Answer

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**Hint:**First we will write the given number as a product of its prime factor , then we will divide the number by the lowest factor such that in the prime product of the quotient the number of each prime factor is even.

**Complete step by step solution:**

Let’s write 9408 as its product of prime factors

$9408=2\times 2\times 2\times 2\times 2\times 2\times 3\times 7\times 7$

So we can write 9408 as ${{2}^{6}}{{3}^{1}}{{7}^{2}}$

Now we can see power of 2 is 6, power 3 is 1 and power of 7 is 2. Power of 2 and 7 is even, but power of 3 is odd so if we divide 9408 by 3 we will get a square number

The quotient when we divide 9408 by 3 we will get 3136 and the square root of 3136 is equal to 56, so the smaller number is 3 , quotient is 3136 and the square root is 56.

**So, the correct answer is “Option B”.**

**Note:**While writing evaluating the prime factor of any number check the divisibility of prime numbers such as if the sum of digits in the number is divisible by 3 the number is divisible by 3. If the last digit of the number from right is 0 or 5 then the number is divisible by 5. Multiply the last digit from right by 5 then add to the remaining number if it is divisible by 7 then the number is divisible by 7.

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