Squares and Square Roots

Finding prime factors of 729

$$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

729 = (3 × 3) × (3 × 3) × (3 × 3)

Since 729 can be expressed as the product of pairs of equal prime factors.

Hence, 729 is a perfect square.

Finding prime factors of 5488

$$\begin{array}{l|r} 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

5488 = (2 × 2) × (2 × 2) × (7 × 7) × 7

Since one 7 is left unpaired, so 5488 cannot be expressed as the product of pairs of equal prime factors.

Hence, 5488 is not a perfect square.

Finding prime factors of 1024

$$\begin{array}{l|r} 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array}$$

1024 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)

Since 1024 can be expressed as the product of pairs of equal prime factors.

Hence, 1024 is a perfect square.

Finding prime factors of 243

$$\begin{array}{l|r} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

243 = (3 × 3) × (3 × 3) × 3

Since one 3 is left unpaired, so 243 cannot be expressed as the product of pairs of equal prime factors.

Hence, 243 is not a perfect square.

Finding prime factors of 1296

$$\begin{array}{l|r} 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

1296 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3)

Since 1296 can be expressed as the product of pairs of equal prime factors, so 1296 is a perfect square.

1296 = (2)² × (2)² × (3)² × (3)²

⇒ 1296 = (2 × 2 × 3 × 3)²

⇒ 1296 = (36)²

Hence, 1296 is a perfect square and 36 is the number whose square is 1296.

Finding prime factors of 1764

$$\begin{array}{l|r} 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1764 = (2 × 2) × (3 × 3) × (7 × 7)

Since 1764 can be expressed as the product of pairs of equal prime factors, so 1764 is a perfect square.

1764 = (2)² × (3)² × (7)²

⇒ 1764 = (2 × 3 × 7)²

⇒ 1764 = (42)²

Hence, 1764 is a perfect square and 42 is the number whose square is 1764.

Finding prime factors of 3025

$$\begin{array}{l|r} 5 & 3025 \\ \hline 5 & 605 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

3025 = (5 × 5) × (11 × 11)

Since 3025 can be expressed as the product of pairs of equal prime factors, so 3025 is a perfect square.

3025 = (5)² × (11)²

⇒ 3025 = (5 × 11)²

⇒ 3025 = (55)²

Hence, 3025 is a perfect square and 55 is the number whose square is 3025.

Finding prime factors of 3969

$$\begin{array}{l|r} 3 & 3969 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

3969 = (3 × 3) × (3 × 3) × (7 × 7)

Since 3969 can be expressed as the product of pairs of equal prime factors, so 3969 is a perfect square.

3969 = (3)² × (3)² × (7)²

⇒ 3969 = (3 × 3 × 7)²

⇒ 3969 = (63)²

Hence, 3969 is a perfect square and 63 is the number whose square is 3969.

Finding prime factors of 1008

$$\begin{array}{l|r} 2 & 1008 \\ \hline 2 & 504 \\ \hline 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

1008 = (2 × 2) × (2 × 2) × (3 × 3) × 7

Since the prime factor 7 is left unpaired, so to make 1008 a perfect square 7 should be paired.

So, the given number should be multiplied by 7.

Hence, the smallest natural number by which 1008 should be multiplied to make it a perfect square is 7.

Finding prime factors of 5808

$$\begin{array}{l|r} 2 & 5808 \\ \hline 2 & 2904 \\ \hline 2 & 1452 \\ \hline 2 & 726 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

5808 = (2 × 2) × (2 × 2) × (11 × 11) × 3

Since the prime factor 3 is left unpaired, so to make 5808 a perfect square it should be divided by 3.

$$\begin{aligned}\Rightarrow 5808 \div 3 = 1936\\[1em] \Rightarrow 1936 = (2 \times 2) \times (2 \times 2) \times (11 \times 11)\\[1em] \Rightarrow \sqrt{1936} = 2 \times 2 \times 11\\[1em] \Rightarrow \sqrt{1936} = 44\end{aligned}$$

Hence, the smallest natural number by which 5808 should be divided to make it a perfect square is 3 and 44 is the number whose square is the resulting number 1936.

According to the property of square numbers, a number having 2, 3, 7 or 8 at its unit place is never a square number.

The numbers 142, 273, 347, 878 and 1273 have 2, 3, 7, 8 and 3 at their unit place respectively.

Hence, 142, 273, 347, 878 and 1273 are five numbers which are not square numbers.

The digit at unit's place of 951 is 1 and 1² = 1.

Hence, the unit digit of the square of 951 is 1.

The digit at unit's place of 502 is 2 and 2² = 4.

Hence, the unit digit of the square of 502 is 4.

The digit at unit's place of 329 is 9 and 9² = 81.

Hence, the unit digit of the square of 329 is 1.

The digit at unit's place of 643 is 3 and 3² = 9.

Hence, the unit digit of the square of 643 is 9.

The digit at unit's place of 5124 is 4 and 4² = 16.

Hence, the unit digit of the square of 5124 is 6.

The digit at unit's place of 7625 is 5 and 5² = 25.

Hence, the unit digit of the square of 7625 is 5.

The digit at unit's place of 68327 is 7 and 7² = 49.

Hence, the unit digit of the square of 68327 is 9.

The digit at unit's place of 95628 is 8 and 8² = 64.

Hence, the unit digit of the square of 95628 is 4.

The digit at unit's place of 99880 is 0 and 0² = 0.

Hence, the unit digit of the square of 99880 is 0.

The digit at unit's place of 12796 is 6 and 6² = 36.

Hence, the unit digit of the square of 12796 is 6.

According to the property of square numbers, a number having 2, 3, 7 or 8 at its unit place is never a square number and a number ending in an odd number of zeros is never a perfect square.

(i) Hence, 567 is not a perfect square as its unit digit is 7.

(ii) Hence, 2453 is not a perfect square as its unit digit is 3.

(iii) Hence, 5298 is not a perfect square as its unit digit is 8.

(iv) Hence, 46292 is not a perfect square as its unit digit is 2.

(v) The number 74000 ends in three zeros.

Hence, 74000 is not a perfect square as it has an odd number of zeros at the end.

According to the property of square numbers, squares of even natural numbers are even and squares of odd natural numbers are odd.

(i) Hence, (573)² is an odd number because 573 is an odd number.

(ii) Hence, (4096)² is an even number because 4096 is an even number.

(iii) Hence, (8267)² is an odd number because 8267 is an odd number.

(iv) Hence, (37916)² is an even number because 37916 is an even number.

We know that there are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Here, n = 12

Natural numbers between 12² and 13² = 2 × 12 = 24

Hence, there are 24 natural numbers between 12² and 13².

We know that there are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Here, n = 90

Natural numbers between 90² and 91² = 2 × 90 = 180

Hence, there are 180 natural numbers between 90² and 91².

As we know that, the sum of first n odd natural numbers = n².

Here, the given numbers are the first 8 odd natural numbers.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8² = 64

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8² = 64.

As we know that, the sum of first n odd natural numbers = n².

Here, the given numbers are the first 15 odd natural numbers.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 15² = 225

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 = 15² = 225.

As we know that, the sum of first n odd natural numbers = n².

64 = 8²

So, 64 is the sum of the first 8 odd natural numbers.

Hence, 64 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15.

As we know that, the sum of first n odd natural numbers = n².

121 = 11²

So, 121 is the sum of the first 11 odd natural numbers.

Hence, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\frac{n^2 - 1}{2}$ and $\frac{n^2 + 1}{2}$.

Since 19² = 361

First number $= \frac{19^2 - 1}{2} = \frac{361 - 1}{2}$

$= \frac{360}{2} = 180$

Second number $= \frac{19^2 + 1}{2} = \frac{361 + 1}{2}$

$= \frac{362}{2} = 181$

Hence, 19² = 180 + 181.

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\frac{n^2 - 1}{2}$ and $\frac{n^2 + 1}{2}$.

Since 33² = 1089

First number $= \frac{33^2 - 1}{2} = \frac{1089 - 1}{2}$

$= \frac{1088}{2} = 544$

Second number $= \frac{33^2 + 1}{2} = \frac{1089 + 1}{2}$

$= \frac{1090}{2} = 545$

Hence, 33² = 544 + 545.

We know that the square of an odd number n can be expressed as the sum of the two consecutive integers $\frac{n^2 - 1}{2}$ and $\frac{n^2 + 1}{2}$.

Since 47² = 2209

First number $= \frac{47^2 - 1}{2} = \frac{2209 - 1}{2}$

$= \frac{2208}{2} = 1104$

Second number $= \frac{47^2 + 1}{2} = \frac{2209 + 1}{2}$

$= \frac{2210}{2} = 1105$

Hence, 47² = 1104 + 1105.

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 45 with a5, we get a = 4.

(45)² = 4(4 + 1) hundreds + 25 = (4 × 5) hundreds + 25 = 20 hundreds + 25 = 2000 + 25 = 2025

Hence, (45)² = 2025.

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 305 with a5, we get a = 30.

(305)² = 30(30 + 1) hundreds + 25 = (30 × 31) hundreds + 25 = 930 hundreds + 25 = 93000 + 25 = 93025

Hence, (305)² = 93025.

We know that (a5)² = a(a + 1) hundreds + 25.

Comparing 525 with a5, we get a = 52.

(525)² = 52(52 + 1) hundreds + 25 = (52 × 53) hundreds + 25 = 2756 hundreds + 25 = 275600 + 25 = 275625

Hence, (525)² = 275625.

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 8

⇒ m = 4

∴ m² - 1 = 16 - 1 = 15 and m² + 1 = 16 + 1 = 17

So, (8, 15, 17) is a Pythagorean triplet.

Again, let m² + 1 = 8

⇒ m² = 7, so m is not a natural number.

Let m² - 1 = 8

⇒ m² = 9

⇒ m = 3

∴ 2m = 2 × 3 = 6 and m² + 1 = 9 + 1 = 10

So, (6, 8, 10) is also a Pythagorean triplet.

Hence, (8, 15, 17) or (6, 8, 10) is a Pythagorean triplet whose one number is 8.

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 15

⇒ m = 7.5, which is not a natural number.

Let m² + 1 = 15

⇒ m² = 14, so m is not a natural number.

Let m² - 1 = 15

⇒ m² = 16

⇒ m = 4

∴ 2m = 2 × 4 = 8 and m² + 1 = 16 + 1 = 17

Hence, (8, 15, 17) is a Pythagorean triplet whose one number is 15.

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 63

⇒ m = 31.5, which is not a natural number.

Let m² + 1 = 63

⇒ m² = 62, so m is not a natural number.

Let m² - 1 = 63

⇒ m² = 64

⇒ m = 8

∴ 2m = 2 × 8 = 16 and m² + 1 = 64 + 1 = 65

Hence, (16, 63, 65) is a Pythagorean triplet whose one number is 63.

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let m² - 1 = 80

⇒ m² = 81

⇒ m = 9

∴ 2m = 2 × 9 = 18 and m² + 1 = 81 + 1 = 82

So, (18, 80, 82) is a Pythagorean triplet.

Again, let 2m = 80

⇒ m = 40

∴ m² - 1 = 1600 - 1 = 1599 and m² + 1 = 1600 + 1 = 1601

So, (80, 1599, 1601) is also a Pythagorean triplet.

Hence, (18, 80, 82) or (80, 1599, 1601) is a Pythagorean triplet whose one number is 80.

Finding prime factors of 784

$$\begin{array}{l|r} 2 & 784 \\ \hline 2 & 392 \\ \hline 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 784 = (2 \times 2) \times (2 \times 2) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{784} = 2 \times 2 \times 7\\[1em] \Rightarrow \sqrt{784} = 28\end{aligned}$$

Hence, $\mathbf{\sqrt{784} = 28}$

Finding prime factors of 441

$$\begin{array}{l|r} 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 441 = (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{441} = 3 \times 7\\[1em] \Rightarrow \sqrt{441} = 21\end{aligned}$$

Hence, $\mathbf{\sqrt{441} = 21}$

Finding prime factors of 1849

$$\begin{array}{l|r} 43 & 1849 \\ \hline 43 & 43 \\ \hline & 1 \\ \end{array}$$

Solving,

⇒ 1849 = (43 × 43)

$\Rightarrow \sqrt{1849} = 43$

Hence, $\mathbf{\sqrt{1849} = 43}$

Finding prime factors of 4356

$$\begin{array}{l|r} 2 & 4356 \\ \hline 2 & 2178 \\ \hline 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 4356 = (2 \times 2) \times (3 \times 3) \times (11 \times 11)\\[1em] \Rightarrow \sqrt{4356} = 2 \times 3 \times 11\\[1em] \Rightarrow \sqrt{4356} = 66\end{aligned}$$

Hence, $\mathbf{\sqrt{4356} = 66}$

Finding prime factors of 6241

$$\begin{array}{l|r} 79 & 6241 \\ \hline 79 & 79 \\ \hline & 1 \\ \end{array}$$

Solving,

⇒ 6241 = (79 x 79)

$\Rightarrow \sqrt{6241} = 79$

Hence, $\mathbf{\sqrt{6241} = 79}$

Finding prime factors of 8836

$$\begin{array}{l|r} 2 & 8836 \\ \hline 2 & 4418 \\ \hline 47 & 2209 \\ \hline 47 & 47 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 8836 = (2 \times 2) \times (47 \times 47)\\[1em] \Rightarrow \sqrt{8836} = 2 \times 47\\[1em] \Rightarrow \sqrt{8836} = 94\end{aligned}$$

Hence, $\mathbf{\sqrt{8836} = 94}$

Finding prime factors of 8281

$$\begin{array}{l|r} 7 & 8281 \\ \hline 7 & 1183 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 8281 = (7 \times 7) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{8281} = 7 \times 13\\[1em] \Rightarrow \sqrt{8281} = 91\end{aligned}$$

Hence, $\mathbf{\sqrt{8281} = 91}$

Finding prime factors of 9025

$$\begin{array}{l|r} 5 & 9025 \\ \hline 5 & 1805 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 9025 = (5 \times 5) \times (19 \times 19)\\[1em] \Rightarrow \sqrt{9025} = 5 \times 19\\[1em] \Rightarrow \sqrt{9025} = 95\end{aligned}$$

Hence, $\mathbf{\sqrt{9025} = 95}$

Converting mixed fraction to improper fraction:

$9\frac{67}{121} = \frac{1156}{121}$

Finding prime factors of 1156 and 121

$$\begin{array}{l|r} 2 & 1156 \\ \hline 2 & 578 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow \sqrt{9\frac{67}{121}}\\[1em] \Rightarrow \sqrt{\frac{1156}{121}}\\[1em] \Rightarrow \frac{\sqrt{1156}}{\sqrt{121}}\\[1em] \Rightarrow \frac{\sqrt{(2 \times 2) \times (17 \times 17)}}{\sqrt{(11 \times 11)}}\\[1em] \Rightarrow \frac{2 \times 17}{11}\\[1em] \Rightarrow \frac{34}{11}\\[1em] \Rightarrow 3\frac{1}{11}\end{aligned}$$

Hence, $\mathbf{\sqrt{9\frac{67}{121}} = 3\frac{1}{11}}$

Converting mixed fraction to improper fraction:

$17\frac{13}{36} = \frac{625}{36}$

Finding prime factors of 625 and 36

$$\begin{array}{l|r} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow \sqrt{17\frac{13}{36}}\\[1em] \Rightarrow \sqrt{\frac{625}{36}}\\[1em] \Rightarrow \frac{\sqrt{625}}{\sqrt{36}}\\[1em] \Rightarrow \frac{\sqrt{(5 \times 5) \times (5 \times 5)}}{\sqrt{(2 \times 2) \times (3 \times 3)}}\\[1em] \Rightarrow \frac{5 \times 5}{2 \times 3}\\[1em] \Rightarrow \frac{25}{6}\\[1em] \Rightarrow 4\frac{1}{6}\end{aligned}$$

Hence, $\mathbf{\sqrt{17\frac{13}{36}} = 4\frac{1}{6}}$

Converting decimal to fraction:

$1.96 = \frac{196}{100}$

Finding prime factors of 196 and 100

$$\begin{array}{l|r} 2 & 196 \\ \hline 2 & 98 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow \sqrt{1.96}\\[1em] \Rightarrow \sqrt{\frac{196}{100}}\\[1em] \Rightarrow \frac{\sqrt{196}}{\sqrt{100}}\\[1em] \Rightarrow \frac{\sqrt{(2 \times 2) \times (7 \times 7)}}{\sqrt{(2 \times 2) \times (5 \times 5)}}\\[1em] \Rightarrow \frac{2 \times 7}{2 \times 5}\\[1em] \Rightarrow \frac{14}{10}\\[1em] \Rightarrow 1.4\end{aligned}$$

Hence, $\mathbf{\sqrt{1.96} = 1.4}$

Converting decimal to fraction:

$0.0064 = \frac{64}{10000}$

Finding prime factors of 64 and 10000

$$\begin{array}{l|r} 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 10000 \\ \hline 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow \sqrt{0.0064}\\[1em] \Rightarrow \sqrt{\frac{64}{10000}}\\[1em] \Rightarrow \frac{\sqrt{64}}{\sqrt{10000}}\\[1em] \Rightarrow \frac{\sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2)}}{\sqrt{(2 \times 2) \times (2 \times 2) \times (5 \times 5) \times (5 \times 5)}}\\[1em] \Rightarrow \frac{2 \times 2 \times 2}{2 \times 2 \times 5 \times 5}\\[1em] \Rightarrow \frac{8}{100}\\[1em] \Rightarrow 0.08\end{aligned}$$

Hence, $\mathbf{\sqrt{0.0064} = 0.08}$

Finding prime factors of 588

$$\begin{array}{l|r} 2 & 588 \\ \hline 2 & 294 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

588 = (2 × 2) × (7 × 7) × 3

Since the prime factor 3 is left unpaired, so 588 should be multiplied by 3.

$$\begin{aligned}\Rightarrow 588 \times 3 = 1764\\[1em] \Rightarrow 1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{1764} = 2 \times 3 \times 7\\[1em] \Rightarrow \sqrt{1764} = 42\end{aligned}$$

Hence, 588 should be multiplied by 3 and $\mathbf{\sqrt{1764} = 42}$.

Finding prime factors of 720

$$\begin{array}{l|r} 2 & 720 \\ \hline 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

720 = (2 × 2) × (2 × 2) × (3 × 3) × 5

Since the prime factor 5 is left unpaired, so 720 should be multiplied by 5.

$$\begin{aligned}\Rightarrow 720 \times 5 = 3600\\[1em] \Rightarrow 3600 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{3600} = 2 \times 2 \times 3 \times 5\\[1em] \Rightarrow \sqrt{3600} = 60\end{aligned}$$

Hence, 720 should be multiplied by 5 and $\mathbf{\sqrt{3600} = 60}$.

Finding prime factors of 2178

$$\begin{array}{l|r} 2 & 2178 \\ \hline 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

2178 = (3 × 3) × (11 × 11) × 2

Since the prime factor 2 is left unpaired, so 2178 should be multiplied by 2.

$$\begin{aligned}\Rightarrow 2178 \times 2 = 4356\\[1em] \Rightarrow 4356 = (2 \times 2) \times (3 \times 3) \times (11 \times 11)\\[1em] \Rightarrow \sqrt{4356} = 2 \times 3 \times 11\\[1em] \Rightarrow \sqrt{4356} = 66\end{aligned}$$

Hence, 2178 should be multiplied by 2 and $\mathbf{\sqrt{4356} = 66}$.

Finding prime factors of 3042

$$\begin{array}{l|r} 2 & 3042 \\ \hline 3 & 1521 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

3042 = (3 × 3) × (13 × 13) × 2

Since the prime factor 2 is left unpaired, so 3042 should be multiplied by 2.

$$\begin{aligned}\Rightarrow 3042 \times 2 = 6084\\[1em] \Rightarrow 6084 = (2 \times 2) \times (3 \times 3) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{6084} = 2 \times 3 \times 13\\[1em] \Rightarrow \sqrt{6084} = 78\end{aligned}$$

Hence, 3042 should be multiplied by 2 and $\mathbf{\sqrt{6084} = 78}$.

Finding prime factors of 6300

$$\begin{array}{l|r} 2 & 6300 \\ \hline 2 & 3150 \\ \hline 3 & 1575 \\ \hline 3 & 525 \\ \hline 5 & 175 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

6300 = (2 × 2) × (3 × 3) × (5 × 5) × 7

Since the prime factor 7 is left unpaired, so 6300 should be multiplied by 7.

$$\begin{aligned}\Rightarrow 6300 \times 7 = 44100\\[1em] \Rightarrow 44100 = (2 \times 2) \times (3 \times 3) \times (5 \times 5) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{44100} = 2 \times 3 \times 5 \times 7\\[1em] \Rightarrow \sqrt{44100} = 210\end{aligned}$$

Hence, 6300 should be multiplied by 7 and $\mathbf{\sqrt{44100} = 210}$.

Finding prime factors of 1872

$$\begin{array}{l|r} 2 & 1872 \\ \hline 2 & 936 \\ \hline 2 & 468 \\ \hline 2 & 234 \\ \hline 3 & 117 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

1872 = (2 × 2) × (2 × 2) × (3 × 3) × 13

Since the prime factor 13 is left unpaired, so 1872 should be divided by 13.

$$\begin{aligned}\Rightarrow 1872 \div 13 = 144\\[1em] \Rightarrow 144 = (2 \times 2) \times (2 \times 2) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{144} = 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt{144} = 12\end{aligned}$$

Hence, 1872 should be divided by 13 and $\mathbf{\sqrt{144} = 12}$.

Finding prime factors of 2592

$$\begin{array}{l|r} 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

2592 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2

Since the prime factor 2 is left unpaired, so 2592 should be divided by 2.

$$\begin{aligned}\Rightarrow 2592 \div 2 = 1296\\[1em] \Rightarrow 1296 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)\\[1em] \Rightarrow \sqrt{1296} = 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt{1296} = 36\end{aligned}$$

Hence, 2592 should be divided by 2 and $\mathbf{\sqrt{1296} = 36}$.

Finding prime factors of 3380

$$\begin{array}{l|r} 2 & 3380 \\ \hline 2 & 1690 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

3380 = (2 × 2) × (13 × 13) × 5

Since the prime factor 5 is left unpaired, so 3380 should be divided by 5.

$$\begin{aligned}\Rightarrow 3380 \div 5 = 676\\[1em] \Rightarrow 676 = (2 \times 2) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{676} = 2 \times 13\\[1em] \Rightarrow \sqrt{676} = 26\end{aligned}$$

Hence, 3380 should be divided by 5 and $\mathbf{\sqrt{676} = 26}$.

Finding prime factors of 16224

$$\begin{array}{l|r} 2 & 16224 \\ \hline 2 & 8112 \\ \hline 2 & 4056 \\ \hline 2 & 2028 \\ \hline 2 & 1014 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

16224 = (2 × 2) × (2 × 2) × (13 × 13) × 2 × 3

Since the prime factors 2 and 3 are left unpaired, so 16224 should be divided by 2 × 3 = 6.

$$\begin{aligned}\Rightarrow 16224 \div 6 = 2704\\[1em] \Rightarrow 2704 = (2 \times 2) \times (2 \times 2) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{2704} = 2 \times 2 \times 13\\[1em] \Rightarrow \sqrt{2704} = 52\end{aligned}$$

Hence, 16224 should be divided by 6 and $\mathbf{\sqrt{2704} = 52}$.

Finding prime factors of 61347

$$\begin{array}{l|r} 3 & 61347 \\ \hline 11 & 20449 \\ \hline 11 & 1859 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

61347 = (11 × 11) × (13 × 13) × 3

Since the prime factor 3 is left unpaired, so 61347 should be divided by 3.

$$\begin{aligned}\Rightarrow 61347 \div 3 = 20449\\[1em] \Rightarrow 20449 = (11 \times 11) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{20449} = 11 \times 13\\[1em] \Rightarrow \sqrt{20449} = 143\end{aligned}$$

Hence, 61347 should be divided by 3 and $\mathbf{\sqrt{20449} = 143}$.

LCM of 3, 6, 10 and 15 = 2 × 3 × 5 = 30.

So, the smallest number divisible by each one of 3, 6, 10 and 15 is 30.

Prime Factorisation of 30:

$$\begin{array}{l|r} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

30 = 2 × 3 × 5

Since the prime factors 2, 3 and 5 are left unpaired, so 30 should be multiplied by 2 × 3 × 5 = 30.

Required square number = 30 × 30

= 900

Hence, the smallest square number divisible by 3, 6, 10 and 15 is 900.

LCM of 6, 9, 27 and 36 = 2 × 2 × 3 × 3 × 3 = 108.

So, the smallest number divisible by each one of 6, 9, 27 and 36 is 108.

Prime Factorisation of 108:

$$\begin{array}{l|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

108 = (2 × 2) × (3 × 3) × 3

Since the prime factor 3 is left unpaired, so 108 should be multiplied by 3.

Required square number = 108 × 3

= 324

Hence, the smallest square number divisible by 6, 9, 27 and 36 is 324.

LCM of 4, 7, 8 and 16 is 2 × 2 × 2 × 2 × 7 = 112.

So, the smallest number divisible by each one of 4, 7, 8 and 16 is 112.

Prime Factorisation of 112:

$$\begin{array}{l|r} 2 & 112 \\ \hline 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

112 = (2 × 2) × (2 × 2) × 7

Since the prime factor 7 is left unpaired, so 112 should be multiplied by 7.

Required square number = 112 × 7

= 784

Hence, the smallest square number divisible by 4, 7, 8 and 16 is 784.

Let the number of rows be $x$.

The number of plants in each row = the number of rows = $x$

Finding prime factors of 4225

$$\begin{array}{l|r} 5 & 4225 \\ \hline 5 & 845 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

Total number of plants = 4225

$$\begin{aligned}\Rightarrow x \times x = 4225\\[1em] \Rightarrow x^2 = 4225\\[1em] \Rightarrow x = \sqrt{4225}\\[1em] \Rightarrow x = \sqrt{(5 \times 5) \times (13 \times 13)}\\[1em] \Rightarrow x = 5 \times 13\\[1em] \Rightarrow x = 65\end{aligned}$$

Hence, the number of rows is 65 and the number of plants in each row is 65.

Let the breadth of the rectangle be $x$ m.

Length of the rectangle = 4 × breadth = $4x$ m

Area of rectangle = length × breadth

Area of rectangle = 1936

$$\begin{aligned}\Rightarrow 4x \times x = 1936\\[1em] \Rightarrow 4x^2 = 1936\\[1em] \Rightarrow x^2 = \frac{1936}{4}\\[1em] \Rightarrow x^2 = 484\\[1em] \Rightarrow x = \sqrt{484}\\[1em]\end{aligned}$$

Finding prime factors of 484

$$\begin{array}{l|r} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow x = \sqrt{(2 \times 2) \times (11 \times 11)}\\[1em] \Rightarrow x = 2 \times 11\\[1em] \Rightarrow x = 22\end{aligned}$$

Breadth of the rectangle = 22 m

Length of the rectangle = 4 × 22 = 88 m

Hence, the length of the rectangle is 88 m and its breadth is 22 m.

Let the number of rows be $x$.

The number of columns = the number of rows = $x$

Students arranged in the P.T. display = 2000 - 64 = 1936

Finding prime factors of 1936

$$\begin{array}{l|r} 2 & 1936 \\ \hline 2 & 968 \\ \hline 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Total number of students accommodated = 1936

$$\begin{aligned}\Rightarrow x \times x = 1936\\[1em] \Rightarrow x^2 = 1936\\[1em] \Rightarrow x = \sqrt{1936}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (2 \times 2) \times (11 \times 11)}\\[1em] \Rightarrow x = 2 \times 2 \times 11\\[1em] \Rightarrow x = 44\end{aligned}$$

Hence, the number of rows is 44.

Let the number of students in the class be $x$.

Amount contributed by each student = number of students = ₹$x$

Finding prime factors of 2304

$$\begin{array}{l|r} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Total amount collected = ₹2304

$$\begin{aligned}\Rightarrow x \times x = 2304\\[1em] \Rightarrow x^2 = 2304\\[1em] \Rightarrow x = \sqrt{2304}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)}\\[1em] \Rightarrow x = 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow x = 48\end{aligned}$$

Hence, the number of students in the class is 48.

Let the first number be $x$.

The second number is 15 times the first number = $15x$

Product of two numbers = 7260

$$\begin{aligned}\Rightarrow x \times 15x = 7260\\[1em] \Rightarrow 15x^2 = 7260\\[1em] \Rightarrow x^2 = \frac{7260}{15}\\[1em] \Rightarrow x^2 = 484\\[1em] \Rightarrow x = \sqrt{484}\\[1em]\end{aligned}$$

Finding prime factors of 484

$$\begin{array}{l|r} 2 & 484 \\ \hline 2 & 242 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow x = \sqrt{(2 \times 2) \times (11 \times 11)}\\[1em] \Rightarrow x = 22\\[1em] \Rightarrow 15x = 15 \times 22 = 330\end{aligned}$$

Hence, the numbers are 22 and 330.

Let the three numbers be $2x$, $3x$ and $5x$.

$$\begin{aligned}\Rightarrow (2x)^2 + (3x)^2 + (5x)^2 = 950\\[1em] \Rightarrow 4x^2 + 9x^2 + 25x^2 = 950\\[1em] \Rightarrow 38x^2 = 950\\[1em] \Rightarrow x^2 = \frac{950}{38}\\[1em] \Rightarrow x^2 = 25\\[1em] \Rightarrow x = \sqrt{25}\\[1em] \Rightarrow x = 5\end{aligned}$$

First number = 2 × 5 = 10

Second number = 3 × 5 = 15

Third number = 5 × 5 = 25

Hence, the three numbers are 10, 15 and 25.

Perimeter of a square = 4 × side

$$\begin{aligned}\Rightarrow \text{Side of first square} = \frac{60}{4} = 15 \text{ m}\\[1em] \Rightarrow \text{Side of second square} = \frac{144}{4} = 36 \text{ m}\end{aligned}$$

Area of first square = 15 × 15 = 225 m²

Area of second square = 36 × 36 = 1296 m²

Sum of the areas of the two squares = 225 + 1296 = 1521 m²

Let the side of the required square be $x$ m.

Finding prime factors of 1521

$$\begin{array}{l|r} 3 & 1521 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow x^2 = 1521\\[1em] \Rightarrow x = \sqrt{1521}\\[1em] \Rightarrow x = \sqrt{(3 \times 3) \times (13 \times 13)}\\[1em] \Rightarrow x = 3 \times 13\\[1em] \Rightarrow x = 39\end{aligned}$$

Perimeter of the required square = 4 × 39 = 156 m

Hence, the perimeter of the required square is 156 m.

Square root of 2401 by division method:

Hence, $\mathbf{\sqrt{2401} = 49}$

Square root of 4489 by division method:

Hence, $\mathbf{\sqrt{4489} = 67}$

Square root of 106929 by division method:

Hence, $\mathbf{\sqrt{106929} = 327}$

Square root of 167281 by division method:

Hence, $\mathbf{\sqrt{167281} = 409}$

Square root of 53824 by division method:

Hence, $\mathbf{\sqrt{53824} = 232}$

Square root of 213444 by division method:

Hence, $\mathbf{\sqrt{213444} = 462}$

Square root of 51.84 by division method:

Hence, $\mathbf{\sqrt{51.84} = 7.2}$

Square root of 42.25 by division method:

Hence, $\mathbf{\sqrt{42.25} = 6.5}$

Square root of 18.4041 by division method:

Hence, $\mathbf{\sqrt{18.4041} = 4.29}$

Square root of 5.774409 by division method:

Hence, $\mathbf{\sqrt{5.774409} = 2.403}$

Square root of 645.8 correct to two decimal places by division method:

$\sqrt{645.8} = 25.412 \approx 25.41$

Hence, $\mathbf{\sqrt{645.8} = 25.41}$

Square root of 107.45 correct to two decimal places by division method:

$\sqrt{107.45} = 10.365 \approx 10.36$

Hence, $\mathbf{\sqrt{107.45} = 10.36}$

Square root of 5.462 correct to two decimal places by division method:

$\sqrt{5.462} = 2.337 \approx 2.34$

Hence, $\mathbf{\sqrt{5.462} = 2.34}$

Square root of 2 correct to two decimal places by division method:

$\sqrt{2} = 1.414 \approx 1.41$

Hence, $\mathbf{\sqrt{2} = 1.41}$

Square root of 3 correct to two decimal places by division method:

$\sqrt{3} = 1.732 \approx 1.73$

Hence, $\mathbf{\sqrt{3} = 1.73}$

Converting mixed fraction into decimal:

$$\begin{aligned}\Rightarrow \sqrt{11\frac{3}{8}}\\[1em] \Rightarrow \sqrt{\frac{91}{8}}\\[1em] \Rightarrow \sqrt{11.375}\end{aligned}$$

Square root of 11.375 correct to two decimal places by division method:

$\sqrt{11.375} = 3.372 \approx 3.37$

Hence, $\mathbf{\sqrt{11\frac{3}{8}} = 3.37}$

Converting mixed fraction into decimal:

$$\begin{aligned}\Rightarrow \sqrt{5\frac{5}{11}}\\[1em] \Rightarrow \sqrt{\frac{60}{11}}\\[1em] \Rightarrow \sqrt{5.4545}\end{aligned}$$

Square root of 5.4545 correct to two decimal places by division method:

$\sqrt{5.4545} = 2.335 \approx 2.34$

Hence, $\mathbf{\sqrt{5\frac{5}{11}} = 2.34}$

Converting mixed fraction into decimal:

$$\begin{aligned}\Rightarrow \sqrt{7\frac{1}{3}}\\[1em] \Rightarrow \sqrt{\frac{22}{3}}\\[1em] \Rightarrow \sqrt{7.3333}\end{aligned}$$

Square root of 7.3333 correct to two decimal places by division method:

$\sqrt{7.3333} = 2.708 \approx 2.71$

Hence, $\mathbf{\sqrt{7\frac{1}{3}} = 2.71}$

Using division method:

Clearly, if 64 is subtracted from 2000, the remainder will be a perfect square.

Since, 2000 - 64 = 1936

and, $\sqrt{1936} = 44$

Hence, 64 should be subtracted from 2000 and $\mathbf{\sqrt{1936} = 44}$.

Using division method:

Clearly, if 23 is subtracted from 984, the remainder will be a perfect square.

Since, 984 - 23 = 961

and, $\sqrt{961} = 31$

Hence, 23 should be subtracted from 984 and $\mathbf{\sqrt{961} = 31}$.

Using division method:

Clearly, if 98 is subtracted from 8934, the remainder will be a perfect square.

Since, 8934 - 98 = 8836

and, $\sqrt{8836} = 94$

Hence, 98 should be subtracted from 8934 and $\mathbf{\sqrt{8836} = 94}$.

Using division method:

Clearly, if 205 is subtracted from 11021, the remainder will be a perfect square.

Since, 11021 - 205 = 10816

and, $\sqrt{10816} = 104$

Hence, 205 should be subtracted from 11021 and $\mathbf{\sqrt{10816} = 104}$.

Using division method:

Clearly, 1750 is greater than 41².

∴ On adding the required number to 1750, we shall be getting 42² i.e, 1764.

So, the required number = 1764 - 1750 = 14.

and, $\sqrt{1764} = 42$

Hence, 14 should be added to 1750 and $\mathbf{\sqrt{1764} = 42}$.

Using division method:

Clearly, 6412 is greater than 80².

∴ On adding the required number to 6412, we shall be getting 81² i.e, 6561.

So, the required number = 6561 - 6412 = 149.

and, $\sqrt{6561} = 81$

Hence, 149 should be added to 6412 and $\mathbf{\sqrt{6561} = 81}$.

Using division method:

Clearly, 6598 is greater than 81².

∴ On adding the required number to 6598, we shall be getting 82² i.e, 6724.

So, the required number = 6724 - 6598 = 126.

and, $\sqrt{6724} = 82$

Hence, 126 should be added to 6598 and $\mathbf{\sqrt{6724} = 82}$.

Using division method:

Clearly, 8000 is greater than 89².

∴ On adding the required number to 8000, we shall be getting 90² i.e, 8100.

So, the required number = 8100 - 8000 = 100.

and, $\sqrt{8100} = 90$

Hence, 100 should be added to 8000 and $\mathbf{\sqrt{8100} = 90}$.

The smallest four digit number = 1000

Using division method:

As we can see 1000 is not a perfect square.

Clearly, 1000 is greater than 31².

∴ On adding the required number to 1000, we shall be getting 32² i.e, 1024.

So, the required number = 1024 - 1000 = 24.

1000 + 24 = 1024 is a perfect square.

Hence, the smallest four digit number which is a perfect square is 1024.

The greatest six digit number = 999999

Using division method:

We get the remainder = 1998

It means (999)² is less than 999999 by 1998.

∴ If we subtract 1998 from 999999, we get (999)² which is a perfect square number.

So, the required perfect square number = 999999 - 1998 = 998001.

Hence, the greatest six digit number which is a perfect square is 998001.

In right triangle ABC, ∠B = 90°.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow AC^2 = AB^2 + BC^2\\[1em] \Rightarrow AC^2 = 14^2 + 48^2\\[1em] \Rightarrow AC^2 = 196 + 2304\\[1em] \Rightarrow AC^2 = 2500\\[1em] \Rightarrow AC = \sqrt{2500}\\[1em] \Rightarrow AC = 50\end{aligned}$$

Hence, AC = 50 cm.

In right triangle ABC, ∠B = 90°.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow AC^2 = AB^2 + BC^2\\[1em] \Rightarrow AB^2 = AC^2 - BC^2\\[1em] \Rightarrow AB^2 = 37^2 - 35^2\\[1em] \Rightarrow AB^2 = 1369 - 1225\\[1em] \Rightarrow AB^2 = 144\\[1em] \Rightarrow AB = \sqrt{144}\\[1em] \Rightarrow AB = 12\end{aligned}$$

Hence, AB = 12 cm.

Using division method:

Clearly, 1400 is greater than 37².

∴ On adding the required number to 1400, we shall be getting 38² i.e, 1444.

So, the required number = 1444 - 1400 = 44.

Hence, the gardener needs 44 more plants.

Using division method:

We get the remainder = 39

It means (31)² is less than 1000 by 39.

∴ If 39 children are left out, the remaining 961 children can be arranged in 31 rows and 31 columns.

Hence, 39 children would be left out in this arrangement.

Let A be the house of Amit, B the point where he turns east and C be his friend's house.

AB = 16 m, BC = 63 m and ∠B = 90°.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow AC^2 = AB^2 + BC^2\\[1em] \Rightarrow AC^2 = 16^2 + 63^2\\[1em] \Rightarrow AC^2 = 256 + 3969\\[1em] \Rightarrow AC^2 = 4225\\[1em] \Rightarrow AC = \sqrt{4225}\\[1em] \Rightarrow AC = 65\end{aligned}$$

Hence, Amit walked a distance of 65 m while returning.

Let AC be the ladder, AB the wall and BC the distance between the wall and the foot of the ladder.

AC = 6 m, AB = 4.8 m and ∠B = 90°.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow AC^2 = AB^2 + BC^2\\[1em] \Rightarrow BC^2 = AC^2 - AB^2\\[1em] \Rightarrow BC^2 = 6^2 - (4.8)^2\\[1em] \Rightarrow BC^2 = 36 - 23.04\\[1em] \Rightarrow BC^2 = 12.96\\[1em] \Rightarrow BC = \sqrt{12.96}\\[1em] \Rightarrow BC = 3.6\end{aligned}$$

Hence, the distance between the wall and the foot of the ladder is 3.6 m.

(i) According to the property of square numbers, a number having 2, 3, 7 or 8 at its unit place is never a square number.

Hence, a number ending in 2, 3, 7 or 8 is never a perfect square.

(ii) We know that 1² = 1 and 9² = 81, so both 1 and 9 give 1 at the unit's place of the square.

Hence, if a number has digits 1 or 9 in the unit's place, then its square ends in 1.

(iii) As we know that, the sum of first n odd natural numbers = n².

Sum of first 10 odd natural numbers = 10²

= 100

Hence, sum of first 10 odd natural numbers is 100.

(iv) We know that there are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Here, n = 11

Non-square numbers between 11² and 12² = 2 × 11

= 22

Hence, number of non-square numbers between 11² and 12² is 22.

(v) According to the property of square numbers, if a number ends with n zeros, its square ends with 2n zeros.

The number 400 ends with 2 zeros, so its square ends with 2 × 2 = 4 zeros.

Hence, number of zeros in the end of the square of 400 is four.

(vi) We can express the square of any odd number greater than 1 as the sum of two consecutive natural numbers.

Hence, square of any odd number can be expressed as the sum of two consecutive natural numbers.

(vii) For any natural number m > 1, (2m)² + (m² - 1)² = (m² + 1)².

Hence, for a natural number m > 1, (2m, m² - 1, m² + 1) is called Pythagorean triplet.

(i) The natural numbers 2, 3, 5, 6, 7, 8, ..... are not perfect squares.

Hence, the given statement is true.

(ii) A perfect square can always be expressed as the product of pairs of equal prime factors.

Hence, the given statement is false.

(iii) A square number always ends with 0, 1, 4, 5, 6 or 9 at its unit's place.

Hence, the given statement is true.

(iv) A square number always ends with 0, 1, 4, 5, 6 or 9 at its unit's place but the converse is not true.

For example, 30, 21, 34, 45, 76 and 89 are not square numbers.

Hence, the given statement is false.

(v) A square number can only have an even number of zeros at the end but the converse is not true.

For example, 300, 60000 and 8000000 end with an even number of zeros but these are not square numbers.

Hence, the given statement is false.

(vi) Squares of even natural numbers are even and squares of odd natural numbers are odd.

Hence, the given statement is true.

(vii) There are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Hence, the given statement is true.

(viii) 4² + 6²

= 16 + 36

= 52

and 8² = 64

Since, 4² + 6² ≠ 8²

Hence, the given statement is false.

We know that there are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Here, n = 25

Natural numbers between 25² and 26² = 2 × 25

= 50

Hence, option 2 is the correct option.

According to the property of square numbers, squares of even natural numbers are even and squares of odd natural numbers are odd.

Hence, option 1 is the correct option.

As we know that, the sum of first n odd natural numbers = n².

Hence, option 4 is the correct option.

Finding prime factors of 2304

$$\begin{array}{l|r} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

2304 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (3 × 3)

$\Rightarrow \sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3 = 48$

$$\begin{aligned}\Rightarrow \sqrt{208 + \sqrt{2304}}\\[1em] \Rightarrow \sqrt{208 + 48}\\[1em] \Rightarrow \sqrt{256}\\[1em] \Rightarrow 16\end{aligned}$$

Hence, option 2 is the correct option.

Converting decimal into proper fraction:

$$\begin{aligned}\Rightarrow \sqrt{0.0016}\\[1em] \Rightarrow \sqrt{\frac{16}{10000}}\\[1em] \Rightarrow \frac{\sqrt{16}}{\sqrt{10000}}\\[1em] \Rightarrow \frac{4}{100}\\[1em] \Rightarrow 0.04\end{aligned}$$

Hence, option 1 is the correct option.

Finding prime factors of 75

$$\begin{array}{l|r} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

75 = (5 × 5) × 3

Since the prime factor 3 is not in pair.

The given number should be divided by 3.

Hence, option 3 is the correct option.

Converting mixed fraction into improper fraction:

$$\begin{aligned}\Rightarrow \sqrt{3\frac{6}{25}}\\[1em] \Rightarrow \sqrt{\frac{81}{25}}\\[1em] \Rightarrow \frac{\sqrt{81}}{\sqrt{25}}\\[1em] \Rightarrow \frac{9}{5}\end{aligned}$$

Hence, option 3 is the correct option.

Finding prime factors of 162

$$\begin{array}{l|r} 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

162 = (3 × 3) × (3 × 3) × 2

Since the prime factor 2 is not in pair.

The given number should be multiplied by 2.

Hence, option 3 is the correct option.

Finding prime factors of 961

$$\begin{array}{l|r} 31 & 961 \\ \hline 31 & 31 \\ \hline & 1 \\ \end{array}$$

Area of square field = (side)²

$$\begin{aligned}\Rightarrow (\text{side})^2 = 961\\[1em] \Rightarrow \text{side} = \sqrt{961}\\[1em] \Rightarrow \text{side} = \sqrt{31 \times 31}\\[1em] \Rightarrow \text{side} = 31\end{aligned}$$

Hence, option 3 is the correct option.

Using division method for 300:

We get the remainder = 11.

It means (17)² is less than 300 by 11.

So, the perfect square just less than 300 is 289.

Required number = 300 - 289 = 11

Hence, option 1 is the correct option.

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 6

⇒ m = 3

∴ m² - 1 = 9 - 1 = 8 and m² + 1 = 9 + 1 = 10

So, (6, 8, 10) is a Pythagorean triplet.

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{0.1521} + \sqrt{15.21}\\[1em] \Rightarrow \sqrt{\frac{1521}{10000}} + \sqrt{\frac{1521}{100}}\\[1em] \Rightarrow \frac{\sqrt{1521}}{\sqrt{10000}} + \frac{\sqrt{1521}}{\sqrt{100}}\\[1em] \Rightarrow \frac{39}{100} + \frac{39}{10}\\[1em] \Rightarrow 0.39 + 3.9\\[1em] \Rightarrow 4.29\end{aligned}$$

Hence, option 2 is the correct option.

Finding prime factors of 100

$$\begin{array}{l|r} 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

100 = (2 × 2) × (5 × 5)

Since all the prime factors are in pair, so 100 is a perfect square.

Also, 100 = 10²

So, statement I is true.

A square number always ends with 0, 1, 4, 5, 6 or 9 at unit's place but the converse is not true.

For example, 30, 21, 34, 45, 76 and 89 are not square numbers.

So, statement II is false.

Hence, option 1 is the correct option.

As, 81 = 9² and 100 = 10², so n = 9

Non-square numbers between 9² and 10² = 2 × 9

= 18

So, statement I is true.

There are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

So, statement II is true.

Hence, option 3 is the correct option.

We know that for any natural number n ≥ 2, (2n, n² - 1, n² + 1) is a Pythagorean triplet.

Finding the Pythagorean triplet having 3 as one of its numbers,

Let n² - 1 = 3

⇒ n² = 4

⇒ n = 2

∴ 2n = 2 × 2 = 4 and n² + 1 = 4 + 1 = 5

So, (4, 3, 5) is a Pythagorean triplet and not (2, 3, 5).

So, statement I is false.

According to statement II, for any natural number n ≥ 2,

$$\begin{aligned}\Rightarrow (2n)^2 + (n^2 - 1)^2\\[1em] \Rightarrow 4n^2 + n^4 - 2n^2 + 1\\[1em] \Rightarrow n^4 + 2n^2 + 1\\[1em] \Rightarrow (n^2 + 1)^2\end{aligned}$$

So, statement II is true.

Hence, option 2 is the correct option.

The sum of first 11 odd natural numbers:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

= 11²

= 121

So, statement I is true.

The sum of first n odd natural numbers = n².

So, statement II is true.

Hence, option 3 is the correct option.

Finding prime factors of 1089

$$\begin{array}{l|r} 3 & 1089 \\ \hline 3 & 363 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$$

1089 = (3 × 3) × (11 × 11)

Since 1089 can be expressed as the product of pairs of equal prime factors, so 1089 is a perfect square.

1089 = (3)² × (11)²

⇒ 1089 = (3 × 11)²

⇒ 1089 = (33)²

Hence, 1089 is a perfect square and 33 is the number whose square is 1089.

Finding prime factors of 3675

$$\begin{array}{l|r} 3 & 3675 \\ \hline 5 & 1225 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

3675 = (5 × 5) × (7 × 7) × 3

Since the prime factor 3 is left unpaired, so 3675 should be multiplied by 3.

$$\begin{aligned}\Rightarrow 3675 \times 3 = 11025\\[1em] \Rightarrow 11025 = (3 \times 3) \times (5 \times 5) \times (7 \times 7)\\[1em] \Rightarrow \sqrt{11025} = 3 \times 5 \times 7\\[1em] \Rightarrow \sqrt{11025} = 105\end{aligned}$$

Hence, 3675 should be multiplied by 3 and $\mathbf{\sqrt{11025} = 105}$.

As we know that, the sum of first n odd natural numbers = n².

144 = 12²

So, 144 is the sum of the first 12 odd natural numbers.

Hence, 144 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23.

We know that there are 2n non-square numbers between the squares of two consecutive numbers n and (n + 1).

Here, n = 99

Numbers between 99² and 100² = 2 × 99

= 198

Hence, there are 198 numbers between 99² and 100².

We know that for any natural number m > 1, (2m, m² - 1, m² + 1) is a Pythagorean triplet.

Let 2m = 17

⇒ m = 8.5, which is not a natural number.

Let m² - 1 = 17

⇒ m² = 18, so m is not a natural number.

Let m² + 1 = 17

⇒ m² = 16

⇒ m = 4

∴ 2m = 2 × 4 = 8 and m² - 1 = 16 - 1 = 15

Hence, (8, 15, 17) is a Pythagorean triplet whose one number is 17.

LCM of 6, 8 and 9 is 2 × 2 × 2 × 3 × 3 = 72.

So, the smallest number divisible by each one of 6, 8 and 9 is 72.

$$\begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

72 = (2 × 2) × (3 × 3) × 2

Since the prime factor 2 is left unpaired, so 72 should be multiplied by 2.

Required square number = 72 × 2

= 144

Hence, the smallest square number divisible by 6, 8 and 9 is 144.

Let the number of chairs in each row be $x$.

The number of rows = the number of chairs in each row = $x$

Finding prime factors of 1764

$$\begin{array}{l|r} 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$$

$$\begin{aligned}\Rightarrow x \times x = 1764\\[1em] \Rightarrow x^2 = 1764\\[1em] \Rightarrow x = \sqrt{1764}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)}\\[1em] \Rightarrow x = 2 \times 3 \times 7\\[1em] \Rightarrow x = 42\end{aligned}$$

Hence, the number of chairs in each row is 42.

Let ABCD be the rectangle with length AB = 12 m and breadth BC = 5 m.

In right triangle ABC, ∠B = 90°.

By Pythagoras theorem,

$$\begin{aligned}\Rightarrow AC^2 = AB^2 + BC^2\\[1em] \Rightarrow AC^2 = 12^2 + 5^2\\[1em] \Rightarrow AC^2 = 144 + 25\\[1em] \Rightarrow AC^2 = 169\\[1em] \Rightarrow AC = \sqrt{169}\\[1em] \Rightarrow AC = 13\end{aligned}$$

Hence, the length of the diagonal of the rectangle is 13 m.

Finding prime factors of 5625

$$\begin{array}{l|r} 3 & 5625 \\ \hline 3 & 1875 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 5625 = (3 \times 3) \times (5 \times 5) \times (5 \times 5)\\[1em] \Rightarrow \sqrt{5625} = 3 \times 5 \times 5\\[1em] \Rightarrow \sqrt{5625} = 75\end{aligned}$$

Hence, $\mathbf{\sqrt{5625} = 75}$

Finding prime factors of 1521

$$\begin{array}{l|r} 3 & 1521 \\ \hline 3 & 507 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow 1521 = (3 \times 3) \times (13 \times 13)\\[1em] \Rightarrow \sqrt{1521} = 3 \times 13\\[1em] \Rightarrow \sqrt{1521} = 39\end{aligned}$$

Hence, $\mathbf{\sqrt{1521} = 39}$

Square root of 21904 by division method:

Hence, $\mathbf{\sqrt{21904} = 148}$

Square root of 108241 by division method:

Hence, $\mathbf{\sqrt{108241} = 329}$

Square root of 17.64 by division method:

Hence, $\mathbf{\sqrt{17.64} = 4.2}$

Square root of 13.3225 by division method:

Hence, $\mathbf{\sqrt{13.3225} = 3.65}$

Converting mixed fraction into improper fraction.

$$\begin{aligned}\Rightarrow \sqrt{1\frac{25}{144}}\\[1em] \Rightarrow \sqrt{\frac{169}{144}}\\[1em] \Rightarrow \frac{\sqrt{169}}{\sqrt{144}}\\[1em] \Rightarrow \frac{\sqrt{13 \times 13}}{\sqrt{12 \times 12}}\\[1em] \Rightarrow \frac{13}{12}\\[1em] \Rightarrow 1\frac{1}{12}\end{aligned}$$

Hence, $\mathbf{\sqrt{1\frac{25}{144}} = 1\frac{1}{12}}$

Converting mixed fraction into improper fraction.

$$\begin{aligned}\Rightarrow \sqrt{11\frac{225}{576}}\\[1em] \Rightarrow \sqrt{\frac{6561}{576}}\\[1em] \Rightarrow \frac{\sqrt{6561}}{\sqrt{576}}\\[1em] \Rightarrow \frac{\sqrt{81 \times 81}}{\sqrt{24 \times 24}}\\[1em] \Rightarrow \frac{81}{24}\\[1em] \Rightarrow \frac{27}{8}\\[1em] \Rightarrow 3\frac{3}{8}\end{aligned}$$

Hence, $\mathbf{\sqrt{11\frac{225}{576}} = 3\frac{3}{8}}$

Using division method:

Clearly, if 7 is subtracted from 2311, the remainder will be a perfect square.

Since, 2311 - 7 = 2304

and, $\sqrt{2304} = 48$

Hence, 7 should be subtracted from 2311.

Using division method:

Clearly, 520 is greater than 22².

∴ On adding the required number to 520, we shall be getting 23² i.e, 529.

So, the required number = 529 - 520 = 9.

Hence, 9 should be added to 520.

The greatest five digit number = 99999

Using division method:

We get the remainder = 143

It means (316)² is less than 99999 by 143.

∴ If we subtract 143 from 99999, we get (316)² which is a perfect square number.

So, the required perfect square number = 99999 - 143 = 99856.

Hence, the greatest five digit number which is a perfect square is 99856.

$$\begin{aligned}\Rightarrow \text{Area of the square field} = \frac{\text{Total cost}}{\text{Rate per sq. m}}\\[1em] \Rightarrow \frac{34560}{15}\\[1em] \Rightarrow 2304 \text{ sq. m}\end{aligned}$$

Let the length of side of the square field be $x$ m.

Finding prime factors of 2304

$$\begin{array}{l|r} 2 & 2304 \\ \hline 2 & 1152 \\ \hline 2 & 576 \\ \hline 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$$

Solving,

$$\begin{aligned}\Rightarrow x^2 = 2304\\[1em] \Rightarrow x = \sqrt{2304}\\[1em] \Rightarrow x = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3)}\\[1em] \Rightarrow x = 2 \times 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow x = 48\end{aligned}$$

Hence, the length of side of the square field is 48 m.

Solving,

$$\begin{aligned}\Rightarrow \sqrt{2025} + \sqrt{0.0612 + x} = 45.25\\[1em] \Rightarrow 45 + \sqrt{0.0612 + x} = 45.25\\[1em] \Rightarrow \sqrt{0.0612 + x} = 45.25 - 45\\[1em] \Rightarrow \sqrt{0.0612 + x} = 0.25\\[1em] \Rightarrow 0.0612 + x = (0.25)^2\\[1em] \Rightarrow 0.0612 + x = 0.0625\\[1em] \Rightarrow x = 0.0625 - 0.0612\\[1em] \Rightarrow x = 0.0013\end{aligned}$$

Hence, x = 0.0013.

Square root of 5 correct to two decimal places by division method:

$\sqrt{5} = 2.236 \approx 2.24$

$$\begin{aligned}\Rightarrow \sqrt{\frac{3 + \sqrt{5}}{3 - \sqrt{5}}}\\[1em] \Rightarrow \sqrt{\frac{(3 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}}\\[1em] \Rightarrow \sqrt{\frac{(3 + \sqrt{5})^2}{3^2 - (\sqrt{5})^2}}\\[1em] \Rightarrow \sqrt{\frac{(3 + \sqrt{5})^2}{9 - 5}}\\[1em] \Rightarrow \sqrt{\frac{(3 + \sqrt{5})^2}{4}}\\[1em] \Rightarrow \frac{3 + \sqrt{5}}{2}\end{aligned}$$

Putting the value of $\sqrt{5}$, we get :

$$\begin{aligned}\Rightarrow \frac{3 + 2.24}{2}\\[1em] \Rightarrow \frac{5.24}{2}\\[1em] \Rightarrow 2.62\end{aligned}$$

Hence, $\mathbf{\sqrt{5} = 2.24}$ and $\mathbf{\sqrt{\frac{3 + \sqrt{5}}{3 - \sqrt{5}}} = 2.62}$.

The chocolates given to the 13 children are 1, 3, 5, 7, ..... which are the first 13 odd natural numbers.

As we know that, the sum of first n odd natural numbers = n².

Total chocolates = 1 + 3 + 5 + ..... + 25

= 13²

= 169

Hence, Aryan had 169 chocolates.