Learn Exercise 6.3 with Free Lessons & Tips

(i) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 65,76,66,025 is 5.

We know that the sum of the first n odd natural numbers is n².

Consider.

(i) 100 − 1 = 99 (ii) 99 − 3 = 96 (iii) 96 − 5 = 91

(iv) 91 − 7 = 84 (v) 84 − 9 = 75 (vi) 75 − 11= 64

(vii) 64 − 13 = 51 (viii) 51 − 15 = 36 (ix) 36 − 17 = 19

(x) 19 − 19 = 0

We have subtracted successive odd numbers starting from 1 to 100, and obtained 0 at 10^th step.

Therefore,

The square root of 169 can be obtained by the method of repeated subtraction as follows.

(i) 169 − 1 = 168 (ii) 168 − 3 = 165 (iii) 165 − 5 = 160

(iv) 160 − 7 = 153 (v) 153 − 9 = 144 (vi) 144 − 11 = 133

(vii) 133 − 13 = 120 (viii) 120 − 15 = 105 (ix) 105 − 17 = 88

(x) 88 − 19 = 69 (xi) 69 − 21 = 48 (xii) 48 − 23 = 25

(xiii)25 − 25 = 0

We have subtracted successive odd numbers starting from 1 to 169, and obtained 0 at 13^th step.

Therefore,

(i) 729 can be factorised as follows.

3	729
3	243
3	81
3	27
3	9
3	3
	1

729 = 3 × 3 × 3 × 3 × 3 × 3

∴ = 27

(ii) 400 can be factorised as follows.

2	400
2	200
2	100
2	50
5	25
5	5
	1

400 = 2 × 2 × 2 × 2 × 5 × 5

∴ = 20

(iii) 1764 can be factorised as follows.

2	1764
2	882
3	441
3	147
7	49
7	7
	1

1764 = 2 × 2 × 3 × 3 × 7 × 7

∴ = 42

(iv) 4096 can be factorised as follows.

2	4096
2	2048
2	1024
2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

∴ = 64

(v) 7744 can be factorised as follows.

2	7744
2	3872
2	1936
2	968
2	484
2	242
11	121
11	11
	1

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

∴ = 88

(vi) 9604 can be factorised as follows.

2	9604
2	4802
7	2401
7	343
7	49
7	7
	1

9604 = 2 × 2 × 7 × 7 × 7 × 7

∴ = 98

(vii) 5929 can be factorised as follows.

7	5929
7	847
11	121
11	11
	1

5929 = 7 × 7 × 11 × 11

∴ = 77

(viii) 9216 can be factorised as follows.

2	9216
2	4608
2	2304
2	1152
2	576
2	288
2	144
2	72
2	36
2	18
3	9
3	3
	1

9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3× 3

∴ = 96

(ix) 529 can be factorised as follows.

23	529
23	23
	1

529 = 23 × 23

(x) 8100 can be factorised as follows.

2	8100
2	4050
3	2025
3	675
3	225
3	75
5	25
5	5
	1

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

∴ = 90

It is given that each student donated as many rupees as the number of students of the class. Number of students in the class will be the square root of the amount donated by the students of the class.

The total amount of donation is Rs 2401.

Number of students in the class =

∴

Hence, the number of students in the class is 49.

It is given that in the garden, each row contains as many plants as the number of rows.

Hence,

Number of rows = Number of plants in each row

Total number of plants = Number of rows × Number of plants in each row

Number of rows × Number of plants in each row = 2025

(Number of rows)² = 2025

∴

Thus, the number of rows and the number of plants in each row is 45.

The number that will be perfectly divisible by each one of 4, 9, and 10 is their LCM. The LCM of these numbers is as follows.

LCM of 4, 9, 10 = 2 × 2 × 3 × 3 × 5 =180

Here, prime factor 5 does not have its pair. Therefore, 180 is not a perfect square. If we multiply 180 with 5, then the number will become a perfect square. Therefore, 180 should be multiplied with 5 to obtain a perfect square.

Hence, the required square number is 180 × 5 = 900