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A Square and A Cube - Square Numbers

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A natural number nn is called a square number or a perfect square if there exists a natural number mm such that n=m2n = m^2.

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Square numbers always end with 0,1,4,5,6, or 90, 1, 4, 5, 6, \text{ or } 9 at the unit's place. Numbers ending in 2,3,7, or 82, 3, 7, \text{ or } 8 are never perfect squares.

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If a number has 11 or 99 in the unit's place, its square ends in 11. If a number has 44 or 66 in the unit's place, its square ends in 66.

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Square numbers can only have an even number of zeros at the end.

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Between the squares of two consecutive numbers nn and n+1n+1, there are 2n2n non-perfect square numbers.

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The sum of the first nn odd natural numbers is n2n^2. For example, 1+3+5=9=321 + 3 + 5 = 9 = 3^2.

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Pythagorean Triplets: For any natural number m>1m > 1, the numbers 2m2m, m2−1m^2 - 1, and m2+1m^2 + 1 form a Pythagorean triplet such that (2m)2+(m2−1)2=(m2+1)2(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2.

📐Formulae

n2=n×nn^2 = n \times n

Number of non-square numbers between n2 and (n+1)2=2n\text{Number of non-square numbers between } n^2 \text{ and } (n+1)^2 = 2n

1+3+5+⋯+(2n−1)=n21 + 3 + 5 + \dots + (2n - 1) = n^2

(2m)2+(m2−1)2=(m2+1)2(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

💡Examples

Problem 1:

How many natural numbers lie between 12212^2 and 13213^2?

Solution:

2×12=242 \times 12 = 24

Explanation:

Using the property that there are 2n2n non-perfect square numbers between n2n^2 and (n+1)2(n+1)^2, where n=12n = 12, we get 2×12=242 \times 12 = 24 numbers.

Problem 2:

Express 8181 as the sum of first nn odd numbers.

Solution:

81=1+3+5+7+9+11+13+15+1781 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

Explanation:

Since 81=9\sqrt{81} = 9, the number 8181 is the sum of the first 99 odd natural numbers.

Problem 3:

Write a Pythagorean triplet whose smallest member is 66.

Solution:

Let 2m=62m = 6, then m=3m = 3. m2−1=32−1=8m^2 - 1 = 3^2 - 1 = 8 m2+1=32+1=10m^2 + 1 = 3^2 + 1 = 10. The triplet is (6,8,10)(6, 8, 10). Check: 62+82=36+64=100=1026^2 + 8^2 = 36 + 64 = 100 = 10^2.

Explanation:

We use the general form 2m,m2−1,m2+12m, m^2 - 1, m^2 + 1 to generate the triplet.

Problem 4:

Calculate 45245^2 using the column method or identity expansion.

Solution:

452=(40+5)2=402+2(40)(5)+52=1600+400+25=202545^2 = (40 + 5)^2 = 40^2 + 2(40)(5) + 5^2 = 1600 + 400 + 25 = 2025

Explanation:

Applying the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 where a=40a=40 and b=5b=5 makes calculating large squares easier.

Problem 5:

Show the vertical calculation for 23×2323 \times 23.

Solution:

23×2369460529\begin{array}{r} 23 \\ \times 23 \\ \hline 69 \\ 460 \\ \hline 529 \end{array}

Explanation:

Multiplying 2323 by itself results in 529529, which is a perfect square.