Question 6

To make a number a perfect square, all its prime factors must appear an even number of times.

Step 1 — Prime Factorization

First, we find the prime factors of 9408. Prime factorization means breaking a number down into its prime number building blocks.

$9408 \div 2 = 4704$

$4704 \div 2 = 2352$

$2352 \div 2 = 1176$

$1176 \div 2 = 588$

$588 \div 2 = 294$

$294 \div 2 = 147$

$147 \div 3 = 49$

$49 \div 7 = 7$

$7 \div 7 = 1$

So, the prime factorization of 9408 is $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 \times 7$. Now, let us group these prime factors into pairs. We have $(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times 3 \times (7 \times 7)$. We can see that the prime factor 3 is not part of a pair. To make 9408 a perfect square, we need to multiply it by 3 to complete the pair for the factor 3.

$\boxed{\text{Smallest number to multiply} = 3}$

Step 2 — Calculate the Product and its Square Root

Now we multiply 9408 by the smallest number we found, which is 3. This will give us a perfect square.

$\text{Product} = 9408 \times 3$

$= 28224$

To find the square root of this product, we use its prime factors. The prime factors of the new product will be $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times (3 \times 3) \times 7 \times 7$. We take one factor from each pair to find the square root.

$\sqrt{28224} = \sqrt{(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)}$

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

$= 8 \times 21$

$\boxed{\text{Square root of the product} = 168}$

Answer

(i) The smallest number by which 9408 must be multiplied is 3. (ii) The square root of the product is 168.