into 2 equal groups
7. Perfect cube: prime factors split into 3 equal groups
8. Hardy–Ramanujan Number: 1729 = 1³+12³ = 9³+10³
Important Exam Questions with Answers
Q1. Which of the following are NOT perfect squares: 2032, 2048, 1027, 1089?
Ans: 2032 (ends in 2), 2048 (ends in 8), 1027 (ends in 7) → NOT perfect squares. 1089 = 33² → perfect square.
Q2. Given 125² = 15625, find 126².
Ans: 126² = 125² + 2×125 + 1 = 15625 + 251 = 15876. Using pattern: 126² = 15625 + 125 + 126 = 15876. (Answer: option iii: 15625 + 253)
Q3. Find the smallest square number divisible by 4, 9, and 10.
Ans: LCM(4,9,10) = 180. 180 = 2²×3²×5. Factor 5 is unpaired, so multiply by 5: 180×5 = 900 = 30². Answer: 900.
Q4. Find the smallest number by which 9408 must be multiplied to get a perfect square.
Ans: 9408 = 2⁶×3×7². Factor 3 is unpaired → multiply by 3. 9408×3 = 28224 = 168².
Q5. How many numbers lie between 99² and 100²?
Ans: Between n² and (n+1)² there are 2n numbers. Here n=99, so there are 2×99 = 198 numbers.
Q6. Find the cube roots of 27000 and 10648.
Ans: 27000 = 27×1000 = 3³×10³ → ∛27000 = 30. 10648 = 2³×11³ = (2×11)³ → ∛10648 = 22.
Q7. True or False: There is no perfect cube that ends with 8.
Ans: FALSE. 2³ = 8 (ends in 8). Also 12³ = 1728 ends in 8.
Q8. True or False: The cube of a 2-digit number may be a 3-digit number.
Ans: FALSE. The smallest 2-digit number is 10, and 10³ = 1000 (4 digits). So a 2-digit cube is always at least 4 digits.
Q9. What number must 1323 be multiplied by to make it a perfect cube?
Ans: 1323 = 3³ × 7². Need one more 7 to complete the triplet. Multiply by 7. 1323 × 7 = 9261 = 21³.
Q10. Which is greatest: 67³−66³, 43³−42³, 67²−66², 43²−42²?
Ans: Use formulas: a²−b² = (a+b)(a−b); a³−b³ = (a−b)(a²+ab+b²).
67²−66² = 133, 43²−42² = 85,
67³−66³ = 67²+67×66+66² = 13267 (largest!),
43³−42³ = 5461. Answer: 67³−66³
67²−66² = 133, 43²−42² = 85,
67³−66³ = 67²+67×66+66² = 13267 (largest!),
43³−42³ = 5461. Answer: 67³−66³
Q11. If 1331 is a perfect cube, guess its cube root without factorisation.
Ans: Split: 1 | 331. Units digit 1 → cube root ends in 1. First part: 1 lies between 1³=1 and 2³=8 → tens digit is 1. ∛1331 = 11.
📚 Class 8 · Ganita Prakash · Chapter 1: A Square and a Cube · NCERT Syllabus
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