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Re: M30-15 [#permalink]
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
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Re: M30-15 [#permalink]
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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]
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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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]
Perfect, now I can see my mistake. Thanks.
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Re: M30-15 [#permalink]
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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Re: M30-15 [#permalink]
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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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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]
I think this is a high-quality question and I agree with explanation.
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Re: M30-15 [#permalink]
I think this is a high-quality question and I agree with the explanation.
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Re: M30-15 [#permalink]
Really great question Bunuel, really had to stretch my brain cells to come to D
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Re: M30-15 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M30-15 [#permalink]
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