问题1.以下每个数字的平方根的可能位数是多少?
一世。 9801
解决方案：
Unit place digit of the number is 1
And we all know 12 = 1 & 92 = 81 whose unit place is 1
Therefore, one’s digit of the square root of 9801 should equal to 1 or 9.
ii。 99856
解决方案：
Unit place digit of the number is 6
And we all know 62 = 36 & 42 = 16, both the squares have unit place 6.
Therefore, one’s digit of the square root of 99856 is equal to 6 or 4.
iii。 998001
解决方案：
Unit place digit of the number is 1
And we all know 12 = 1 & 92 = 81 whose unit place is 1
Therefore, one’s digit of the square root of 998001 should equal to 1 or 9.
iv。 657666025
解决方案：
Unit place digit of the number is 5
And we all know 52 = 25 whose unit place is 5
Therefore, one’s digit of the square root of 657666025 should equal to 5.
问题2。不做任何计算，找出肯定不是理想平方的数字。
一世。 153
解决方案：
Unit place digit of the number is 3.
Therefore, 153 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].
ii。 257
解决方案：
Unit place digit of the number is 7.
Therefore, 257 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].
iii。 408
解决方案：
Unit place digit of the number is 8.
Therefore, 408 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].
iv。 441
解决方案：
Unit place digit of the number is 1.
Therefore, 441 is a perfect square
问题3.通过反复相减的方法求出100和169的平方根。
解决方案：
For 100
100 – 1 = 99 [1]
99 – 3 = 96 [2]
96 – 5 = 91 [3]
91 – 7 = 84 [4]
84 – 9 = 75 [5]
75 – 11 = 64 [6]
64 – 13 = 51 [7]
51 – 15 = 36 [8]
36 – 17 = 19 [9]
19 -19 = 0 [10]
Here, subtraction has been performed for ten times.
Therefore, √100 = 10
For 169
169 – 1 = 168 [1]
168 – 3 = 165 [2]
165 – 5 = 160 [3]
160 – 7 = 153 [4]
153 – 9 = 144 [5]
144 – 11 = 133 [6]
133 – 13 = 120 [7]
120 – 15 = 105 [8]
105 – 17 = 88 [9]
88 – 19 = 69 [10]
69 – 21 = 48 [11]
48 – 23 = 25 [12]
25 – 25 = 0 [13]
Here, subtraction has been performed for thirteen times.
Therefore, √169 = 13
问题4.通过素因数分解法找到以下数字的平方根。
一世。 729
解决方案：
729 = 1 × 3 × 3 × 3 × 3 × 3 × 3
729 = (3 × 3) × (3 × 3) × (3 × 3)
729 = (3 × 3 × 3) × (3 × 3 × 3)
729 = (3 × 3 × 3)2
Therefore, √729 = 3 × 3 × 3 = 27
ii。 400
解决方案：
400 = 1 × 5 × 5 × 2 × 2 × 2 × 2
400 = (2 × 2) × (2 × 2) × (5 × 5)
400 = (2 × 2 × 5) × (2 × 2 × 5)
400 = (2 × 2 × 5)2
Therefore, √400 = 2 × 2 × 5 = 20
iii。 1764年
解决方案：
1764 = 2 × 2 × 3 × 3 × 7 × 7 × 1
1764 = (2 × 2) × (3 × 3) × (7 × 7)
1764 = (2 × 3 × 7) × (2 × 3 × 7)
1764 = (2 × 3 × 7)2
Therefore, √1764 = 2 × 3 × 7 = 42
iv。 4096
解决方案：
4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 1
4096 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)
4096 = (2 × 2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2 ×2)
4096 = (2 × 2 × 2 × 2 × 2 × 2)2
Therefore, √4096 = 2 × 2 × 2 × 2 × 2 × 2 = 64
第7744号
解决方案：
7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 × 1
7744 = (2 × 2) × (2 × 2) × (2 × 2) × (11 × 11)
7744 = (2 × 2 × 2 × 11) ×( 2 × 2 × 2 × 11)
7744 = (2 × 2 × 2 × 11)2
Therefore, √7744 = 2 × 2 × 2 × 11 = 88
vi。 9604
解决方案：
9604 = 2 × 2 × 7 × 7 × 7 × 7× 1
9604 = (2 × 2) × (7 × 7) × (7 × 7)
9604 = (2 × 7 × 7) × (2 × 7 ×7)
9604 = (2 × 7 × 7)2
Therefore, √9604 = 2 × 7 × 7 = 98
七。 5929
解决方案：
5929 = 7 × 7 × 11 × 11
5929 = (7 × 7) × (11 × 11)
5929 = (7 × 11) × (7 × 11)
5929 = (7 × 11)2
Therefore, √5929 = 7 × 11 = 77
八。 9216
解决方案：
9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 1
9216 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)
9216 = (2 × 2 × 2 × 2 × 2 × 3) × (2 × 2 × 2 × 2 × 2 × 3)
9216 = 96 × 96
9216 = (96)2
Therefore, √9216 = 96
ix。 529
解决方案：
529 = 23 × 23 × 1
529 = (23)2
Therefore, √529 = 23
X。 8100
解决方案：
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 1
8100 = (2 × 2) × (3 × 3) × (3 × 3) × (5 × 5)
8100 = (2 × 3 × 3 × 5) × (2 × 3 × 3 × 5)
8100 = 90 × 90
8100 = (90)2
Therefore, √8100 = 90
问题5.对于以下每个数字，找到最小的整数，然后将其乘以一个完美的平方数。同样，找到如此获得的平方数的平方根。
一世。 252
解决方案：
252 = 2 × 2 × 3 × 3 × 7
= (2 × 2) × (3 × 3) × 7
7 cannot be paired.
Therefore, multiply by 7 to get perfect square.
New number obtained = 252 × 7 = 1764
1764 = 2 × 2 × 3 × 3 × 7 × 7
1764 = (2 × 2) × (3 × 3) × (7 × 7)
1764 = (2 × 3 × 7)2
Therefore, √1764 = 2×3×7 = 42
ii。 180
解决方案：
180 = 2 × 2 × 3 × 3 × 5
= (2 × 2) × (3 × 3) × 5
5 cannot be paired.
Therefore, multiply by 5 to get perfect square.
New number obtained = 180 × 5 = 900
900 = 2 × 2 × 3 × 3 × 5 × 5 × 1
900 = (2 × 2) × (3 × 3) × (5 × 5)
900 = (2 × 3 × 5)2
Therefore, √900 = 2 × 3 × 5 = 30
iii。 1008
解决方案：
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7
= (2 × 2) × (2 × 2) × (3 × 3) × 7
7 cannot be paired.
Therefore, multiply by 7 to get perfect square.
New number obtained = 1008 × 7 = 7056
7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7
7056 = (2 × 2) × (2 × 2) × (3 × 3) × (7 × 7)
7056 = (2 × 2 × 3 × 7)2
Therefore, √7056 = 2 × 2 × 3 × 7 = 84
iv。 2028年
解决方案：
2028 = 2 × 2 × 3 × 13 × 13
= (2 × 2) × (13 × 13) × 3
3 cannot be paired.
Therefore, multiply by 3 to get perfect square.
New number obtained = 2028 × 3 = 6084
6084 = 2 × 2 × 3 × 3 × 13 ×13
6084 = (2 × 2) × (3 × 3) × (13 × 13)
6084 = (2 × 3 × 13)2
Therefore, √6084 = 2×3×13 = 78
1458号
解决方案：
1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3
= (3 × 3) × (3 × 3) × (3 × 3) × 2
2 cannot be paired.
Therefore, multiply by 2 to get perfect square.
New number obtained = 1458 × 2 = 2916
2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
2916 = (3 × 3) × (3 × 3) × (3 × 3) × (2 × 2)
2916 = (3×3×3×2)2
Therefore, √2916 = 3×3×3×2 = 54
vi。 768
解决方案：
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × 3
3 cannot be paired.
Therefore, multiply 768 by 3 to get perfect square.
New number obtained = 768×3 = 2304
2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
2304 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)
2304 = (2 × 2 × 2 × 2 × 3)2
√2304 = 2 × 2 × 2 × 2 × 3 = 48
问题6.对于以下每个数字，找到应该除以最小的整数，以获得一个完美的平方。同样，找到如此获得的平方数的平方根。
一世。 252
解决方案：
252 = 2 × 2 × 3 × 3 × 7
= (2 × 2) × (3 × 3) × 7
7 cannot be paired.
Divide 252 by 7 to get perfect square.
Therefore, New number obtained = 252 ÷ 7 = 36
36 = 2 × 2 × 3 × 3
36 = (2 × 2) × (3 × 3)
36 = (2 × 3)2
Therefore, √36 = 2 × 3 = 6
ii。 2925
解决方案：
252 = 2 × 2 × 3 × 3 × 7
= (2 × 2) × (3 × 3) × 7
7 cannot be paired.
Divide by 7 to get perfect square.
Therefore, New number obtained = 252 ÷ 7 = 36
36 = 2 × 2 × 3 × 3
36 = (2 × 2) × (3 × 3)
36 = (2 × 3)2
Therefore, √36 = 2 × 3 = 6
iii。 396
解决方案：
396 = 2 × 2 × 3 × 3 × 11
= (2 × 2) × (3 × 3) × 11
11 cannot be paired.
Divide by 11 to get perfect square.
Therefore, New number obtained = 396 ÷ 11 = 36
36 = 2 × 2 × 3 × 3
36 = (2 × 2) × (3 × 3)
36 = (2 × 3)2
Therefore, √36 = 2 × 3 = 6
iv。 2645
解决方案：
2645 = 5 × 23 × 23
2645 = (23 × 23) × 5
5 cannot be paired.
Divide by 5 to get perfect square.
Therefore, New number obtained = 2645 ÷ 5 = 529
529 = 23 × 23
529 = (23)2
Therefore, √529 = 23
v。2800
解决方案：
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7
= (2 × 2) × (2 × 2) × (5 × 5) × 7
7 cannot be paired.
Divide by 7 to get perfect square.
Therefore, New number obtained = 2800 ÷ 7 = 400
400 = 2 × 2 × 2 × 2 × 5 × 5
400 = (2 × 2) × (2 × 2) × (5 × 5)
400 = (2 × 2 × 5)2
Therefore, √400 = 20
vi。 1620
解决方案：
1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5
= (2 × 2) × (3 × 3) × (3 × 3) × 5
5 cannot be paired.
Divide by 5 to get perfect square.
Therefore, New number obtained = 1620 ÷ 5 = 324
324 = 2 × 2 × 3 × 3 × 3 × 3
324 = (2 × 2) × (3 × 3) × (3 × 3)
324 = (2 × 3 × 3)2
√324 = 18
问题7：一所学校八班的学生总共为总理的国家救济基金捐款2401卢比。每个学生捐赠的卢比与班上学生的数量一样多。找到班上的学生人数。
解决方案：
Let as assume number of students be, a
So, Each Student has donated Rs a.
Therefore, Total amount donated = a x a
That mean’s a x a = 2401
a2 = 2401
a2 = 7 × 7 × 7 × 7
a2 = (7 × 7) × (7 × 7)
a2 = 49 × 49
a = √(49 × 49)
a = 49
Therefore, The number of students = 49
问题8：要在花园中种植2025种植物，使每一行包含的植物数量与行数一样多。查找行数和每行中的植物数。
解决方案：
Let as assume number of rows be, a
So, Each row has number of plants = a.
Therefore, Total number of plants = a x a
That mean’s a x a = 2025
a2 = 3 × 3 × 3 × 3 × 5 × 5
a2 = (3 × 3) × (3 × 3) × (5 × 5)
a2 = (3 × 3 × 5) × (3 × 3 × 5)
a2 = 45 × 45
a = √(45 × 45)
a = 45
Therefore, The number of rows = 45 and also number of plants in each rows = 45.
问题9：找到可被数字4、9和10整除的最小平方数。
解决方案：
First, we have to find L.C.M of 4, 9 and 10
4 = 2 x 2 x 1
9 = 3 x 3 x 1
5 = 1 x 5
Therefore, L.C.M = (2 × 2 × 3 x 3 × 5) = 180.
Now we have to find the smallest whole number divisible by 180
180 = 2 × 2 × 9 × 5
= (2 × 2)× 3 × 3 × 5
= (2 × 2) × (3 × 3) × 5
5 cannot be paired.
Therefore, multiply 180 by 5 to get perfect square.
The smallest square number divisible by 180 and also by 4, 9 and 10 = 180 × 5
= 900
问题10。找到可被数字8、15和20整除的最小平方数。
解决方案：
First, we have to find L.C.M of 8, 15 and 20
8 = 1 x 2 x 2 x 2
15 = 1 x 5 x 3
20 = 1 x 2 x 5 x 2
Therefore, L.C.M = (2 × 2 × 5 × 2 × 3) = 120.
Now we have to find the smallest whole number divisible by 120
120 = 2 × 2 × 3 × 5 × 2
= (2 × 2) × 3 × 5 × 2
3, 5 and 2 cannot be paired.
Therefore, multiply 120 by (3 × 5 × 2) i.e 30 to get perfect square.
The smallest square number divisible by 120 and also by 8, 15 and 20 = 120 × 30
= 3600