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# If a and b are positive integers and 90a = b^3, which of the following

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Joined: 02 Sep 2009
Posts: 58421
If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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20 Sep 2018, 01:03
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45% (medium)

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57% (01:48) correct 43% (02:03) wrong based on 106 sessions

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If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^3*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

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Posts: 8003
Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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20 Sep 2018, 03:22
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1
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^2*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

Now $$90a=b^3.......2*3^2*5*a=b^3$$
So a has to be a multiple of 2^2*3*5^2...
Thus a divided by a factor of 2^2*3*5^2 will be an integer..

I. $$\frac{a}{2^2*3*5}$$
2^2*3*5 is a factor of 2^2*3*5^2....so YES

II. $$\frac{a}{2*3^2*5}$$
2*3^2*5 is NOT a factor of 2^2*3*5^2....so Not necessary

III. $$\frac{a}{2^2*3 *5^2}$$
2^2*3*5^2 is a factor of 2^2*3*5^2....so YES

I and III

Bunuel, no choice matches the answer, so there must be a typo in choice I
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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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20 Sep 2018, 03:35
chetan2u wrote:
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^2*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

Now $$90a=b^3.......2*3^2*5*a=b^3$$
So a has to be a multiple of 2^2*3*5^2...
Thus a divided by a factor of 2^2*3*5^2 will be an integer..

I. $$\frac{a}{2^2*3*5}$$
2^2*3*5 is a factor of 2^2*3*5^2....so YES

II. $$\frac{a}{2*3^2*5}$$
2*3^2*5 is NOT a factor of 2^2*3*5^2....so Not necessary

III. $$\frac{a}{2^2*3 *5^2}$$
2^2*3*5^2 is a factor of 2^2*3*5^2....so YES

I and III

Bunuel, no choice matches the answer, so there must be a typo in choice I

You are right. Edited option I. Thank you.
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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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20 Sep 2018, 03:54
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^3*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

$$90a = b^3$$. So, 90a is a perfect cube.

90 = 3*3*5*2*a.

$$a = 3*5^2*2^2$$ = 300

scan each option:

I. $$\frac{a}{2^3*3*5}$$ = 300 / 120 = no integer.

II. $$\frac{a}{2*3^2*5}$$ = 300 / 90 = no integer.

III. $$\frac{a}{2^2*3 *5^2}$$ = 300/ 300 = integer .

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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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27 Sep 2018, 11:49
chetan2u wrote:
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^2*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

Now $$90a=b^3.......2*3^2*5*a=b^3$$
So a has to be a multiple of 2^2*3*5^2...
Thus a divided by a factor of 2^2*3*5^2 will be an integer..

Don't get the step highlighted in red. Could someone be so kind and explain it to me?
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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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27 Sep 2018, 12:15
1
fallenx wrote:
chetan2u wrote:
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^2*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

Now $$90a=b^3.......2*3^2*5*a=b^3$$
So a has to be a multiple of 2^2*3*5^2...
Thus a divided by a factor of 2^2*3*5^2 will be an integer..

Don't get the step highlighted in red. Could someone be so kind and explain it to me?

$$90a = b^3$$

90a has to be a perfect cube. How can we make a perfect cube?

If we break down 90 what we get:

90 = 3 * 3*5 * 2 .

how many 3 u have and what's about 5 and 2. u have 3^2 and a 5 and a 2. Still we don't know about a.

when u say something is a perfect cube it means all the prime factors of that number must have to be expressed through cube (3)

$$3^2*5*2$$is not perfect cube. So rest of the elements come from a. remember that 90a is perfect cube. but 90 alone is not a perfect cube.

$$3^2$$..........need another 3

2 .....need $$2^2$$

5.......need $$5^2$$

Therefore this additional properties is the part of a.

$$a = 3 * 2^2 * 5^2 = 300$$

now see: $$90a = 90*300 = b^3$$
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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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27 Sep 2018, 12:26
Since 90a=b^3 -> 3*3*2*5a=b^3
Since b is a positive integer then LHS of above has to be a perfect cube.

so a=2*5*2*5*3=300 ,Now if we put the value of a=300 in each of the options I,II and III, only III returns a integral value.
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Re: If a and b are positive integers and 90a = b^3, which of the following  [#permalink]

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02 Oct 2018, 19:33
1
1
Bunuel wrote:
If a and b are positive integers and $$90a = b^3$$, which of the following must be an integer?

I. $$\frac{a}{2^3*3*5}$$

II. $$\frac{a}{2*3^2*5}$$

III. $$\frac{a}{2^2*3 *5^2}$$

(A) Only I
(B) Only II
(C) Only III
(D) Only I and II
(E) I, II and III

We see that 90a must be a perfect cube. Thus 90a must have prime factors whose exponents are multiples of 3. Let’s break 90 into prime factors:

90 = 9 x 10 = 2^1 x 3^2 x 5^1

So in order for 90a to be a perfect cube, the number “a” must contain the product 2^2 x 3^1 x 5^2.

Thus, only III must be an integer.

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Re: If a and b are positive integers and 90a = b^3, which of the following   [#permalink] 02 Oct 2018, 19:33
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