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M30-15

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New post 16 Sep 2014, 00:45
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The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36

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New post 16 Sep 2014, 00:45
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Official Solution:


The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36


\(f(b)*g(a) = 0\): any positive integer has at least one odd factor: 1. Thus, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, on the other hand, means that \(b\) is an odd integer (odd integers does not have even factors).

\(f(a) = 1\): \(a\) has 1 even factor. This means that \(a\) must be 2 (the only positive integer which has only 1 even factor is 2).

The least common multiple of \(a=2\) and \(b=odd\) is \(2*odd\). Only option D could be represented this way: \(30=2*15\).


Answer: D
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M30-15  [#permalink]

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New post 30 Sep 2014, 15:17
Good question Bunuel - Definitely makes you work your logic...actually I kind of used my Critical Reasoning skills for this question. As R.Purewal rightly says - CR is not something that can be learned, it could only be thought through....
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Re: M30-15  [#permalink]

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New post 20 May 2016, 16:46
Bunuel wrote:
Official Solution:


The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36


\(f(b)*g(a) = 0\): any positive integer has at least one odd factor: 1. Thus, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, on the other hand, means that \(b\) is an odd integer (odd integers does not have even factors).

\(f(a) = 1\): \(a\) has 1 even factor. This means that \(a\) must be 2 (the only positive integer which has only 1 even factor is 2).

The least common multiple of \(a=2\) and \(b=odd\) is \(2*odd\). Only option D could be represented this way: \(30=2*15\).


Answer: D


What if b = 3? The least common multiple would be 6, and A is a multiple of 6?
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Re: M30-15  [#permalink]

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New post 20 May 2016, 23:05
redfield wrote:
Bunuel wrote:
Official Solution:


The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36


\(f(b)*g(a) = 0\): any positive integer has at least one odd factor: 1. Thus, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, on the other hand, means that \(b\) is an odd integer (odd integers does not have even factors).

\(f(a) = 1\): \(a\) has 1 even factor. This means that \(a\) must be 2 (the only positive integer which has only 1 even factor is 2).

The least common multiple of \(a=2\) and \(b=odd\) is \(2*odd\). Only option D could be represented this way: \(30=2*15\).


Answer: D


What if b = 3? The least common multiple would be 6, and A is a multiple of 6?


What do you mean by "A is a multiple of 6"?

If b=3, then the LCM of 2 and 3 would be 6. Do we have 6 as an option? No.
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Re: M30-15  [#permalink]

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New post 21 May 2016, 06:20
Bunuel wrote:
redfield wrote:
Bunuel wrote:
Official Solution:


The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36


\(f(b)*g(a) = 0\): any positive integer has at least one odd factor: 1. Thus, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, on the other hand, means that \(b\) is an odd integer (odd integers does not have even factors).

\(f(a) = 1\): \(a\) has 1 even factor. This means that \(a\) must be 2 (the only positive integer which has only 1 even factor is 2).

The least common multiple of \(a=2\) and \(b=odd\) is \(2*odd\). Only option D could be represented this way: \(30=2*15\).


Answer: D


What if b = 3? The least common multiple would be 6, and A is a multiple of 6?


What do you mean by "A is a multiple of 6"?

If b=3, then the LCM of 2 and 3 would be 6. Do we have 6 as an option? No.


Right but then 12 would be the next LCM on that list between 2 and 3.

Is it that the wording of the question means the number HAS to be the LCM and I"m approaching it as "of the answer choices, which could be the lowest common multiple"?
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Re: M30-15  [#permalink]

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New post 21 May 2016, 07:53
redfield wrote:
Right but then 12 would be the next LCM on that list between 2 and 3.

Is it that the wording of the question means the number HAS to be the LCM and I"m approaching it as "of the answer choices, which could be the lowest common multiple"?


Yes, you are reading the question in a wrong way.
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Re: M30-15  [#permalink]

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New post 05 Sep 2016, 13:47
Hi Bunuel,

I was able to solve this question, but my question is Why is 0 , an even number, not considered a even factor here? f(b) will at least be 1 in that case. Please advice.

Thanks,
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New post 05 Sep 2016, 21:27
amariappan wrote:
Hi Bunuel,

I was able to solve this question, but my question is Why is 0 , an even number, not considered a even factor here? f(b) will at least be 1 in that case. Please advice.

Thanks,
Arun


0 is not a factor of any number.
0 is a multiple of every integer.
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Re: M30-15  [#permalink]

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New post 05 Nov 2016, 12:37
While I arrived at the same answer as the OA, I don't agree with the part of the explanation stated below.
"This means that a must be 2 (the only positive integer which has only 1 even factor is 2)."

We know that a has only one even factor, but we don't know how many odd factors it has. So A could easily have been, 2*5 or 2*7 or..
Can some one tell me how my reasoning is wrong?

Now the way I arrived at the answer is that LCM (a, b) has 2 and one or more odd numbers as its factors. Now all the answers apart from option D contained higher powers of 2 so D was the correct answer
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Re: M30-15  [#permalink]

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New post 06 Nov 2016, 03:26
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siddharthharsh wrote:
While I arrived at the same answer as the OA, I don't agree with the part of the explanation stated below.
"This means that a must be 2 (the only positive integer which has only 1 even factor is 2)."

We know that a has only one even factor, but we don't know how many odd factors it has. So A could easily have been, 2*5 or 2*7 or..
Can some one tell me how my reasoning is wrong?

Now the way I arrived at the answer is that LCM (a, b) has 2 and one or more odd numbers as its factors. Now all the answers apart from option D contained higher powers of 2 so D was the correct answer


No, you are not correct.

f(a)=1 means that a has 1 even factor.

2*5 = 10 --> 10 has 2 even factors 2 and 10.
2*7 = 14 --> 14 has 2 even factors 2 and 14.
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Re: M30-15  [#permalink]

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New post 06 Nov 2016, 04:42
Perfect, now I can see my mistake. Thanks.
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Re: M30-15  [#permalink]

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New post 10 Jun 2017, 08:26
1
Explanation says,
f(a)=1 : a has 1 even factor. This means that a must be 2 (the only positive integer which has only 1 even factor is 2).

Saying "only 1 even factor" does not mean a must be 2 right? For e.g. "a" could be 6 as well since it has one even factor i.e. 2.
I agree that a=2 and b=15 is correct but I actually reached ans choice 30 by a=6 and b=5. No other choice could give any valid values of a and b. Is my understanding wrong in assuming that "a=6 and b=5" are valid values?
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New post 25 Dec 2018, 09:36
One question: Since 'a' can have only one even factor (2) why cant it be something like a=6, 10, 14, 18 .. etc? All these numbers satisfy the condition.
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New post 25 Dec 2018, 09:41
kallu wrote:
One question: Since 'a' can have only one even factor (2) why cant it be something like a=6, 10, 14, 18 .. etc? All these numbers satisfy the condition.


6 has two even factors 2 and 6.
10 has two even factors 2 and 10.
14 has two even factors 2 and 14.
18 has two even factors 2 and 18.
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Re: M30-15  [#permalink]

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New post 27 Dec 2018, 18:41
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Bunuel wrote:
The function \(f(x)\) is defined for all positive integers \(x\) as the number of even factors of \(x\) and the function \(g(x)\) is defined for all positive integers \(x\) as the number of odd factors of \(x\). For positive integers \(a\) and \(b\) if \(f(b)*g(a) = 0\) and \(f(a) = 1\), which of the following could be the least common multiple of \(a\) and \(b\)?


A. 12
B. 16
C. 20
D. 30
E. 36


Given:- f(b)*g(a)=0 and f(a)=1.

f(a)=1 means that there is an only even factor for 'a' and that even factor will be 2. Now, either f(b)=0 or g(a)=0 which means either b has 0 even factors or a has 0 odd factors.

So, when b has 0 even factors, least common multiple would be of odd factors. i.e. it would be odd. No such option given.

When a has 0 odd factors, least common multiple would be of at least 1 even factor because f(a)=1, That culminates into that least common multiple would be a factor of 2.

Now, say, lcm is a multiple of 4, then f(a) has to be 2, whereas f(a) is only 1. The result being that required lcm is a product of 2 and other odd factors. The only option that fits the criteria is 30=2*5*3. So, 30, Option D is our answer.
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Re: M30-15  [#permalink]

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New post 21 Jan 2019, 01:56
Exactly my doubt. Thanks Bunuel! This is a brilliant question which really tests HOTS.
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Re: M30-15 &nbs [#permalink] 21 Jan 2019, 01:56
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