# NCERT Class 8 Mathematics Solutions: Chapter 6 – Squares and Square Roots Exercise 6.3 Part 4 (For CBSE, ICSE, IAS, NET, NRA 2022)

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**Question: 5** For each of the following numbers find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also, find the square root of the square number so obtained:

(i) 252

(ii) 180

(iii) 1008

(iv) 2028

(v) 1458

(vi) 768

**Answer**:

**(i)** 252

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

Here, prime factor 7 has no pair.

Therefore 252 must be multiplied by 7 to make it a perfect square.

**(ii)** 180

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Here, prime factor 5 has no pair.

Therefore 180 must be multiplied by 5 to make it a perfect square.

**(iii)** 1008

2 | 1008 |

2 | 504 |

2 | 252 |

2 | 126 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

Here, prime factor 7 has no pair.

Therefore 1008 must be multiplied by 7 to make it a perfect square.

**(iv)** 2028

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 | 13 |

1 |

Here, prime factor 3 has no pair.

Therefore 2028 must be multiplied by 3 to make it a perfect square.

**(v)** 1458

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Here, prime factor 2 has no pair.

Therefore 1458 must be multiplied by 3 to make it a perfect square.

**(vi)** 768

2 | 768 |

2 | 384 |

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

1 |

Here, prime factor 3 has no pair.

Therefore 768 must be multiplied by 3 to make it a perfect square.