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# If a and b are integers, and 240a = b 3 , which of the following must

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If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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26 Oct 2017, 22:28
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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

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Re: If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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26 Oct 2017, 22:32
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Option D
a shall be 2^2*5^2*3^2

Both I and III fits this. Hence I and III are proven. Option D

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Re: If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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27 Oct 2017, 14:48
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Given : $$240*a = b^3$$ and a and b are integers.

as $$240 = 2^4 * 3* 5$$

For b to be an integer and given $$b^3= 2^4 * 3* 5 *a$$ => value of a should be at least => $$a= 2^2*3^2*5^2$$ to form cube of a number.

so minimum value of a = $$a= 2^2*3^2*5^2 = 900$$. And now "a" can be a multiple of this number with any other cube of an integer
=> a= 900x => where x is cube of an integer

lets check given options now

1. $$\frac{a}{300} = \frac{900x}{300} = 3x$$ => integer True
2. $$\frac{a}{600} = \frac{900x}{600} = \frac{3x}{2}$$ => may or may not be an integer
3. $$\frac{a}{450} = \frac{900x}{450} = 2x$$ => integer True

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If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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28 Oct 2017, 09:51
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Hi Nikkb,

I understood that we need to break 240 down into its primes, but I'm having a hard time trying to figure out where to go from there; particularly this part in your working:

Nikkb wrote:

For b to be an integer and given $$b^3= 2^4 * 3* 5 *a$$ => value of a should be at least => $$a= 2^2*3^2*5^2$$ to form cube of a number.

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If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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28 Oct 2017, 10:08
2
calappa1234 wrote:
Hi Nikkb,

I understood that we need to break 240 down into its primes, but I'm having a hard time trying to figure out where to go from there; particularly this part in your working:

Nikkb wrote:

For b to be an integer and given $$b^3= 2^4 * 3* 5 *a$$ => value of a should be at least => $$a= 2^2*3^2*5^2$$ to form cube of a number.

Hi calappa1234

it is mentioned that $$b$$ is an integer so $$b^3$$ must be an integer

as $$b^3=2^4*3*5*a => b=(2^4*3*5*a)^{\frac{1}{3}}$$

for $$b$$ to be an integer $$a$$ has to be of the form $$= 2^2*3^2*5^2*k^3$$, where $$k$$ is any integer

so $$a=900k^3$$
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If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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28 Oct 2017, 13:47
2
calappa1234 wrote:
Hi Nikkb,
I understood that we need to break 240 down into its primes, but I'm having a hard time trying to figure out where to go from there; particularly this part in your working:
Nikkb wrote:
For b to be an integer and given $$b^3= 2^4 * 3* 5 *a$$ => value of a should be at least => $$a= 2^2*3^2*5^2$$ to form cube of a number.

We have $$b^3= 2^4 * 3* 5 *a$$

=>$$b = \sqrt[3] 2^4 * 3* 5 *a = 2 \sqrt[3] 2 * 3* 5 *a$$

Now as b is an integer, above cube root value need to be an integer. to make it an integer we should have cube of all the integers present inside the root.

so minimum value of "a" required to make "b" as an integer = $$2^2*3^2*5^2 = 900$$

this will be minimum value of a . other possible values of "a" => above value * cube of an integer => as b need to be an integer so we need to have cubes of integers inside the root.

so we can write $$a = 2^2*3^2*5^2* x$$ , where x is cube of some integer.

=> we can write $$a = 900x$$ , where x is cube of some integer.
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If a and b are integers, and 240a = b 3 , which of the following must  [#permalink]

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28 Oct 2017, 15:27
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1
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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If a and b are integers, and 240a = b 3 , which of the following must &nbs [#permalink] 28 Oct 2017, 15:27
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