Question 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

Accepted Answer

To find the unit digit of the square of a number, we only need to consider the unit digit of the number itself and then find the unit digit of its square.

(i) 81 ends with 1; 1^2 = 1 → unit digit 1
(ii) 272 ends with 2; 2^2 = 4 → unit digit 4
(iii) 799 ends with 9; 9^2 = 81 → unit digit 1
(iv) 3853 ends with 3; 3^2 = 9 → unit digit 9
(v) 1234 ends with 4; 4^2 = 16 → unit digit 6
(vi) 26387 ends with 7; 7^2 = 49 → unit digit 9
(vii) 52698 ends with 8; 8^2 = 64 → unit digit 4
(viii) 99880 ends with 0; 0^2 = 0 → unit digit 0
(ix) 12796 ends with 6; 6^2 = 36 → unit digit 6
(x) 55555 ends with 5; 5^2 = 25 → unit digit 5 — The unit digit of the square depends only on the unit digit of the original number. Squaring the unit digit and taking its unit digit gives the answer.

Question 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

Accepted Answer

A number is a perfect square only if its unit digit is 0,1,4,5,6 or 9. Also, perfect squares have specific patterns.

(i) 1057 ends with 7; 7^2 ends with 9 or 3, so 1057 cannot be a perfect square.
(ii) 23453 ends with 3; no perfect square ends with 3.
(iii) 7928 ends with 8; no perfect square ends with 8.
(iv) 222222 ends with 2; no perfect square ends with 2.
(v) 64000 ends with 0 but the number of zeros is not even (perfect squares ending with 0 have even number of zeros), so not a perfect square.
(vi) 89722 ends with 2; no perfect square ends with 2.
(vii) 222000 ends with 0 but number of zeros is odd, so not a perfect square.
(viii) 505050 ends with 0 but number of zeros is odd, so not a perfect square. — Unit digit rules and zero count rules help identify if a number can be a perfect square.

Question 3

The squares of which of the following would be odd numbers?

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Accepted Answer

Squares of odd numbers are odd; squares of even numbers are even.

(i) 431 is odd → square is odd
(ii) 2826 is even → square is even
(iii) 7779 is odd → square is odd
(iv) 82004 is even → square is even — Odd numbers squared remain odd; even numbers squared remain even.

Question 4

Observe the following pattern and find the missing digits.

11^2 = 121

101^2 = 10201

1001^2 = 1002001

100001^2 = 1 ... 2 ... 1

10000001^2 = ...

Accepted Answer

The pattern shows that:

11^2 = 121
101^2 = 10201
1001^2 = 1002001

Notice the pattern: For 1 followed by n zeros and 1, the square is 1 followed by n-1 zeros, then 2, then n-1 zeros, then 1.

For 100001^2 (which is 1 followed by 4 zeros and 1):

Square = 1 0000 2 0000 1 = 10000200001

For 10000001^2 (1 followed by 6 zeros and 1):

Square = 1 000000 2 000000 1 = 100000020000001 — The square of numbers of the form 1...01 follows a pattern where the square has 1, then zeros, then 2, then zeros, then 1.

Question 5

Observe the following pattern and supply the missing numbers.

11^2 = 121

101^2 = 10201

10101^2 = 102030201

1010101^2 = ...

... 2 = 10203040504030201

Accepted Answer

The pattern is:

11^2 = 121
101^2 = 10201
10101^2 = 102030201

The numbers are of the form 1 0 1 0 ... 1, and their squares have a pattern of increasing digits from 1 up to n and then decreasing.

For 1010101^2 (which has 4 ones separated by zeros):

Square = 102030405060504030201

The last missing number is 10203040504030201, which corresponds to 101010101^2.

So,

1010101^2 = 102030405060504030201

and

101010101^2 = 10203040504030201 — The squares of numbers with alternating 1s and 0s follow a palindromic pattern with increasing and then decreasing digits.

Question 6

Using the given pattern, find the missing numbers.

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

Accepted Answer

Observing the pattern:

1^2 + 2^2 + 2^2 = 9 = 3^2
2^2 + 3^2 + 6^2 = 49 = 7^2
3^2 + 4^2 + 12^2 = 169 = 13^2

For the next:
4^2 + 5^2 + x = 21^2 = 441
16 + 25 + x = 441
41 + x = 441
x = 441 - 41 = 400

Similarly,
5^2 + y + 30^2 = 31^2 = 961
25 + y + 900 = 961
925 + y = 961
y = 36

Finally,
6^2 + 7^2 + z = w
36 + 49 + z = w
Since the last line is incomplete, assuming the pattern continues similarly:

The missing numbers are:

4^2 + 5^2 + 20^2 = 21^2 (since 20^2=400)
5^2 + 6^2 + 30^2 = 31^2 (6^2=36)
6^2 + 7^2 + 42^2 = 43^2 (42^2=1764, 43^2=1849)

So the missing numbers are 20, 6, and 42 respectively. — The pattern shows sums of squares of three numbers equal to the square of another number. Using algebra, missing numbers are found by subtracting known squares from the square on the right side.

Question 7

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

Accepted Answer

Sum of first n odd numbers = n^2

(i) There are 5 odd numbers, so sum = 5^2 = 25

(ii) There are 10 odd numbers, so sum = 10^2 = 100

(iii) There are 12 odd numbers, so sum = 12^2 = 144 — The sum of first n odd numbers is always n squared.

Question 8

(i) Express 49 as the sum of 7 odd numbers.

(ii) Express 121 as the sum of 11 odd numbers.

Accepted Answer

(i) Since sum of first n odd numbers = n^2, 49 = 7^2, so sum of first 7 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49

(ii) 121 = 11^2, so sum of first 11 odd numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121 — Using the formula for sum of first n odd numbers, express the given numbers as sums of consecutive odd numbers.

Question 9

How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

Accepted Answer

(i) Squares: 12^2 = 144, 13^2 = 169
Numbers between 144 and 169 are 169 - 144 - 1 = 24

(ii) Squares: 25^2 = 625, 26^2 = 676
Numbers between 625 and 676 are 676 - 625 - 1 = 50

(iii) Squares: 99^2 = 9801, 100^2 = 10000
Numbers between 9801 and 10000 are 10000 - 9801 - 1 = 198 — The count of numbers between squares of n and (n+1) is (n+1)^2 - n^2 - 1 = 2n

Question 10

Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Accepted Answer

Solution:
(i) 32² = 32 × 32 = 1024
(ii) 35² = 35 × 35 = 1225
(iii) 86² = 86 × 86 = 7396
(iv) 93² = 93 × 93 = 8649
(v) 71² = 71 × 71 = 5041
(vi) 46² = 46 × 46 = 2116 — To find the square of a number, multiply the number by itself.
For example, 32² means 32 × 32 = 1024.

Question 11

Write a Pythagorean triplet whose one member is.

(i) 6

(ii) 14

(iii) 16

(iv) 18

Accepted Answer

Solution:
A Pythagorean triplet consists of three positive integers a, b, c such that a² + b² = c².

(i) For 6:
One known triplet is (6, 8, 10) because 6² + 8² = 36 + 64 = 100 = 10².

(ii) For 14:
One triplet is (14, 48, 50) because 14² + 48² = 196 + 2304 = 2500 = 50².

(iii) For 16:
One triplet is (16, 30, 34) because 16² + 30² = 256 + 900 = 1156 = 34².

(iv) For 18:
One triplet is (18, 24, 30) because 18² + 24² = 324 + 576 = 900 = 30². — To find a Pythagorean triplet with a given member, find two other integers such that the sum of the squares of the smaller two equals the square of the largest.
Use known triplets or generate by formulas.

Question 12

TRY THESE

(i) 11² = 121. What is the square root of 121?

(ii) 14² = 196. What is the square root of 196?

Accepted Answer

Solution:
(i) Since 11² = 121, the square root of 121 is 11.
(ii) Since 14² = 196, the square root of 196 is 14. — Square root is the inverse operation of squaring. Since 11 squared is 121, √121 = 11.
Similarly, √196 = 14.

Question 13

TRY THESE

By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square then find its square root.

(i) 121
(ii) 55
(iii) 36
(iv) 49
(v) 90

Accepted Answer

Solution:
Use repeated subtraction of successive odd numbers starting from 1.

(i) 121:
121 - 1 = 120
120 - 3 = 117
117 - 5 = 112
112 - 7 = 105
105 - 9 = 96
96 - 11 = 85
85 - 13 = 72
72 - 15 = 57
57 - 17 = 40
40 - 19 = 21
21 - 21 = 0
Since remainder is 0 at 11th step, 121 is a perfect square and √121 = 11.

(ii) 55:
55 - 1 = 54
54 - 3 = 51
51 - 5 = 46
46 - 7 = 39
39 - 9 = 30
30 - 11 = 19
19 - 13 = 6
6 - 15 = -9 (negative)
Since remainder becomes negative, 55 is not a perfect square.

(iii) 36:
36 - 1 = 35
35 - 3 = 32
32 - 5 = 27
27 - 7 = 20
20 - 9 = 11
11 - 11 = 0
Remainder zero at 6th step, so 36 is perfect square and √36 = 6.

(iv) 49:
49 - 1 = 48
48 - 3 = 45
45 - 5 = 40
40 - 7 = 33
33 - 9 = 24
24 - 11 = 13
13 - 13 = 0
Remainder zero at 7th step, so 49 is perfect square and √49 = 7.

(v) 90:
90 - 1 = 89
89 - 3 = 86
86 - 5 = 81
81 - 7 = 74
74 - 9 = 65
65 - 11 = 54
54 - 13 = 41
41 - 15 = 26
26 - 17 = 9
9 - 19 = -10 (negative)
Remainder negative, so 90 is not a perfect square. — Repeated subtraction of successive odd numbers from a perfect square will reduce it to zero exactly after n steps, where n is the square root.
If remainder becomes negative before zero, number is not a perfect square.

Question 14

What could be the possible ‘one’s’ digits of the square root of each of the following numbers?

(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025

Accepted Answer

To find the possible one's digit of the square root, observe the one's digit of the given number and recall the possible one's digits of perfect squares:

- Squares ending with 1 have roots ending with 1 or 9.
- Squares ending with 6 have roots ending with 4 or 6.
- Squares ending with 5 have roots ending with 5.
- Squares ending with 0 have roots ending with 0.
- Squares ending with 9 have roots ending with 3 or 7.
- Squares ending with 4 have roots ending with 2 or 8.

(i) 9801 ends with 1, so root ends with 1 or 9.
(ii) 99856 ends with 6, so root ends with 4 or 6.
(iii) 998001 ends with 1, so root ends with 1 or 9.
(iv) 657666025 ends with 5, so root ends with 5. — The one's digit of a perfect square determines the possible one's digit of its square root. Using the known pattern of squares:

- 0²=0 ends with 0
- 1²=1 ends with 1
- 2²=4 ends with 4
- 3²=9 ends with 9
- 4²=16 ends with 6
- 5²=25 ends with 5
- 6²=36 ends with 6
- 7²=49 ends with 9
- 8²=64 ends with 4
- 9²=81 ends with 1

Thus, the possible one's digits of the root correspond to the digits whose squares end with the given digit.

Question 15

Without doing any calculation, find the numbers which are surely not perfect squares.

(i) 153
(ii) 257
(iii) 408
(iv) 441

Accepted Answer

Check the one's digit of each number and see if it can be a perfect square:

(i) 153 ends with 3 — no square ends with 3, so not a perfect square.
(ii) 257 ends with 7 — no square ends with 7, so not a perfect square.
(iii) 408 ends with 8 — no square ends with 8, so not a perfect square.
(iv) 441 ends with 1 — possible for a perfect square. Indeed, 21² = 441.

Therefore, 153, 257, and 408 are surely not perfect squares; 441 is a perfect square. — The one's digit of a perfect square can only be 0,1,4,5,6,9. Since 3,7,8 are not possible one's digits of any perfect square, numbers ending with these digits cannot be perfect squares.

Chapter 5

Chapter 5 — Study Notes

5.1 Introduction

5.2 Properties of Square Numbers

5.3 Some More Interesting Patterns

Practice Questions — Chapter 5

All 13 Chapters in Mathematics