问题1：以下数字的平方的单位位数是多少?
解决方案：
要找到任何数字的平方的单位位数，请找到该数字的平方的位数，然后
其单位数字将为实际数字平方的单位数字。
例如：-如果数字为2007，则2007单位数字的平方为49(7平方)。并且49的单位数字为9。因此，2007年平方的单位数字为9。
(i)81
Ans: 1
Reason: If a number has 1 or 9 in the units place, then its square ends in 1.
(or) Unit digit of 81 is 1. and 12 = 1, therefore unit digit of square of 81 is 1.
(ii)272
Ans: 4
Reason: Unit digit of 272 is 2 and 22 = 4. Therefore unit digit of the square of 272 is 4.
(iii)799
Ans: 1
Reason: If a number has 1 or 9 in the units place, then its square ends in 1. (or) Unit digit of 799 is 9 and 92 = 81, and unit digit of 81 is 1. Therefore unit digit of square of 799 is 1.
(iv)3853
Ans: 9
Reason: Unit digit of 3853 is 3 and 32 = 9, therefore unit digit of square of 3853 is 9.
(v)1234
Ans: 6
Reason: Unit digit of 1234 is 4 and 42 = 16, and the unit digit of 16 is 6. Therefore the unit digit of the square of 1234 is 6.
(vi)26387
Ans: 9
Reason: Unit digit of 26387 is 7 and 72 = 49, and the unit digit of 49 is 9. Therefore the unit digit of the square of 26387 is 9.
(vii)52698
Ans: 4
Reason: Unit digit of 52698 is 8 and 82 = 64, and the unit digit of 64 is 4. Therefore the unit digit of the square of 52698 is 4.
(viii)99880
Ans: 0
Reason: If a number has 0 in its unit place then square of its number will also have 0 in its units place (since 02 = 0).
(ix)12796
Ans: 6
Reason: Unit digit of 12796 is 6 and 62 = 36, and the unit digit of 36 is 6. Therefore the unit digit of the square of 12796 is 6.
(x)55555
Ans: 5
Reason: The unit digit of 55555 is 5 and 52 = 25, and the unit digit of 25 is 5. Therefore the unit digit of the square of 55555 is 5.
范式2：以下数字显然不是完美的平方。给出理由。
(i)1057(ii)23453(iii)7928(iv)222222
(v)64000(vi)89722(vii)222000(viii)505050
解决方案：
- If a number ends with 2 (or) 3 (or) 7 (or) 8 at units place then we can tell it is not a perfect square. From the above rule, we can tell 1057, 23453, 7928, 222222, 89722 are not perfect squares.
- For a number to be a perfect square it should have an even number of zeros at the end. From rule 2 we can tell 64000, 222000, 505050 are not perfect squares as they are having an odd number of zeros at the end.
问题3：以下哪个平方是奇数?
(i)431(ii)2826(iii)7779(iv)82004
Ans: 431 and 7779
Reason: The square of an odd number will be an odd number, and the square of an even number will be an even number.
问题4：观察以下模式并找到丢失的数字。
11 2 = 121
101 2 = 10201
1001 2 = 1002001
100001 2 = 1………2………1
10000001 2 =…………………………
Ans:
1000012 = 10000200001
100000012 = 100000020000001
问题5：观察以下模式并提供缺少的数字。
11 2 = 121
101 2 = 10201
10101 2 = 102030201
1010101 2 =…………………………
………… 2 = 10203040504030201
Ans:
10101012 = 1020304030201
1010101012 = 10203040504030201
问题6：使用给定的模式，找到丢失的数字。
1 2 + 2 2 + 2 2 = 3 2
2 2 + 3 2 + 6 2 = 7 2
3 2 + 4 2 + 12 2 = 13 2
4 2 + 5 2 + __ 2 = 21 2
5 2 + __ 2 + 30 2 = 31 2
6 2 + 7 2 + __ 2 = __ 2
解决方案：
To find pattern third number is related to first and second number. How?
If we multiply the first and second numbers we will get the third number.
The fourth number is related to the third number. How?
Forth number = third number + 1
42 + 52 + __2 = 212
Ans: 4 * 5 = 20
52 + __2 + 302 = 312
Ans: 5 * x = 30
=> x = (30 / 5) = 6
62 + 72 + x2 = y2
Ans: x = 6 * 7 = 42
y = x + 1 = 43
问题7：不加总和即可找到总和。
解决方案：
前n个奇数自然数的总和为n 2
(i)1 + 3 + 5 + 7 + 9
Ans: 25
Sum of first 5 odd natural numbers is 52, and 52 = 25.
(ii)1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Ans: 100
Sum of first 10 odd natural numbers is 102, and 102 = 100.
(iii)1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Ans: 144
Sum of first 12 odd natural numbers is 122, and 122 = 144.
问题8
(i)将49表示为7个奇数之和。
Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13
Explanation: 49 = 72 therefore the sum of the first 7 odd natural numbers is 49.
(ii)将121表示为11个奇数之和
Ans: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
Explanation: 121 = 112 therefore sum of first 11 odd natural numbers is 121.
问题9：以下数字的平方之间有多少个数字?
在n 2和(n + 1) 2之间，将有2n个自然数
(i)12和13
Ans: 24 (as n is 12, 2n will be 24)
(ii)25和26
Ans: 50 (as n is 25, 2n will be 50)
(iii)99和100
Ans: 198 (as n is 99, 2n will be 198)