问题1.写下以下数字的平方根的可能单位数字。这些数字中的哪一个是奇数平方根?
(i)9801
解决方案：
Unit digit of 9801 is 1
Unit digit of square root is 1 or 9
9801 is an odd number. Therefore, square root is also odd
(ii)99856
解决方案：
Unit digit of 99856 = 6
Unit digit of square root is 4 or 6
99856 is an even number. Therefore, square root is also even
(iii)998001
解决方案：
Unit digit of 998001 = 1
Unit digit of square root is 1 or 9
998001 is an odd number. Therefore, square root is also odd
(iv)657666025
解决方案：
Unit digit of 657666025 = 5
Unit digit of square root is 5
657666025 is an odd number. Therefore, square root is also odd
问题2。通过质因子分解找到以下各项的平方根。
(i)441
解决方案：
Prime factorization of
441 = 3×3×7×7 (Pairing of 3 and 7)
= 32×72
√441 = 3×7
= 21
(ii)196
解决方案：
Prime factorization of
196 = 2×2×7×7 (Pairing of 2 and 7)
= 22×72
√196 = 2×7
= 14
(iii)529
解决方案：
Prime factorization of
529 = 23×23 (Pairing of 23)
= 232
√529 = 23
(iv)1764年
解决方案：
Prime factorization of
1764 = 2×2×3×3×7×7 (Pairing of 2, 3 and 7)
= 22×32×72
√1764 = 2×3×7
= 42
(v)1156
解决方案：
Prime factorization of
1156 = 2×2×17×17 (Pairing of 2 and 17)
= 22×172
√1156 = 2×17
= 34
(vi)4096
解决方案：
Prime factorization of
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 (Pairing of 2)
= 212
√4096 = 26
= 64
(vii)7056
解决方案：
Prime factorization of
7056 = 2×2×2×2×21×21 (Pairing of 2 and 21)
= 22×22×212
√7056 = 2×2×21
= 84
(viii)8281
解决方案：
Prime factorization of
8281 = 91×91 (Pairing of 91)
= 912
√8281=91
(ix)11664
解决方案：
Prime factorization of
11664 = 2×2×2×2×3×3×3×3×3×3 (Pairing of 2 and 3)
= 22×22×32×32×32
√11664 = 2×2×3×3×3
= 108
(x)47089
解决方案：
Prime factorization of
47089 = 217×217 (Pairing of 217)
= 2172
√47089 = 217
(xi)24336
解决方案：
Prime factorization of
24336 = 2×2×2×2×3×3×13×13 (Pairing of 2,3 and 13)
= 22×22×32×132
√24336 = 2×2×3×13
= 156
(xii)190969
解决方案：
Prime factorization of
190969 = 23×23×19×19 (Pairing of 23 and 19)
= 232×192
√190969 = 23×19
= 437
(十三)586756
解决方案：
Prime factorization of
586756 = 2×2×383×383 (Pairing of 2 and 383)
= 22×3832
√586756 = 2×383
= 766
(十四)27225
解决方案：
Prime factorization of
27225 = 5×5×3×3×11×11 (Pairing of 5,3 and 11)
= 52×32×112
√27225 = 5×3×11
= 165
(xv)3013696
解决方案：
Prime factorization of
3013696 = 2×2×2×2×2×2×217×217 (Pairing of 2 and 17)
= 26×2172
√3013696 = 23×217
= 1736
问题3.找出必须乘以180的最小数，以使其成为一个完美的平方。另外，找到如此获得的理想平方的平方根。
解决方案：
Prime factorization of
180 = (2 × 2) × (3 × 3) × 5 (Pairing of 2 and 3)
=22 × 32 × 5
5 is left out
Therefore, multiplying the number with 5
180 × 5 = 22 × 32 × 52
Therefore, square root of √ (180 × 5) = 2 × 3 × 5
= 30
问题4。找出最小的数字，必须乘以147，以使其成为一个完美的平方。另外，找到如此获得的数字的平方根。
解决方案：
Prime factorization of
147 = (7 × 7) × 3 (Pairing of 7)
=72 × 3
3 is left out
Therefore, multiplying the number with 3
147 × 3 = 72 × 32
Therefore, square root of √ (147 × 3) = 7 × 3
= 21
问题5.找到3645必须除以的最小数，以使其成为一个完美的平方。另外，找到所得数字的平方根。
解决方案：
Prime factorization of
3645 = (3 × 3) × (3 × 3) × (3 × 3) × 5 (Pairing of 3)
=32 × 32 × 32 × 5
5 is left out
Therefore, by dividing with 5
3645 ÷ 5 = 32 × 32 × 32
Therefore, square root of √ (3645 ÷ 5) = 3 × 3 × 3
= 27
问题6.找到必须除以1152的最小数，以使其成为正方形。另外，找到如此获得的数字的平方根。
解决方案：
Prime factorization of
1152 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3) (Pairing of 2 and 3)
=22 × 22 × 22 × 32 × 2
2 is left out
Therefore, by dividing with number 2
1152 ÷ 2 = 22 × 22 × 22 × 32
Therefore, square root of √ (1152 ÷ 2) = 2 × 2 × 2 × 3
= 24
问题7.两个数字的乘积是1296。如果一个数字是另一个的16倍，请找到这些数字。
解决方案：
Let us consider two numbers x and y
y =16x
x × y= 1296
x × 16x = 1296
16x2 = 1296
x2 = 1296/16 = 81
x = 9
y = 16x
= 16(9)
= 144
Therefore, y =144 and x =9
问题8.一家福利协会从居民那里收取了202500卢比的捐款。如果每个人支付的卢比与当地居民的卢比一样多，请找到居民数。
解决方案：
Let us consider total residents as x
So, each paid Rs. x
Total collection = x (x) = x2
Total Collection = 202500
x = √ 202500
x = √(2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5 × 5)
= 2 × 3 × 3 × 5 × 5
= 450
Therefore, total residents = 450
问题9：一个社会收取了92.16卢比。每个成员收集的比萨饼数量与成员人数一样多。那里有多少成员，每个成员贡献了多少?
解决方案：
Let there be x members and x paisa collected by each member
Therefore,
x2=Total amount= 9216 paisa
x2 = 9216
x = √9216
= 2 × 2 × 2 × 12
= 96
Therefore, there were 96 members in the society and each contributed 96 paisas.
问题10：一个社会向其学生收取了2304卢比的费用。如果每个学生支付的比萨饼与学校里的学生一样多，那么学校里有多少个学生?
解决方案：
Let there be x number of students and each contributed Rs.x
Total money obtained = x2 = 2304 paisa
x = √2304
x = √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
x = 2 × 2 × 2 × 2 × 3
x = 48
Therefore, there were 48 students in the school