Squares and Square Roots

Q1: Multiple Choice Type:

i. The value of \({18}^2-{17}^2\) is :

Step 1: Use the identity: \[ a^2 – b^2 = (a – b)(a + b) \]Step 2: Apply the identity: \[ 18^2 – 17^2 = (18 – 17)(18 + 17) = 1 \times 35 = 35 \]Answer: c. 35

ii. The sum of first four odd natural numbers is

Step 1: First four odd natural numbers are: 1, 3, 5, 7
Step 2: Add them: \[ 1 + 3 + 5 + 7 = 16 = 2^4 \]Answer: b. \(2^4\)

iii. \({24}^2\) has n at its unit place, the value of n is:

Step 1: Calculate \(24^2\) \[ 24^2 = 576 \]Step 2: Look at unit digit of 576 → it’s 6Answer: d. 6

iv. A number ends with 5 zeros, the number of zeros in its square will be:

Step 1: If a number ends in 5 zeros, it is divisible by \(10^5 = 100000\)
So, the number is of the form: \[ n = k \times 10^5 \]Then: \[ n^2 = k^2 \times 10^{10} \]So, the square will end with 10 zeros.
Answer: c. 10

Q2: Seeing value of digit at unit’s place, state which of the following can be square of a number?

Rule:
Reminder: A perfect square number always ends with the digit: 0, 1, 4, 5, 6, or 9

It can never end with: 2, 3, 7, or 8

i. 3051

Unit digit = 1 (may be a perfect square)
Answer: Can be a perfect square

ii. 2332

Unit digit = 2 (a square number cannot end in 2)
Answer: Cannot be a perfect square

iii. 5684

Unit digit = 4 (may be a perfect square)
Answer: Can be a perfect square

iv. 6908

Unit digit = 8 (a square number cannot end in 8)
Answer: Cannot be a perfect square

Q3: Squares of which of the following numbers will have 1(one) at their unit’s place?

Rule:
Reminder: If a number ends in 1 or 9, its square ends in 1.Let us check each:

i. 57

Unit digit = 7
⇒ \(7^2 = 49\) ⇒ Ends with 9
Answer: Will not have 1 at unit’s place

ii. 81

Unit digit = 1
⇒ \(1^2 = 1\) ⇒ Ends with 1
Answer: Will have 1 at unit’s place

iii. 139

Unit digit = 9
⇒ \(9^2 = 81\) ⇒ Ends with 1
Answer: Will have 1 at unit’s place

iv. 73

Unit digit = 3
⇒ \(3^2 = 9\) ⇒ Ends with 9
Answer: Will not have 1 at unit’s place

v. 64

Unit digit = 4
⇒ \(4^2 = 16\) ⇒ Ends with 6
Answer: Will not have 1 at unit’s place

Final Answer:

81 and 139 will have 1 at their unit’s place.

Q4: Which of the following numbers will not have 1(one) at their unit’s place?

Rule:
Step: A number will have 1 at its unit’s place only if it ends in 1 or 9. We can also compute the square of the given numbers and observe the unit digit.

i. \(32^2\)

Unit digit of 32 = 2
\(2^2 = 4\), so square ends in 4
Answer: Will not have 1 at unit’s place

ii. \(57^2\)

Unit digit of 57 = 7
\(7^2 = 49\), square ends in 9
Answer: Will not have 1 at unit’s place

iii. \(69^2\)

Unit digit of 69 = 9
\(9^2 = 81\), square ends in 1
Answer: Will have 1 at unit’s place

iv. \(321^2\)

Unit digit of 321 = 1
\(1^2 = 1\), square ends in 1
Answer: Will have 1 at unit’s place

v. \(265^2\)

Unit digit of 265 = 5
\(5^2 = 25\), square ends in 5
Answer: Will not have 1 at unit’s place

Q5: Square of which of the following numbers will not have 6 at their unit’s place?

Concept:
Observation: A number will have 6 at its unit’s place in the square if it ends with 4 or 6.
Because:

• \(4^2 = 16\) → ends with 6
• \(6^2 = 36\) → ends with 6

i. 35

Unit digit = 5 → \(5^2 = 25\) → ends with 5
Answer: Does NOT end with 6

ii. 23

Unit digit = 3
→ \(3^2 = 9\) → ends with 9
Answer: Does NOT end with 6

iii. 64

Unit digit = 4
→ \(4^2 = 16\) → ends with 6
Answer: Ends with 6

iv. 76

Unit digit = 6
→ \(6^2 = 36\) → ends with 6
Answer: Ends with 6

v. 98

Unit digit = 8
→ \(8^2 = 64\) → ends with 4
Answer: Does NOT end with 6

Q6: Which of the following numbers will have 6 at the unit’s place:

Concept:
Observation: The square of a number ends with 6 if the number ends with 4 or 6.
Because:

• \(4^2 = 16\) → ends with 6
• \(6^2 = 36\) → ends with 6

i. \({26}^2\)

Unit digit = 6
→ \(6^2 = 36\) → ends with 6
Answer: Ends with 6

ii. \({49}^2\)

Unit digit = 9
→ \(9^2 = 81\) → ends with 1
Answer: Does NOT end with 6

iii. \({34}^2\)

Unit digit = 4
→ \(4^2 = 16\) → ends with 6
Answer: Ends with 6

iv. \({43}^2\)

Unit digit = 3 → \(3^2 = 9\) → ends with 9
Answer: Does NOT end with 6

v. \({244}^2\)

Unit digit = 4
→ \(4^2 = 16\) → ends with 6
Answer: Ends with 6

Q7: If a number ends with 3 zeroes, how many zeroes its square have?

Concept:
Step 1: A number ending with 3 zeroes can be written as: \[ N = k \times 10^3 \] where \(k\) is an integer not ending with zero.
Step 2: The square of the number is: \[ N^2 = (k \times 10^3)^2 = k^2 \times 10^{6} \]Explanation: When we square the number, the exponent of 10 doubles (from 3 to 6), so the square ends with \(6\) zeroes.
Answer: The square of a number ending with 3 zeroes will have 6 zeroes at the end.

Q8: If the square of a number ends with 10 zeroes, how many zeroes number have?

Concept:
Step 1: Let the number end with \(n\) zeroes: \[ N = k \times 10^n \] where \(k\) is an integer not ending with zero.
Step 2: Then the square of the number is: \[ N^2 = (k \times 10^n)^2 = k^2 \times 10^{2n} \]Explanation: The number of zeroes at the end of the square is twice the number of zeroes at the end of the number.
Step 3: Given, the square ends with 10 zeroes: \[ 2n = 10 \] \[ \Rightarrow n = \frac{10}{2} = 5 \]Answer: The number ends with 5 zeroes.

Q9: Is it possible for the square of a number to end with 5 zeroes? Give reason.

Explanation:
Step 1: Let the number be \(N = k \times 10^n\), where \(k\) is an integer not ending with zero.
Step 2: The square of the number is: \[ N^2 = (k \times 10^n)^2 = k^2 \times 10^{2n} \]Observation: The number of zeroes at the end of \(N^2\) is always even, since it is \(2n\).
Step 3: If \(N^2\) ends with 5 zeroes, then the number of zeroes at the end of \(N\) must be: \[ 2n = 5 \] \[ ⇒ n = \frac{5}{2} = 2.5 \] But \(n\) must be a whole number (number of zeros in the original number).
Conclusion: Since \(n\) is not an integer, it is impossible for the square of a number to end with an odd number of zeroes like 5.
Answer: No, it is not possible for the square of a number to end with 5 zeroes because the number of zeroes at the end of a square is always an even number.

Q10: Give reason to show that none of the numbers given below is a perfect square.

• 2162
• 6843
• 9637
• 6598

Explanation:
Step 1: Check the unit digit of each number:

• 2162 ends with 2
• 6843 ends with 3
• 9637 ends with 7
• 6598 ends with 8

Step 2: Recall the possible unit digits of perfect squares:
The unit digit of a perfect square can only be 0, 1, 4, 5, 6, or 9.
Observation: None of the given numbers end with 0, 1, 4, 5, 6, or 9.
Step 3: Since none of these numbers have a valid unit digit for a perfect square, none of these numbers can be perfect squares.
Answer: None of the given numbers are perfect squares because their unit digits (2, 3, 7, and 8 respectively) are not possible unit digits of any perfect square.

Q11: State. whether the square of following numbers is even or odd?

• 23
• 54
• 76
• 75

Explanation:
Step 1: Determine whether the number is even or odd.
– If the number is even, its square is also even.
– If the number is odd, its square is also odd.
Step 2: Check each number:

Answer: i. 23 → Odd number, so square is Odd.
ii. 54 → Even number, so square is Even.
iii. 76 → Even number, so square is Even.
iv. 75 → Odd number, so square is Odd.

Q12: Give to show that none of the numbers 640, 81000 and 3600000 is a perfect square.

Explanation:
Step 1: For a number to be a perfect square, all prime factors must have even powers.
Step 2: Prime factorize each number and check powers of the prime factors.
(i) 640:
\(640 = 64 \times 10 = 2^6 \times (2 \times 5) = 2^7 \times 5^1\)
– Powers are: \(2^7\) (odd), \(5^1\) (odd)
– Since powers are not all even, 640 is not a perfect square.

(ii) 81000:
\(81000 = 81 \times 1000 = 3^4 \times 10^3 = 3^4 \times (2 \times 5)^3 = 3^4 \times 2^3 \times 5^3\)
– Powers are: \(3^4\) (even), \(2^3\) (odd), \(5^3\) (odd)
– Since powers of 2 and 5 are odd, 81000 is not a perfect square.

(iii) 3600000:
\(3600000 = 36 \times 100000 = 6^2 \times (10^5) = (2 \times 3)^2 \times (2 \times 5)^5 = 2^2 \times 3^2 \times 2^5 \times 5^5 = 2^{7} \times 3^2 \times 5^5\)
– Powers are: \(2^7\) (odd), \(3^2\) (even), \(5^5\) (odd)
– Since powers of 2 and 5 are odd, 3600000 is not a perfect square.

Answer: None of the numbers 640, 81000, and 3600000 is a perfect square because their prime factorization contains primes with odd powers.

Q13: Evaluate:

i. \({37}^2 – {36}^2\)

Step 1: Use the difference of squares formula: \(a^2 – b^2 = (a+b)(a-b)\)
\(a = 37, b = 36\) \[ 37^2 – 36^2 = (37 + 36)(37 – 36) = 73 \times 1 = 73 \] Answer: 73

ii. \({85}^2 – {84}^2\)

Step 1: Using the difference of squares formula:
\(a = 85, b = 84\) \[ 85^2 – 84^2 = (85 + 84)(85 – 84) = 169 \times 1 = 169 \] Answer: 169

iii. \({101}^2 – {100}^2\)

Step 1: Using the difference of squares formula:
\(a = 101, b = 100\) \[ 101^2 – 100^2 = (101 + 100)(101 – 100) = 201 \times 1 = 201 \] Answer: 201

Q14: Without doing the actual addition, find the sum of:

i. \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23\)

Step 1: Count the number of terms. The sequence is of odd numbers starting from 1.
The nth odd number = \(2n – 1\).
To find number of terms for 23: \[ 2n – 1 = 23 \Rightarrow 2n = 24 \Rightarrow n = 12 \] There are 12 terms.
Step 2: Sum of first n odd numbers = \(n^2\). \[ \text{Sum} = 12^2 = 144 \] Answer: 144

ii. \(1 + 3 + 5 + 7 + 9 + \ldots + 39 + 41\)

Step 1: Find n for the last term 43: \[ 2n – 1 = 41 \Rightarrow 2n = 42 \Rightarrow n = 21 \] Number of terms = 21
Step 2: Sum = \(n^2 = 21^2 = 441\)
Answer: 441

iii. \(1 + 3 + 5 + 7 + 9 + \ldots + 51 + 53\)

Step 1: Find n for last term 53: \[ 2n – 1 = 53 \Rightarrow 2n = 54 \Rightarrow n = 27 \] Number of terms = 27
Step 2: Sum = \(n^2 = 27^2 = 729\)
Answer: 729

Q15: Write three sets of Pythagorean triplets such that each set has numbers less than 30.

Definition: A Pythagorean triplet consists of three positive integers \(a\), \(b\), and \(c\) such that: \[ a^2 + b^2 = c^2 \]Three sets of Pythagorean triplets (all less than 30):
Set 1: (3, 4, 5)
Check: \[ 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \]Set 2: (5, 12, 13)
Check: \[ 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \]Set 3: (7, 24, 25)
Check: \[ 7^2 + 24^2 = 49 + 576 = 625 = 25^2 \]Answer: (3, 4, 5), (5, 12, 13), (7, 24, 25)