Find the best tutors and institutes for Class 8 Tuition
Search in
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
We know that if a number has its unit’s place digit as a, then its square will end with the unit digit of the multiplication a × a.
(i) 81
Since the given number has its unit’s place digit as 1, its square will end with the unit digit of the multiplication (1 ×1 = 1) i.e., 1.
(ii) 272
Since the given number has its unit’s place digit as 2, its square will end with the unit digit of the multiplication (2 × 2 = 4) i.e., 4.
(iii) 799
Since the given number has its unit’s place digit as 9, its square will end with the unit digit of the multiplication (9 × 9 = 81) i.e., 1.
(iv) 3853
Since the given number has its unit’s place digit as 3, its square will end with the unit digit of the multiplication (3 × 3 = 9) i.e., 9.
(v) 1234
Since the given number has its unit’s place digit as 4, its square will end with the unit digit of the multiplication (4 × 4 = 16) i.e., 6.
(vi) 26387
Since the given number has its unit’s place digit as 7, its square will end with the unit digit of the multiplication (7 × 7 = 49) i.e., 9.
(vii) 52698
Since the given number has its unit’s place digit as 8, its square will end with the unit digit of the multiplication (8 × 8 = 64) i.e., 4.
(viii) 99880
Since the given number has its unit’s place digit as 0, its square will have two zeroes at the end. Therefore, the unit digit of the square of the given number is 0.
(xi) 12796
Since the given number has its unit’s place digit as 6, its square will end with the unit digit of the multiplication (6 × 6 = 36) i.e., 6.
(x) 55555
Since the given number has its unit’s place digit as 5, its square will end with the unit digit of the multiplication (5 × 5 = 25) i.e., 5.
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
The square of numbers may end with any one of the digits 0, 1, 5, 6, or 9. Also, a perfect square has even number of zeroes at the end of it.
(i) 1057 has its unit place digit as 7. Therefore, it cannot be a perfect square.
(ii) 23453 has its unit place digit as 3. Therefore, it cannot be a perfect square.
(iii) 7928 has its unit place digit as 8. Therefore, it cannot be a perfect square.
(iv) 222222 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(v) 64000 has three zeros at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.
(vi) 89722 has its unit place digit as 2. Therefore, it cannot be a perfect square.
(vii) 222000 has three zeroes at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.
(viii) 505050 has one zero at the end of it. However, since a perfect square cannot end with odd number of zeroes, it is not a perfect square.
The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826
(iii) 7779 (iv) 82004
The square of an odd number is odd and the square of an even number is even. Here, 431 and 7779 are odd numbers.
Thus, the square of 431 and 7779 will be an odd number.
Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1…2…1
100000012 = …
In the given pattern, it can be observed that the squares of the given numbers have the same number of zeroes before and after the digit 2 as it was in the original number. Therefore,
1,00,0012 = 10,00,02,00,001
1,00,00,0012 = 10,00,00,02,00,00,001
Observe the following pattern and supply the missing number.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = …
…2 = 10203040504030201
By following the given pattern, we obtain
10,10,1012 = 10,20,30,40,30,201
10,10,10,1012 = 10,20,30,40,50,40,30,201
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
From the given pattern, it can be observed that,
(i) The third number is the product of the first two numbers.
(ii) The fourth number can be obtained by adding 1 to the third number.
Thus, the missing numbers in the pattern will be as follows.
42 + 52 + 
= 212
52 + 
+ 302 = 312
62 + 72 + 
= 
Without adding find the sum
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
We know that the sum of first n odd natural numbers is n2.
(i) Here, we have to find the sum of first five odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = (5)2 = 25
(ii) Here, we have to find the sum of first ten odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100
(iii) Here, we have to find the sum of first twelve odd natural numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = (12)2 = 144
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11odd numbers.
We know that the sum of first n odd natural numbers is n2.
(i) 49 = (7)2
Therefore, 49 is the sum of first 7 odd natural numbers.
49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) 121 = (11)2
Therefore, 121 is the sum of first 11 odd natural numbers.
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
We know that there will be 2n numbers in between the squares of the numbers n and (n + 1).
(i) Between 122 and 132, there will be 2 × 12 = 24 numbers
(ii) Between 252 and 262, there will be 2 × 25 = 50 numbers
(iii) Between 992 and 1002, there will be 2 × 99 = 198 numbers
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 8 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 8 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
