Question

Least positive integer whose product is 9408 gives perfect square

Answer 1

Solution :-

To find the smallest number by which 9408 must be divided so that the quotient is a perfect square, we have to find the prime factors of 9408.

9408 = 2*2*2*2*2*2*3*7*7

Prime factors of 9408 are 2, 2, 2, 2, 2, 2. 3, 7, 7

Out of the prime factors of 9408, only 3 is without pair.

So, 3 is the number by which 9408 must be divided to make the quotient a perfect square.

9408/3 = 3136

Square root of 3136

              56

        _____________

   5   |    3136

   5   |    25

___  |______

106  |      636

   6   |      636

        |_______

        |      000

       

So, √3136 = 56

hope it's helps

Answer 2

The least positive integer whose product is 9408 gives perfect square is 3.

Step-by-step explanation:

To find : Least positive integer whose product is 9408 gives perfect square?

Solution :

First we factories the number,

$9408=2\times2 \times2\times2\times2\times2\times 3\times 7\times 7$

$9408=2^2\times2^2\times2^2\times 3\times 7^2$

We have to multiply 3 to 9408 so as to get the perfect square.

i.e. $9408\times 3=2^2\times2^2\times2^2\times 3^2\times 7^2$

$28224=(2\times2\times2\times 3\times 7)^2$

$28224=(168)^2$

Therefore, the least positive integer whose product is 9408 gives perfect square is 3.

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