Example 7 - Chapter 5 Class 8 Squares and Square Roots
Last updated at April 16, 2024 by Teachoo
Smallest number divided to get perfect square
Chapter 5 Class 8 Squares and Square Roots
Concept wise
- Square numbers
- Properties of square numbers
- Sum of consecutive odd numbers
- Numbers between square numbers
- Pattern Solving
- Finding square of large numbers
- Pythagorean triplets
- Square root
- Finding Square root through repeated subtraction
- Finding Square root through prime factorisation
- Checking if perfect square by prime factorisation
- Smallest number multiplied to get perfect square
-
Smallest number divided to get perfect square
- Smallest square number divisible by numbers
- Finding number of digits in square root (without calculation)
- Finding square root by division method - Integers
- Least number subtracted to get a perfect square
- Least number added to get a perfect square
- Finding square root by division method - Decimals
- Statement Questions
Transcript
Example 7 Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient. Prime factorizing 9408 Prime factorizing 9408 We see that , 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7 Since 3 does not occur in pairs, we divide by 3 to make it a pair So, our number becomes 9408 × 𝟏/𝟑 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7 × 𝟏/𝟑 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 Square root of 3136 ∴ √3136 = 2 × 2 × 2 × 7 = 56 ∴ The smallest whole number to be divided = 3 and square root of new number = 56
