1. What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
vi. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
i.Unit digit of given number 81 = 1
Hence, unit digit of square of 1 = 1
ii. Unit digit of given number 272 = 2
Hence, unit digit square of 2 = 4
iii. Unit digit of given number 799 = 9
square of 9 = 81
Hence, its unit digit = 1
iv. Unit digit of given number 3853 =3
Hence, unit digit of suare of 3 = 9
v. Unit digit of given number 1234 =4
square of 4 = 16
Hence, unit digit = 6
vi. Unit digit of given number 26387 =7
square of 7 = 49
Hence, unit digit number = 9
vii. Unit digit of given number 52698 = 8
square of 8 = 64
Hence, its unit digit = 4
viii. The unit digit of given number 99880 = 0
Hence, Unit digit of the square of number 99880 is equal to 0.
ix. The unit digit of given number 12796 = 6
Square of 6 = 36
Hence, its unit digit = 6
x. The unit digit of given number 55555 = 5
square of 5 = 25
Hence, its unit digit = 5
2. The following numbers are obviously not perfect squares. Give reason.
i. 1057
ii. 23453
iii. 7928
iv. 222222
v. 64000
vi. 89722
vii. 222000
viii. 505050
Solution:
We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.
i. 1057 =Not a perfect square because it ends with 7
ii. 23453 = Not a perfect square because it ends with 3
iii. 7928 = Not a perfect square because it ends with 8
iv. 222222 = Not a perfect square because it ends with 2
v. 64000 = Not a perfect square because it ends with 0
vi. 89722 = Not a perfect square because it ends with 2
vii. 222000 = Not a perfect square because it ends with 0
viii. 505050 = Not a perfect square because it ends with 0
3. The squares of which of the following would be odd numbers?
i. 431
ii. 2826
iii. 7779
iv. 82004
Solution:
We know that the square of an odd number is odd and the square of an even number is even.
i. The square of 431 is an odd number.
ii. The square of 2826 is an even number.
iii. The square of 7779 is an odd number.
iv. The square of 82004 is an even number.
4. Observe the following pattern and find the missing numbers.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …….2……1
100000012 = ………………
Solution:
According to the above pattern:
1000012 = 10000200001
100000012 = 100000020000001
5. Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Solution:
According to the above pattern:
10101012 =1020304030201
1010101012 =10203040505030201
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
Solution:
According to the given pattern, we have
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432
7. Without adding, find the sum.
We know that the sum of n odd numbers = n2
i. 1 + 3 + 5 + 7 + 9
Solution:
Sum of first five odd number = (5)2 = 25
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Solution:
Sum of first ten odd number = (10) 2 = 100
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Sum of first thirteen odd number = (12) 2 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Solution:
We know, sum of first n odd natural numbers = n2
Since,49 = 72
Here n = 7
Hence, 49 = sum of first 7 odd natural numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
Since, 121 = 112
Here n = 11
Hence, 121 = sum of first 11 odd natural numbers
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Solution:
We know that between n2 and (n+1)2, there are 2n non–perfect square numbers.
Now,
i. 122 and 132 there are 2×12 = 24 natural numbers.
ii. 252 and 262 there are 2×25 = 50 natural numbers.
iii. 992 and 1002 there are 2×99 =198 natural numbers.
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