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If a is an integer and (a^2)/(12^3) is odd, which of the fol

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If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post Updated on: 31 Dec 2013, 04:23
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If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

A. a/4
B. a/12
C. a/27
D. a/36
E. a/72

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Originally posted by teabecca on 30 Dec 2013, 11:22.
Last edited by Bunuel on 31 Dec 2013, 04:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 06 Jul 2014, 16:13
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maggie27 wrote:
Hi Bunuel,

Can you provide clear insight on this question?


If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

A. a/4
B. a/12
C. a/27
D. a/36
E. a/72

\(\frac{a^2}{12^3}=odd\);

\(a^2=12^3*odd=12^2*(2^2*3)*odd\);

\(a=24\sqrt{3*odd}\).

Notice that the least positive value of a is \(a=24\sqrt{3*3}=72\) (if that odd integer is 3) and if a is 72, then only option E gives and odd integer, hence it must be correct.

Answer: E.
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Re: If a is an integer and (a2)/(123) is odd...  [#permalink]

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New post 21 Mar 2014, 15:49
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My approach:

\(\frac{a^2}{12^3} = \frac{a^2}{2^6*3^3} = \frac{a^2}{2^6}/3^3\) is odd --> so \(\frac{a^2}{2^6}\) must be odd --> so \(\frac{a}{2^3}=\frac{a}{8}\) must be odd

Therefore, to be odd, the solution must have a multiple of 8 in the denominator.
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Re: If a is an integer and (a2)/(123) is odd...  [#permalink]

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New post 30 Dec 2013, 12:21
2
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?
1. a/4
2. a/12
3. a/27
4. a/36
5. a/72

Can anybody explain?

ans: 5


I'll take a stab at explaining this one, let me know if this helps.

In order for the result of the above equation to result in an odd integer, there must be zero even prime integers of the number after canceling out common factors (this is because any even number times an even/odd number will always result in an even number). Therefore, we need to target an answer that will only have odd prime factors.

With this said, my first step in tackling this problem was to break down 12 into its prime factors and then distribute the exponent to those prime factors. This left me with a denominator of --> 2^6 * 3^3

My next step was to look for an integer in the answer choice that had a prime factor of 2^3 (since the ^2 would distribute to that number making it 2^6 --> allowing me to cancel the even numbers and focus on the 3s left over)

Beginning with E, a tip that is touted in the Kaplan Advanced GMAT 800 books, I broke 72 down into its primes: 2^3 * 3^2 --> distributing this to A^2 resulted in 2^6*3^4 / 2^6 * 3^3.

Canceling out common bases leaves me with 3^1 = odd integer

Let me know if this helps and/or if you have any follow up questions.
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Re: If a is an integer and (a2)/(123) is odd...  [#permalink]

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New post 06 Jan 2014, 11:08
bparrish89 wrote:
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?
1. a/4
2. a/12
3. a/27
4. a/36
5. a/72

Can anybody explain?

ans: 5


I'll take a stab at explaining this one, let me know if this helps.

In order for the result of the above equation to result in an odd integer, there must be zero even prime integers of the number after canceling out common factors (this is because any even number times an even/odd number will always result in an even number). Therefore, we need to target an answer that will only have odd prime factors.

With this said, my first step in tackling this problem was to break down 12 into its prime factors and then distribute the exponent to those prime factors. This left me with a denominator of --> 2^6 * 3^3

My next step was to look for an integer in the answer choice that had a prime factor of 2^3 (since the ^2 would distribute to that number making it 2^6 --> allowing me to cancel the even numbers and focus on the 3s left over)

Beginning with E, a tip that is touted in the Kaplan Advanced GMAT 800 books, I broke 72 down into its primes: 2^3 * 3^2 --> distributing this to A^2 resulted in 2^6*3^4 / 2^6 * 3^3.

Canceling out common bases leaves me with 3^1 = odd integer

Let me know if this helps and/or if you have any follow up questions.


Do you recommend that book? Was it helpful in terms of tips? or basic?

Cheers!
J :)
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 06 Jan 2014, 11:54
2
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

A. a/4
B. a/12
C. a/27
D. a/36
E. a/72


I think "If a is an integer and (a^2)/(12^3) is odd," should be odd integer.
If not, all of the options can be odd. Consider a = 1.

First and foremost, simplify. a^2/ (2^6 * 3^3) so if we need a^2 to be 2^6 to eliminate any 2s and also we need to eliminate 3^3.
so a^2 can be 2^6 * 3^6.
Now just test the options :-D
Option E) 2^6 * 3^6 / (2^3 * 3^2) leaves one 3 in numerator. Hence the right option.
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Re: If a is an integer and (a2)/(123) is odd...  [#permalink]

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New post 06 Jan 2014, 16:22
jlgdr wrote:
bparrish89 wrote:
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?
1. a/4
2. a/12
3. a/27
4. a/36
5. a/72

Can anybody explain?

ans: 5


I'll take a stab at explaining this one, let me know if this helps.

In order for the result of the above equation to result in an odd integer, there must be zero even prime integers of the number after canceling out common factors (this is because any even number times an even/odd number will always result in an even number). Therefore, we need to target an answer that will only have odd prime factors.

With this said, my first step in tackling this problem was to break down 12 into its prime factors and then distribute the exponent to those prime factors. This left me with a denominator of --> 2^6 * 3^3

My next step was to look for an integer in the answer choice that had a prime factor of 2^3 (since the ^2 would distribute to that number making it 2^6 --> allowing me to cancel the even numbers and focus on the 3s left over)

Beginning with E, a tip that is touted in the Kaplan Advanced GMAT 800 books, I broke 72 down into its primes: 2^3 * 3^2 --> distributing this to A^2 resulted in 2^6*3^4 / 2^6 * 3^3.

Canceling out common bases leaves me with 3^1 = odd integer

Let me know if this helps and/or if you have any follow up questions.


Do you recommend that book? Was it helpful in terms of tips? or basic?

Cheers!
J :)


Hey J,

I would recommend the Manhattan Advanced GMAT Quant books over the Kaplan GMAT 800 book. Felt as if the Kaplan book just provided additional set of questions and explanations of 700 level question, of which you can get just as much out of reviewing on this website.

Let me know if this helps. Or feel free to PM if you have any follow-up questions.
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 06 Jul 2014, 15:24
Hi Bunuel,

Can you provide clear insight on this question?
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If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 08 Oct 2015, 18:16
1
Here's my approach.

\(\frac{a^2}{3^3 * 4^3}=\frac{a^2}{3^3 * 2^6}=odd\)

We know that \(a^2\) needs to be a multiple of \(2^6\) otherwise \(\frac{a^2}{12}\) wouldn't be an odd integer. Thus, \(a\) must be at least a multiple of \(2^3\). In order to get an odd number when dividing \(a\) by some integer, we need to get rid of \(2^3\). The only answer choice that got three 2s as factors is E) 72.
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 18 Mar 2016, 20:39
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

A. a/4
B. a/12
C. a/27
D. a/36
E. a/72


we can rewrite: a^2 / 2^6 x 3^3
so a^2 must be at least 2^6*3^4
or a is at least 2^3 * 3^2

A. if we divide by 4, or 2^2, we still have a factor of 2 left out, which will make an even integer..so no
B. if we divide by 12 (3x2^2), we are still left with a factor of 2- so no
C. 27=3^3. it might be even a non-integer, so no. moreover, we are left with 2^3 ..which will make the number even.
D. 36 = 2^2 * 3^2. still we have at least one factor of 2, so no.
E. by poe, E is left..72=2^3 * 3^2. since we know that a is at least this number (other variations would include another odd factor)..then dividing it by this will always yield an odd integer.
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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New post 12 Jul 2017, 17:00
teabecca wrote:
If a is an integer and (a^2)/(12^3) is odd, which of the following must be an odd integer?

A. a/4
B. a/12
C. a/27
D. a/36
E. a/72


If (a^2)/(12^3) is odd and 12^3 is even, we see that a^2 must cancel out all even factors of 12^3. Let’s break down 12^3:

12^3 = (2^2 x 3^1)^3 = 2^6 x 3^3

Thus, a^2 must cancel out 2^6 and also be divisible by 3^3 = 27. The smallest value a could be is 2^3 x 3^2, since then a^2 = 2^6 x 3^4 and (a^2)/(12^3) = 3. Moreover, a cannot have more than 3 factors of 2, because otherwise a^2/12^3 would have been even.

The only answer choice in which the denominator has 3 factors of 2 is E, and when a is divided by 72, all the factors of 2 are cancelled out; therefore, the result must be an odd integer.

Answer: E
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Re: If a is an integer and (a^2)/(12^3) is odd, which of the fol  [#permalink]

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