Jaka EduTech
Jaka EduTech
Parent Sign In

© 2026 Jaka EduTech

Jaka EduTech

📖 Mathematics — Part 1

Std 8
Curriculum Book Guide
1 2 3 4 5 6 7
Chapters
1 A Square and a Cube 2 Power Play (Exponents) 3 A Story of Numbers 4 Quadrilaterals 5 Number Play 6 We Distribute Yet Things Multiply 7 Proportional Reasoning - I
1
Chapter 1
Skill: 30%

A Square and a Cube

A Square and a Cube

Chapter at a Glance

This chapter introduces the concepts of perfect squares and perfect cubes, exploring their properties, geometric significance, and arithmetic patterns. Through a narrative puzzle involving lockers, it demonstrates the relationship between factors, primes, and square numbers. The chapter teaches key methods to find and estimate square and cube roots, including successive subtraction of odd numbers, prime factorisation, and interval estimation. It also highlights notable mathematical themes, such as Ramanujan’s taxicab numbers ($1729$) and historical terms used by ancient Indian mathematicians like Aryabhata and Brahmagupta.

Key Definitions & Terminology

  • Square Number (Perfect Square): A number that can be expressed as the product of an integer multiplied by itself (i.e., $n^2 = n \times n$).
  • Perfect Cube: A number obtained by multiplying an integer by itself three times (i.e., $n^3 = n \times n \times n$).
  • Square Root: The inverse operation of squaring. If $y = x^2$, then $x$ is the square root of $y$, denoted by $x = \sqrt{y}$.
  • Cube Root: The inverse operation of cubing. If $y = x^3$, then $x$ is the cube root of $y$, denoted by $x = \sqrt[3]{y}$.
  • Prime Factorisation: The decomposition of a composite number into a product of prime numbers.
  • Taxicab Number: A number that can be expressed as the sum of two positive cubes in two different ways. The smallest taxicab number is $1729$ (known as the Hardy-Ramanujan Number).
  • Varga: An ancient Sanskrit mathematical term representing a square figure, its area, or a squared quantity ($n^2$).
  • Ghana: An ancient Sanskrit term representing a solid geometric cube or a cubed quantity ($n^3$).
  • Mula: The Sanskrit word for "root" of a plant, used historically in India to denote the root of a number (leading to the Arabic jidhr and Latin radix).
  • Pada: An ancient Sanskrit term meaning "foot" or "basis," also used to refer to a root of a square.

Formulas, Rules & Properties

  • Locker Divisor Rule:
  • The number of times a locker is toggled is equal to the count of its factors. Since factors occur in pairs, only perfect squares have an odd number of factors. Thus, in a sequence of open/close toggles, only perfect square lockers remain open.
  • Prime lockers are toggled exactly twice because they have only two factors ($1$ and themselves).
  • Properties of Perfect Squares:
  • Every perfect square ends in the units digit $0, 1, 4, 5, 6,$ or $9$.
  • A number ending in $2, 3, 7,$ or $8$ is never a perfect square.
  • If a number ends in $1$ or $9$, its square ends in $1$. If it ends in $4$ or $6$, its square ends in $6$.
  • Perfect squares can only have an even number of zeros at the end.
  • Sum of Successive Odd Numbers:
  • The sum of the first $n$ consecutive odd natural numbers starting from $1$ is equal to $n^2$:
    $$1 + 3 + 5 + 7 + \dots + (2n - 1) = n^2$$
  • The $n$-th odd number is given by the formula:
    $$\text{n-th odd number} = 2n - 1$$
  • Spacing Between Squares:
  • The quantity of non-square numbers lying between the squares of two consecutive natural numbers $n^2$ and $(n+1)^2$ is:
    $$\text{Non-squares} = 2n$$
  • Triangular and Square Numbers:
  • The sum of two consecutive triangular numbers ($1, 3, 6, 10, 15, \dots$) forms a perfect square:
    $$1 + 3 = 4 = 2^2$$
    $$3 + 6 = 9 = 3^2$$
    $$6 + 10 = 16 = 4^2$$
  • Perfect Cubes as Sums of Odd Numbers:
  • Cubes can be represented as sums of consecutive odd numbers grouped sequentially:
    $$1^3 = 1$$
    $$2^3 = 3 + 5 = 8$$
    $$3^3 = 7 + 9 + 11 = 27$$
    $$4^3 = 13 + 15 + 17 + 19 = 64$$
  • Root prime factor grouping rules:
  • A number is a perfect square if its prime factors can be split into two identical groups.
  • A number is a perfect cube if its prime factors can be split into three identical groups.
  • Successive Difference Constant:
  • For squares, the differences between consecutive terms are odd numbers (Level 1), and their second-level differences are constant at $2$.
  • For cubes, the successive differences become constant at $6$ at the third level.

Core Concepts & Topics

  • Square and Cube Root Inverses:
  • Real numbers have two square roots (e.g., $(+8)^2 = 64$ and $(-8)^2 = 64$). However, the symbol $\sqrt{}$ is defined to represent the positive square root.
  • Estimating Roots: To find the square root of $250$, check the bounding squares: $15^2 = 225$ and $16^2 = 256$. Since $250$ is much closer to $256$ than to $225$, $\sqrt{250}$ is estimated to be approximately $16$ (specifically, slightly less than $16$).
  • Taxicab Numbers (Hardy-Ramanujan Numbers):
  • $1729$ is the smallest number that can be written as the sum of two positive cubes in two different ways:
    $$1729 = 1^3 + 12^3 = 9^3 + 10^3$$
  • The next two taxicab numbers are $4104$ and $13832$:
    $$4104 = 2^3 + 16^3 = 9^3 + 15^3$$
    $$13832 = 2^3 + 24^3 = 18^3 + 20^3$$
  • Etymology of Roots:
  • The Sanskrit term mula (plant root or origin) was used in India for root operations since the 1st century BCE. Arabic translators adopted jidhr (plant root) and Latin translators chose radix, giving rise to the modern word "root" and the radical sign $\sqrt{}$.

Worked Examples

  • Units Digit Determination:
  • Problem: Which of these numbers have $6$ in the units place when squared?
    (i) $38^2$, (ii) $34^2$, (iii) $46^2$, (iv) $56^2$, (v) $74^2$, (vi) $82^2$.
  • Solution: A square ends in $6$ if the base number ends in $4$ or $6$.
    • $34^2$ ends in $4^2 = 16 \rightarrow$ unit digit $6$.
    • $46^2$ ends in $6^2 = 36 \rightarrow$ unit digit $6$.
    • $56^2$ ends in $6^2 = 36 \rightarrow$ unit digit $6$.
    • $74^2$ ends in $4^2 = 16 \rightarrow$ unit digit $6$.
    • Answers: (ii), (iii), (iv), and (v).
  • Finding $36^2$ via Odd Sums:
  • Problem: Given $35^2 = 1225$, find the value of $36^2$ using the consecutive odd sums rule.
  • Solution: The sum of the first $35$ odd numbers is $1225$. To find $36^2$, add the $36$-th odd number.
    • $\text{36-th odd number} = 2(36) - 1 = 71$.
    • $36^2 = 1225 + 71 = 1296$.
  • Identifying Non-squares (Figure it Out Q1):
  • Problem: Determine which of $2032, 2048, 1027, 1089$ are not perfect squares.
  • Solution: Perfect squares never end in $2, 3, 7,$ or $8$.
    • $2032$ ends in $2 \rightarrow$ Not a perfect square.
    • $2048$ ends in $8 \rightarrow$ Not a perfect square.
    • $1027$ ends in $7 \rightarrow$ Not a perfect square.
    • $1089$ ends in $9 \rightarrow$ Can be a perfect square ($33^2 = 1089$).
    • Answer: $2032$, $2048$, and $1027$ are not perfect squares.
  • Smallest Divisible Square (Figure it Out Q5):
  • Problem: Find the smallest square number divisible by $4, 9,$ and $10$.
  • Solution: First, find the LCM of $4, 9,$ and $10$.
    • $4 = 2^2$, $9 = 3^2$, $10 = 2 \times 5$.
    • $\text{LCM} = 2^2 \times 3^2 \times 5 = 180$.
    • In the prime factorisation of $180$, the factor $5$ is not paired. Multiply by $5$ to make it a perfect square.
    • $\text{Smallest square number} = 180 \times 5 = 900$.
  • Finding Square Root by Multiplication (Figure it Out Q6):
  • Problem: Find the smallest number by which $9408$ must be multiplied to make it a perfect square. Find the square root of the product.
  • Solution: Find the prime factorisation of $9408$.
    • $9408 = 2^6 \times 3 \times 7^2$.
    • The prime factor $3$ is unpaired. Multiply the number by $3$.
    • $\text{New product} = 9408 \times 3 = 28224$.
    • $\sqrt{28224} = 2^3 \times 3 \times 7 = 8 \times 21 = 168$.
  • Numbers Between Squares (Figure it Out Q7):
  • Problem: How many numbers lie between: (i) $16^2$ and $17^2$, (ii) $99^2$ and $100^2$.
  • Solution: The number of non-square values between $n^2$ and $(n+1)^2$ is $2n$.
    • (i) For $16^2$ and $17^2$: $2 \times 16 = 32$.
    • (ii) For $99^2$ and $100^2$: $2 \times 99 = 198$.
  • Pattern Puzzle (Figure it Out Q8):
  • Problem: Fill in:
    (i) $4^2 + 5^2 + 20^2 = (__)^2$
    (ii) $9^2 + 10^2 + (__)^2 = (__)^2$
  • Solution: The pattern is $a^2 + b^2 + c^2 = d^2$ where $b = a+1$, $c = ab$, and $d = c+1$.
    • (i) $a=4, b=5 \rightarrow c=20 \rightarrow d=21$. Thus: $4^2 + 5^2 + 20^2 = 21^2$.
    • (ii) $a=9, b=10 \rightarrow c=90 \rightarrow d=91$. Thus: $9^2 + 10^2 + 90^2 = 91^2$.
  • Cubes True or False Verification (Figure it Out Q3, Page 16):
  • (i) The cube of any odd number is even $\rightarrow$ False ($3^3 = 27$, which is odd).
  • (ii) There is no perfect cube that ends with $8 \rightarrow$ False ($2^3 = 8$).
  • (iii) The cube of a 2-digit number may be a 3-digit number $\rightarrow$ False (the smallest 2-digit cube is $10^3 = 1000$, which has 4 digits).
  • (iv) The cube of a 2-digit number may have 7 or more digits $\rightarrow$ False (the largest 2-digit cube is $99^3 = 970299$, which has 6 digits).
  • (v) Cube numbers have an odd number of factors $\rightarrow$ False ($8$ is a cube with factors ${1, 2, 4, 8}$, which is 4 factors—an even number).
  • Guessing Cube Roots (Figure it Out Q4, Page 16):
  • Problem: Find the cube roots of $4913$ and $32768$ without factorisation.
  • Solution:
    • For $4913$: Since $10^3 = 1000 < 4913 < 8000 = 20^3$, the root is a 2-digit number starting with $1$. The number ends in $3$, which means the units digit of its root must be $7$ (since $7^3 = 343$ ends in $3$). Thus, $\sqrt[3]{4913} = 17$.
    • For $32768$: Since $27000 = 30^3 < 32768 < 64000 = 40^3$, the root starts with $3$. The number ends in $8$, so the units digit of the root must be $2$ (since $2^3 = 8$). Thus, $\sqrt[3]{32768} = 32$.

Practical Activities & Experiments

  • Geometric Odd-Sum Grid: Build a square grid of $n \times n$ tiles. Add successive L-shaped borders (gnomons) made of contrasting colored tiles representing successive odd numbers ($1, 3, 5, 7, \dots$) to visually demonstrate that the sum of the first $n$ odd numbers always forms the perfect square $n^2$.
  • Successive Difference Tree: Create a difference chart on a poster. Write down consecutive perfect squares ($1, 4, 9, 16, 25, 36$) and compute their first differences (revealing the odd numbers: $3, 5, 7, 9, 11$). Compute their second differences to show they are constant at $2$. Repeat the activity for perfect cubes ($1, 8, 27, 64, 125, 216$) to show their differences become constant at $6$ at the third level.
Chapter 1 of 7 Next