**Chapter 3: Squares and Square Roots Exercise – 3.1**

**Question: 1**

Which of the following numbers are perfect squares?

(i) 484

(ii) 625

(iii) 576

(iv) 941

(v) 961

(vi) 2500

**Solution:**

(i) 484

484 = 22^{2}

(ii) 625

625 = 25^{2}

(iii) 576

576 = 24^{2}

(iv) 941

Perfect squares closest to 941 are 900 (30^{2}) and 961 (31^{2}). Since 30 and 31 are consecutive numbers, there are no perfect squares between 900 and 961. Hence, 941 is not a perfect square.

(v) 961

961 = 31^{2}

(vi) 2500

2500 = 50^{2}

Hence, all numbers except that in (iv), i.e. 941, are perfect squares.

**Question: 2**

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number in each case:

(i) 1156 = 2 x 2 x 17 x 17

(ii) 2025 = 3 x 3 x 3 x 3 x 5 x 5

(iii) 14641 = 11 x 11 x 11 x 11

(iv) 4761 = 3 x 3 x 23 x 23

In each problem, factorize the number into its prime factors.

**Solution:**

(i) 1156 = 2 x 2 x 17 x 17

Grouping the factors into pairs of equal factors, we obtain:

1156 = (2 x 2) x (17 x 17) No factors are left over. Hence, 1156 is a perfect square. Moreover, by grouping 1156 into equal factors:

1156 = (2 x 17) x (2 x 17) = (2 x 17)^{2}

Hence, 1156 is the square of 34, which is equal to 2 x 17.

(ii) 2025 = 3 x 3 x 3 x 3 x 5 x 5

Grouping the factors into pairs of equal factors, we obtain:

2025 = (3 x 3) x (3 x 3) x (5 x 5)

No factors are left over. Hence, 2025 is a perfect square. Moreover, by grouping 2025 into equal factors:

2025 = (3 x 3 x 5) x (3 x 3 x 5) = (3 x 3 x 5)^{2}

Hence, 2025 is the square of 45, which is equal to 3 x 3 x 5.

(iii) 14641 = 11 x 11 x 11 x 11

Grouping the factors into pairs of equal factors, we obtain:

14641 = (11 x 11) x (11 x 11)

No factors are left over. Hence, 14641 is a perfect square. The above expression is already grouped into equal factors:

14641 = (11 x 11) x (11 x 11) = (11 x 11)2 Hence, 14641 is the square of 121, which is equal to 11 x 11.

(iv) 4761 = 3 x 3 x 23 x 23

Grouping the factors into pairs of equal factors, we obtain:

4761 = (3 x 3) x (23 x 23)

No factors are left over. Hence, 4761 is a perfect square. The above expression is already grouped into equal factors:

4761 = (3 x 23) x (3 x 23) = (3 x 23)^{2}

Hence, 4761 is the square of 69, which is equal to 3 x 23.

**Question: 3**

Find the smallest number by which of the following number must be multiplied so that the product is a perfect square:

Factorize each number into its factors

(i) 23805 = 3 x 3 x 5 x 23 x 23

(ii) 12150 = 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5

(iii) 7688 = 2 x 2 x 2 x 31 x 31

**Solution:**

(i) 23805 = 3 x 3 x 5 x 23 x 23

3 | 23805 |

3 | 7935 |

5 | 2645 |

23 | 529 |

23 | 23 |

1 |

Grouping 23805 into pairs of equal factors:

23805 = (3 x 3) x (23 x 23) x 5

Here, the factor 5 does not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence, the smallest number by which 23805 must be multiplied is 5.

(ii) 12150 = 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5

2 | 12150 |

3 | 6075 |

3 | 2025 |

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Grouping 12150 into pairs of equal factors:

12150 = (3 x 3 x 3 x 3) x (5 x 5) x 2 x 3

Here, 2 and 3 do not occur in pairs. To be a perfect square, every prime factor has to be in pairs.

Hence, the smallest number by which 12150 must be multiplied is 2 x 3, i.e. by 6.

(iii) 7688 = 2 x 2 x 2 x 31 x 31

2 | 7688 |

2 | 3844 |

2 | 1922 |

31 | 961 |

31 | 31 |

1 |

Grouping 7688 into pairs of equal factors:

7688 = (2 x 2) x (31 x 31) x 2

Here, 2 do not occur in pairs. To be a perfect square, every prime factor has to be in pairs. Hence the smallest number by which 7688 must be multiplied is 2.

**Question: 4**

Find the smallest number by which the given number must be divided so that the resulting number is a perfect square:

(i) 14283 = 3 x 3 x 3 x 23 x 23

(ii) 1800 = 2 x 2 x 2 x 3 x 3 x 5 x 5

(iii) 2904 = 2 x 2 x 2 x 3 x 11 x 11

**Solution:**

(i) 14283 = 3 x 3 x 3 x 23 x 23

3 | 14283 |

3 | 4761 |

3 | 1587 |

23 | 529 |

23 | 23 |

1 |

Grouping the factors into pairs:

14283 = (3 x 3) x (23 x 23) x 3

Here, the factor 3 does not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 14283 must be divided for it to be a perfect square is 3.

(ii) 1800 = 2 x 2 x 2 x 3 x 3 x 5 x 5

2 | 1800 |

2 | 900 |

2 | 450 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Grouping the factors into pairs:

1800 = (2 x 2) x (3 x 3) x (5 x 5) x 2

Here the factor 2 does not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 1800 must be divided for it to be a perfect square is 2.

(iii) 2904 = 2 x 2 x 2 x3 x 11 x 11

2 | 2904 |

2 | 1452 |

2 | 726 |

3 | 363 |

11 | 121 |

11 | 11 |

1 |

Grouping the factors into pairs:

2904 = (2 x 2) x (11 x 11) x 2 x 3

Here the factor 2 and 3 does not occur in pairs. To be a perfect square, all the factors have to be in pairs. Hence, the smallest number by which 2304 must be divided for it to be a perfect square is 2 x 3, i.e. 6.

**Question: 5**

Which of the following numbers are perfect squares?

(i) 11

(ii) 12

(iii) 32

(iv) 50

(v) 79

(vi) 111

**Solution:**

(i) 11

11: The perfect squares closest to 11 are 9 (9 = 3^{2}) and 16 (16 = 4^{2}). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16, which mean that 11 is not a perfect square.

(ii) 12

12: The perfect squares closest to 12 are 9 (9 =3^{2}) and 16 (16 = 4^{2}). Since 3 and 4 are consecutive numbers, there are no perfect squares between 9 and 16, which mean that 12 is not a perfect square.

16 = 4^{2}

(iii) 32

32: The perfect squares closest to 32 are 25 (25 = 52) and 36 (36 = 62). Since 5 and 6 are consecutive numbers, there are no perfect squares between 25 and 36, which means that 32 is not a perfect square.

36 = 6^{2}

(iv) 36

50: The perfect squares closest to 50 are 49 (49 = 72) and 64 (64 = 82). Since 7 and 8 are consecutive numbers, there are no perfect squares between 49 and 64, which means that 50 is not a perfect square. 64 = 8^{2}

(v) 79

79: The perfect squares closest to 79 are 64 (64 = 82) and 81 (81 = 92). Since 8 and 9 are consecutive numbers, there are no perfect squares between 64 and 81, which mean that 79 is not a perfect square.

81 = 92

(vi) 111

111: The perfect squares closest to 111 are 100 (100 = 102) and 121 (121 = 112). Since 10 and 11 are consecutive numbers, there are no perfect squares between 100 and 121, which means that 111Is not a perfect square.

121 = 11^{2}

Hence, the perfect squares are 16, 36, 64, 81 and 121.

**Question: 6**

Using prime factorization method, find which of the following numbers are perfect squares?

(i) 189 = 3 x 3 x 3 x 7

(ii) 225 = 3 x 3 x 5 x 5

(iii) 2048 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

(iv) 343 = 7 x 7 x 7

(v) 441 = 3 x 3 x 7 x 7

(vi) 2916 = 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 x 3

(vii) 11025= 3 x 3 x5 x7 x 7

(viii) 3549 = 3 x 7 x 13 x 13

**Solution:**

(i) 189 = 3 x 3 x 3 x 7

3 | 189 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

189 = (3 x 3) x 3 x 7

The factors 3 and 7 cannot be paired. Hence, 189 is not a perfect square.

(ii) 225 = 3 x 3 x 5 x 5

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Grouping them into pairs of equal factors:

225 = (3 x 3) x (5 x 5)

There are no left out of pairs. Hence, 225 is a perfect square.

(iii) 2048 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

2 | 2048 |

2 | 1024 |

2 | 512 |

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

Grouping them into pairs of equal factors:

2048 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x 2

The last factor, 2 cannot be paired. Hence, 2048 is a perfect square.

(iv) 343 = 7 x 7 x 7

7 | 343 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

343 = (7 x 7) x 7

The last factor, 7 cannot be paired. Hence, 343 is not a perfect square.

(v) 441 = 3 x 3 x 7 x 7

3 | 441 |

3 | 147 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

441 = (3 x 3) x (7 x 7)

There are no left out of pairs. Hence, 441 is a perfect square.

(vi) 2916 = 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 x 3

2 | 2916 |

2 | 1458 |

3 | 729 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Grouping them into pairs of equal factors:

2916 = (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3)

There are no left out of pairs. Hence, 2916 is a perfect square.

(vii) 11025 = 3 x 3 x 5 x 7 x 7

3 | 11025 |

3 | 3675 |

5 | 1225 |

5 | 245 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

11025= (3 x 3) x (5 x 5) x (7 x 7)

There are no left out of pairs. Hence, 11025 is a perfect square.

(viii) 3549 = 3 x 7 x 13 x 13

3 | 3549 |

7 | 1183 |

13 | 169 |

13 | 13 |

1 |

Grouping them into pairs of equal factors:

3549 = (13 x 13) x 3 x 7

The last factors, 3 and 7 cannot be paired. Hence, 3549 is not a perfect square.

Hence, the perfect squares are 225, 441, 2916 and 11025.

**Question: 7**

By what number should each of the following numbers be multiplied to get a perfect square in each case? Also, find the number whose square is the new number.

Factorizing each number

(i) 8820 = 2 x 2 x 2 x 3 x 5 x 7 x 7

(ii) 3675 = 3 x 5 x 5 x 7 x 7

(iii) 605 = 5 x 11 x 11

(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5

(v) 4056 =2 x 2 x 2 x 3 x 13 x 13

(vi) 3468 = 2 x 2 x 3 x 17 x 17

(vii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

**Solution:**

(i) 8820 = 2 x 2 x 2 x 3 x 5 x 7 x 7

2 | 8820 |

2 | 4410 |

3 | 2205 |

3 | 735 |

5 | 245 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal of equal factors:

8820 = (2 x 2) x (3 x 3) x (7 x 7) x 5

The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired.

Hence, 8820 must be multiplied by 5 for it to be a perfect square.

The new number would be (2 × 2) x (3 x 3) x (7 x7) x (5 x 5).

Furthermore, we have:

(2 x 2) x (3 x 3) x (7 x 7) x (5 x 5) = (2 x 3 x 5 x 7) x ( 2 x 3 x 5 x 7)

Hence, the number whose square is the new number is:

2 x 3 x 5 x 7 = 210

(ii) 3675 = 3 x 5 x 5 x 7 x 7

3 | 3675 |

5 | 1225 |

5 | 245 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

3675 = (5 x 5) x (7 x 7) x 3

The factor 3 is not paired. For a number to be the perfect square, each prime factor has to be paired.

Hence, 3675 must be multiplied by 3 for it to be a perfect square.

The new number would be (5 x 5) x (7 x 7) x (3 x 3).

Furthermore, we have:

(5 x 5) x (7 x 7) x (3 x 3) = (3 x 5 x 7) x (3 x 5 x 7)

Hence, the number whose square is the new number is:

3 x 5 x 7 = 105

(iii) 605 = 5 x 11 x 11

5 | 605 |

11 | 121 |

11 | 11 |

1 |

Grouping them into pairs of equal factors:

605 = 5 x (11 x 11)

The factor 5 is not paired. For a number to be perfect square, each prime factor has to be paired.

Hence, 605 must be multiplied by 5 for it to be a perfect square.

The new number would be (5 x 5) x (11 x 11)

Furthermore, we have:

(5 x 5) x (11 x 11) = (5 x 11) x (5x 11)

Hence, the number whose square is the new number is:

5 x 11 = 55

(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5

2 | 2880 |

2 | 1440 |

2 | 720 |

2 | 360 |

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Grouping them into pairs of equal factors:

2880 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 5

There is a 5 as the leftover. For a number to be a perfect square, each prime factor has to be paired.

Hence, 2880 must be multiplied by 5 to be a perfect square.

The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5).

Furthermore, we have:

(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5) = (2 x 2 x 2 x 3 x 5) x (2 x 2 x 2 x 3 x 5) Hence, the number whose square is the new number is:

2 x 2 x 2 x 3 x 5 = 120

(v) 4056 = 2 x 2 x 2 x 3 x 13 x 13

2 | 4056 |

2 | 2028 |

2 | 1014 |

3 | 507 |

13 | 169 |

13 | 13 |

Grouping them into pairs of equal factors:

4056 = (2 x 2) x (13 x 13) x 2 x 3

The factors at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 4056 must me multiplied by 6 (2 x 3) for it to be a perfect square.

The new number would be (2 x 2) x (2 x2) x (3 x 3) x (13 x 13).

Furthermore, we have

(2 x 2) x (2 x 2) x (3 x 3) x (13 x 13) = (2 x 2 x 3 x 13) x (2 x 2 x 3 x13)

Hence, the number whose square is the new number is:

2 x 2 x 3 x 13 = 156

(vi) 3468 = 2 x 2 x 3 x 17 x 17

2 | 3468 |

2 | 1734 |

3 | 864 |

17 | 289 |

17 | 17 |

1 |

Grouping them into pairs of equal factors:

3468 = (2 x 2) x (17 x 17) x 3

The factor at the end, 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3468 must me multiplied by 3 for it to be a perfect square.

The new number would be (2 x 2) x (17 x 17) x (3 x 3).

Furthermore, we have

(2 x 2) x (17 x17) x (3 x3) = (2 x 3 x 17) x (2 x 3 x17)

Hence, the number whose square is the new number is:

2 x 3 x17 = 102

(vii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3

2 | 7776 |

2 | 3888 |

2 | 1944 |

2 | 972 |

2 | 486 |

3 | 243 |

3 | 81 |

3 | 27 |

3 | 9 |

3 | 3 |

1 |

Grouping them into pairs of equal factors:

7776 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x 2 x 3

The factor at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 7776 must me multiplied by 6 (2 x 3) for it to be a perfect square.

The new number would be (2 x 2) x (2 x 2) x (2 x 2) (3 x 3) x (3 x 3) x (3 x 3).

Furthermore, we have

(2 x 2) x (2 x 2) x (2 x 2) (3 x 3) x (3 x 3) x (3 x 3) = (2 x 2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 3 x 3 x 3)

Hence, the number whose square is the new number is:

2 x 2 x 2 x 3 x 3 x 3 = 216

**Question: 8**

By what numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.

Factorizing each number

(i) 16562 = 2 x 7 x 7 x 13 x13

(ii) 3698 = 2 x 43 x 43

(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7

(iv) 3174 = 2 x 3 x 23 x 23

(v) 1575 = 3 x 3 x 5 x 7

**Solution:**

(i) 16562 = 2 x 7 x 7 x 13 x 13

2 | 16562 |

7 | 8281 |

7 | 1183 |

13 | 169 |

13 | 13 |

1 |

Grouping them into pairs of equal factors:

16562 = 2 x (7 x 7) x (13 x 13)

The factor at the end, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 16652 must me multiplied by 2 for it to be a perfect square.

The new number would be (7 x 7) x (13 x 13).

Furthermore, we have

(7x 7) x (13 x 13) = (7 x 13) x (7 x 13)

Hence, the number whose square is the new number is:

7 x 13 = 91

(ii) 3698 = 2 x 43 x 43

2 | 3698 |

43 | 1849 |

43 | 43 |

1 |

Grouping them into pairs of equal factors:

3698 = 2 x (43 x 43)

The factor at the end, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3698 must me multiplied by 2 for it to be a perfect square.

The new number would be (43 x 43)

Hence, the number whose square is the new number is 43.

(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7

3 | 5103 |

3 | 1701 |

3 | 567 |

3 | 189 |

3 | 63 |

3 | 21 |

7 | 7 |

1 |

Grouping them into pairs of equal factors: 5103 = (3 x 3) x (3 x 3) x (3 x 3) x 7

The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired.

Hence, 5103 must be divided by 7 for it to be a perfect square. The new number would be (3 x 3) x (3 x 3) x (3 x 3).

Furthermore, we have: (3 x 3) x (3 x 3) x (3 x 3) = (3 x 3 x 3) x (3 x 3 x 3) Hence, the number whose square is the new number is:

3 x 3 x 3 = 27

(iv) 3174 = 2 x 3 x 23 x 23

2 | 3174 |

1587 | |

529 | |

23 | |

1 |

Grouping them into pairs of equal factors:

3174 = 2 x 3 x (23 x 23)

The factors, 2 and 3 are not paired.

For a number to be a perfect square, each prime factor has to be paired.

Hence, 3174 must be divided by 6 (2 x 3) for it to be a perfect square.

The new number would be (23 x 23).

Hence, the number whose square is the new number is 23.

(v) 1575 = 3 x 3 x 5 x 7

3 | 1575 |

3 | 525 |

5 | 175 |

5 | 35 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

1575 = (3 x 3) x (5 x 5) x 7

The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired.

Hence, 1575 must be divided by 7 for it to be a perfect square.

The new number would be (3 x 3) x (5 x 5).

Furthermore, we have:

(3 x 3) x (5 x 5) = (3 x 5) x (3 x 5)

Hence, the number whose square is the new number is: 3 x 5 = 15

**Question: 9**

Find the greatest number of two digits which is a perfect square.

**Solution:**

We know that 10^{2} is equal to 100 and 9^{2} is equal to 81.

Since 10 and 9 are consecutive numbers, there is no perfect square between 100 and 81.

Since 100 is the first perfect square that has more than two digits, 81 is the greatest two-digit perfect square.

**Question: 10**

Find the least number of three digits which is a perfect square.

**Solution:**

Let us make a list of the squares starting from 1.

1^{2 }= 1

2^{2} = 4

3^{2} = 9

4^{2} = 16

5^{2}=25

6^{2} = 36

7^{2}= 49

8^{2} = 64

9^{2} = 81

10^{2 }= 100

The square of 10 has three digits. Hence, the least three-digit perfect square is 100.

**Question: 11**

Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.

**Solution:**

4581 = 3 x 3 x 7 x 7 x 11

3 | 4851 |

3 | 1617 |

7 | 539 |

7 | 77 |

11 | 11 |

1 |

Grouping them into pairs of equal factors:

4851 = (3 x 3) x (7 x 7) x 11

The factor, 11 is not paired. The smallest number by which 4851 must be multiplied such that the resulting number is a perfect square is 11.

**Question: 12**

Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.

**Solution:**

Prime factorization of 28812:

28812 = 2 x 2 x 3 x 7 x 7 x 7 x 7

2 | 22812 |

2 | 14406 |

3 | 7203 |

7 | 2401 |

7 | 343 |

7 | 49 |

7 | 7 |

1 |

Grouping them into pairs of equal factors:

28812 = (2 x 2) x (7 x 7) x (7 x 7) x 3

The factor, 3 is not paired. Hence, the smallest number by which 28812 must be divided such that the resulting number is a perfect square is 3.

**Question:** **13**

Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the number whose square is the resulting number.

**Solution:**

Prime factorization of 1152:

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

2 | 1152 |

2 | 576 |

2 | 288 |

2 | 144 |

2 | 72 |

2 | 36 |

2 | 18 |

3 | 9 |

3 | 3 |

1 |

Grouping them into pairs of equal factors:

1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2

The factor, 2 at the end is not paired.

For a number to be a perfect square, each prime factor has to be paired.

Hence, 1152 must be divided by 2 for it to be a perfect square.

The resulting number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3).

Furthermore, we have:

(2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = (2 x 2 x 2 x 3) x (2 x 2 x 2 x 3)

Hence, the number whose square is the resulting number is: 2 x 2 x 2 x 3 = 24