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**Hint:**To solve this question, we will use the concept of prime factorization. The fundamental theorem of arithmetic states that every composite number can be expressed or we can say that it is factorized as a product of prime numbers, and this factorization is unique except for the order in which the prime factors occur.

**Complete step-by-step answer:**

The numbers that are divisible by only two numbers that are 1 and themselves are called prime numbers.

(i) 390625.

First, we have to make the prime factors of 390625.

\[

\Rightarrow 390625 = 5 \times 78125 \\

\Rightarrow 390625 = 5 \times 5 \times 15625 \\

\Rightarrow 390625 = 5 \times 5 \times 5 \times 3125 \\

\Rightarrow 390625 = 5 \times 5 \times 5 \times 5 \times 625 \\

\Rightarrow 390625 = 5 \times 5 \times 5 \times 5 \times 5 \times 125 \\

\Rightarrow 390625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 25 \\

\]

\[ \Rightarrow 390625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5\] ……. (i)

These are the prime factors of 390625.

Now, we will find out its square root.

Taking square root on both sides of equation (i), we will get

\[ \Rightarrow \sqrt {390625} = \sqrt {5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5} \]

\[ \Rightarrow \sqrt {390625} = \sqrt {\left( {5 \times 5} \right) \times \left( {5 \times 5} \right) \times \left( {5 \times 5} \right) \times \left( {5 \times 5} \right)} \]

\[ \Rightarrow \sqrt {390625} = 5 \times 5 \times 5 \times 5\]

\[ \Rightarrow \sqrt {390625} = 625\]

Here, we can see that the square root of 390625 by the prime factorization method is 625.

(ii) 119025.

First, we have to make the prime factors of 119025.

\[

\Rightarrow 119025 = 5 \times 23805 \\

\Rightarrow 119025 = 5 \times 5 \times 4761 \\

\Rightarrow 119025 = 5 \times 5 \times 3 \times 1587 \\

\Rightarrow 119025 = 5 \times 5 \times 3 \times 3 \times 529 \\

\]

\[ \Rightarrow 119025 = 5 \times 5 \times 3 \times 3 \times 23 \times 23\] …….. (ii)

These are the prime factors of 119025.

Now, we will find out its square root.

Taking square root on both sides of equation (ii), we will get

\[ \Rightarrow \sqrt {119025} = \sqrt {5 \times 5 \times 3 \times 3 \times 23 \times 23} \]

\[ \Rightarrow \sqrt {119025} = 5 \times 3 \times 23\]

\[ \Rightarrow \sqrt {119025} = 345\]

Hence, the square root of 119025 by prime factorisation is 345.

(iii) 193600.

First, we have to make the prime factors of 193600.

\[

\Rightarrow 193600 = 5 \times 38720 \\

\Rightarrow 193600 = 5 \times 5 \times 7744 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 3872 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 1936 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 2 \times 968 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 2 \times 2 \times 484 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 2 \times 2 \times 2 \times 242 \\

\Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 121 \\

\]

\[ \Rightarrow 193600 = 5 \times 5 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11\] ……… (iii)

These are the prime factors of 193600.

Now, we will find out its square root.

Taking square root on both sides of equation (iii), we will get

\[ \Rightarrow \sqrt {193600} = \sqrt {5 \times 5 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11} \]

\[ \Rightarrow \sqrt {193600} = \sqrt {\left( {5 \times 5} \right) \times \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {2 \times 2} \right) \times \left( {11 \times 11} \right)} \]

\[ \Rightarrow \sqrt {193600} = 5 \times 2 \times 2 \times 2 \times 11\]

\[ \Rightarrow \sqrt {193600} = 440\]

Hence, the square root of 193600 by prime factorisation is 440.

**Note:**Whenever we ask this type of questions, first, we have to remember what are prime numbers? And also, we should know the prime factorization method. Then we have to find out the prime factors of the given number and after that we will do the square root of that number and its prime factors. By solving this, we will get the required answer.

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