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Question 1. What will be the unit digit of the squares of the following numbers?
81
272
799
3853
1234
26387
52698
99880
12796
55555
Answer.
i. 81
The unit digit (1) of the square of the following number is 1 x 1 = 1
ii. 272
The unit digit is 2 of the square of the following number is 2 x 2 = 4
iii. 799
The unit digit is 9 of the square of the following number is 9 x 9 = 81
iv. 3853
The unit digit is 3 of the square of the following number is 3 x 3 = 9
v. 1234
The unit digit is 4 of the square of the following number is 4 x 4 = 16
vi. 26387
The unit digit is 7 of the square of the following number is 7 x 7 = 49
vii. 52698
The unit digit is 8 of the square of the following number is 8 x 8 = 64
viii. 99880
The unit digit is 0 of the square of the following number is 0 x 0 = 0
ix. 12796
The unit digit is 6 of the square of the following number is 6 x 6 = 36
x. 55555
The unit digit is 5 of the square of the following number is 5 x 5 = 25
Question 2. The following numbers are not perfect squares. Give a reason.
1057
23453
7928
222222
64000
89722
222000
505050
Answer.
i. 1057
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 1057, and the unit digit is 7. Thus, the following numbers are not perfect squares.
ii. 23453
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 23453, and the unit digit is 3. Thus, the following numbers are not perfect squares.
iii. 7928
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 1057, and the unit digit is 7. Thus, the following numbers are not perfect squares.
iv. 222222
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 7928, and the unit digit is 2. Thus, the following numbers are not perfect squares.
v. 64000
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 64000, and the unit digit is 0. Thus, the following numbers are not perfect squares.
vi. 89722
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 89722, and the unit digit is 0. Thus, the following numbers are not perfect squares.
vii. 222000
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 222000, and the unit digit is 0. Thus, the following number is not perfect squares.
viii. 505050
Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.
The following number is 505050, and the unit digit is 0. Thus, the following number is not perfect squares.
Question 3. The squares of which of the following would be odd numbers?
431
2826
7779
82004
Answer.
i. 431
We know that an odd number's square is odd, while an even number's square is even. The square of 431 is an odd number.
ii. 2826
We know that an odd number's square is odd, while an even number's square is even. The square of 2826 is an even number.
iii. 7779
We know that an odd number's square is odd, while an even number's square is even. The square of 7779 is an odd number.
iv. 82004
We know that an odd number's square is odd, while an even number's square is even. The square of 82004 is an even number.
Question 4. Observe the following pattern and find the missing numbers. 112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …….2………1
100000012 = ……………………..
Answer.
The square on the R.H.S of the equality contains an odd number of digits, with the first and last digits being 1 and the middle digit being 2. And the number of zeros between the left and middle digits 1 and 2 and the right and middle digits 1 and 2 is the same as the number of zeros in the supplied integer.
∴, 1000012 = 100020001
100000012 = 100000020000001
Question 5. Observe the following pattern and supply the missing numbers. 112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Answer.
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
We see that the square of the R.H.S. number of the equality has an odd number of digits, with the first and last digits both being 1. Furthermore, the square is symmetrical around the central digit. If the middle digit is 4, the squared number is 10101, and the squared number is 102030201.
∴, 10101012 = 1020304434
1010101012 = 10203040504030201
Question 6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + _2 = 212
52 + _ 2 + 302 = 312
62 + 72 + _ 2 = _ 2
Answer.
Given, 12 + 22 + 22 = 32
i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2
22 + 32 + 62 = 72
∴, 22 + 32 + (2 x 3)2 = 72
= (22 + 32 -2 x 3)2
32 + 42 + 122 = 132
∴, 32 + 42 + (3 x 4)2 = 132
= (32 + 42 -3 x 4)2
42 + 52 + _2 = 212
∴, 42 + 52 + (4 x 5)2 = 212
= (42 + 52 -4 x 5)2
52 + _ 2 + 302 = 312
∴, 52 + 62 + (5 x 6)2 = 312
= (52 + 62 -5 x 6)2
6 + 7 + _ 2 = _ 2
∴, 62 + 72 + (6 x 7)2 = 432
= (62 + 72 -6 x 7)2
Question 7. Without adding, find the sum.
1 + 3 + 5 + 7 + 9
1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Answer.
i. 1 + 3 + 5 + 7 + 9
(5)2 = 25 as the sum of the first five odd numbers.
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
(10)2 = 100 as the sum of the first ten odd numbers added together.
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
(12)2 = 144, which is the total of the first thirteen odd numbers.
Question 8. (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Answer.
(i) Express 49 as the sum of 7 odd numbers.
We already know that the sum of the first n odd natural numbers is n2.
Because 49 is 72, 1+3+5+7+9+11+13 = sum of the first 7 odd natural numbers
(ii) Express 121 as the sum of 11 odd numbers.
We already know that the sum of the first n odd natural numbers is n2.
Because 121 is 112, 1+3+5+7+9+11+13+15+17+19+21 = sum of the first 11 odd natural numbers
Question 9. How many numbers lie between squares of the following numbers?
12 and 13
25 and 26
99 and 100
Answer.
i. 12 and 13
Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 212 = 24 natural numbers) between 122 and 132.
ii. 25 and 26
Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 225 = 50 natural numbers between 252 and 262.
iii. 99 and 100
Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 299 =198 natural numbers between 992 and 1002.