## Book: NCERT - Mathematics

### Chapter: 6. Squares and Square Roots

#### Subject: Maths - Class 8th

##### Q. No. 5 of Exercises 6.3

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5
##### For each of the following numbers, find the smallest whole number by which it shouldbe multiplied so as to get a perfect square number. Also find the square root of thesquare number so obtained.(i) 252 (ii) 180(iii) 1008 (iv) 2028(v) 1458 (vi) 768

(i)The prime factorization of 252 is follows:

252 = 2 × 2 × 3 × 3 × 7

Here,

Prime factor 7 does not have its pair.

If 7 gets a pair, then the number will become a perfect square.

Therefore, 252 has to be multiplied with 7 to obtain a perfect square

252 × 7 = 2 × 2 × 3 × 3 × 7 × 7

Hence,

252 × 7 = 1764 is a perfect square

2 * 3 * 7

= 42

(ii) The prime factorization of 180 is as follows:

180 = 2 × 2 × 3 × 3 × 5

Here, prime factor 5 does not have its pair.

If 5 gets a pair, then the number will become a perfect square.

Therefore, 180 has to be multiplied with 5 to obtain a perfect square.

180 × 5 = 900 = 2 × 2 × 3 × 3 × 5 × 5

Therefore,

180 × 5 = 900 is a perfect square

= 2 * 3 * 5

= 30

(iii) The prime factorization of 1008 isas follows:

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, prime factor 7 does not have its pair.

If 7 gets a pair, then the number will become a perfect square.

Therefore, 1008 can be multiplied with 7 to obtain a perfect square.

1008 × 7 = 7056 = 2 × 2 ×2 × 2 × 3 × 3 × 7 × 7

Therefore,

1008 × 7 = 7056 is a perfect square

= 2 * 2 * 3 * 7

= 84

(iv) The prime factorization of 2028 is as follows:

2028 = 2 * 2 * 3 * 13 * 13

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 2028 can be multiplied with 3 to obtain a perfect square.

2028 × 3 = 6084 = 2 × 2 ×3 × 3 × 13 × 13

Therefore,

2028 × 3 = 6084 is a perfect square

= 2 * 3 * 13

= 78

(v) The prime factorization of 1458 is as follows:

1458 = 2 * 3 * 3 * 3 * 3 * 3 * 3

Here, prime factor 2 does not have its pair.

If 2 gets a pair, then the number will become a perfect square.

Therefore, 1458 can be multiplied with 2 to obtain a perfect square.

1458 × 2 = 2916 = 2 × 2 ×3 × 3 × 3 × 3 × 3 × 3

Therefore,

1458 × 2 = 2916 is a perfect square

= 2 * 3 * 3 * 3

= 54

(vi) The prime factorization of 768 is as:

768 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3

Here, prime factor 3 does not have its pair.

If 3 gets a pair, then the number will become a perfect square.

Therefore, 768 can be multiplied with 3 to obtain a perfect square.

768 × 3 = 2304 = 2 × 2 ×2 × 2 × 2 × 2 × 2 × 2 ×3 ×3

Therefore,

768 × 3 = 2304 is a perfect square

= 2 * 2 * 2 * 2 * 3

= 48

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