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1 vote
(iii) Find the smallest whole number h such that
4704/h
is a cube number.


User Themiurgo
by
5.0k points

2 Answers

7 votes

Final answer:

To make 4704 a cube number when divided by h, we first factorize 4704 and then adjust h so that all prime factors of 4704 have exponents that are multiples of three. The smallest whole number h is 126.

Step-by-step explanation:

To find the smallest whole number h such that 4704 divided by h is a cube number, we start by prime factorizing 4704. After factorization, we ensure that each prime factor's exponent is a multiple of three, which is required for it to be a cube number. If a prime factor's exponent is not a multiple of three, we multiply h by the necessary power of that prime number to make the exponent divisible by three.

After factorizing 4704, we find that 4704 = 25 × 31 × 72. To make each exponent a multiple of 3, we need to multiply by 21 to make the power of 2 into 26, and by 32 × 71 to make the powers of 3 and 7 into 33 and 73 respectively. Thus, h will be 2 × 32 × 7 = 2 × 9 × 7 = 126.

The smallest whole number h such that 4704/h is a cube number is 126.

User Kevin Tianyu Xu
by
5.0k points
4 votes

Answer:


h=588

Step-by-step explanation:

Given the fraction
(4,704)/(h)

Consider its numerator and factor it:


4,704=2^5\cdot 3\cdot 7^2

You cann not take the cube root from
3, 7^2, 2^5, you can only take the cube root from
2^3, then the smallest whole number
h such that
(4,704)/(h) is a cube number is


h=2^2\cdot 3\cdot 7^2=4\cdot 3\cdot 49=588

When
h=588,


(4,704)/(588)=8=2^3

User Valery Kozlov
by
4.4k points