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2 years ago

What is the smallest number by which 9408 must be divided so that the quotient is perfect cube?

Mathematics

2 answers:

Mars2501 [29]2 years ago

8 0

Step-by-step explanation:

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

_____________

5 | 3136

5 | 25

___ |______

106 | 636

6 | 636

|_______

| 000

So, √3136 = 56

Answer.

Mandarinka [93]2 years ago

7 0

Answer:

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.

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What is the smallest number by which 9408 must be divided so that the quotient is perfect cube?​

What is the smallest number by which 9408 must be divided so that the quotient is perfect cube?