nkmungila wrote:

If \(a\) and \(b\) are integers, and \(240a = b^3\), which of the following must be an integer?

I. \(\frac{a}{300}\)

II. \(\frac{a}{600}\)

III.\(\frac{a}{450}\)

A. None

B. I only

C. III only

D. I and III only

E. I, II, and II

If \(240a = b^3\), where \(a\) and \(b\) are integers and \(b^3\) is a perfect cube, then \(240a\) is also a perfect cube.

We must find prime factors to make \(240a\) a perfect cube.

The term

\(240a\) must contain "triplets" of prime factors, that is, prime factors in groups of three.

A perfect square, for example, must have prime factors that are paired evenly, that can be grouped in twos or couplets.

Thus, perfect square

\(144 = (2^43^2) = 2^22^23^2\)Similarly, a perfect cube's prime factors must be in groups of three, or in triplets. Thus

\(1000 = 2^35^3\)\(240a\) must contain enough factors to form the cube of an integer. But 240 is not a perfect cube; so \(a\) must contain the "missing" perfect cube factors.

Find the prime factors that 240 is "missing" to be a perfect cube. Multiply those missing factors. That product = least value of

a. Then assess.

1) Prime factorize:

\(240 = 2^43^15^1\): None of the factors is in a group of three.

2) Calculate what is needed to make each factor a triplet. That tells you what \(a\) must be at a minimum*:

\(240 = 2^43^15^1\). Each factor lacks two copies of itself to create a triplet.

--We need

\(2^2\) to get a triplet

\(2^6\) --We need

\(3^2\) to get a triplet

\(3^3\)--We need

\(5^2\) to get a triplet

\(5^3\)So we need

\(2^23^25^2 = a = 900\)3) No need to multiply 900 by 240. The answers ask only about \(a\). Assess: Which

must be integers?

I. \(\frac{a}{300}\)

YES. 900 is the minimum value of \(a\). See notes below

900 divided by ANY factor of itself will be an integer.

\(\frac{900}{300} = 3\) KEEP

II. \(\frac{a}{600}\)

NO. We need only ONE "not integer" to reject "must be integer."

600 is not a factor of 900. And

\(\frac{900}{600} = 1.5\) REJECT

III.\(\frac{a}{450}\)

YES. 450 is a factor of 900. KEEP.

ANSWER D, I and III only

*At a "minimum"? \(2^23^25^2\) are the minimum missing factors needed to create a perfect cube.

But we do not know how great the actual number is (240a and \(b^3\)).

a could contain another perfect cube as a factor.

As long as the factors are (240)(a)(C), where (C) is a perfect cube, the equation will work.

Example: Let (C) = (1), a perfect cube.

So (240)(900)(1) = 216,000. Perfect cube.

216,000 = 60\(^3\). Maybe \(b^3 = 60^3\)

But let (C) = (8), also a perfect cube.

(240)(900)(8) = 1,728,000 = 120\(^3\)

So maybe \(b^3 = 120^3\)

Thus \(2^23^25^2 = 900 =\) the least value of a.
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