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Bunuel
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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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amariappan
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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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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siddharthharsh
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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Perfect, now I can see my mistake. Thanks.
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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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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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Bunuel
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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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with the explanation.
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Really great question Bunuel, really had to stretch my brain cells to come to D
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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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O great Bunuel, Math Genius birthed by Harvard and MIT scholars who found the cure to the quadratic equation, do you have any more similar problems?
Bunuel
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


\(f(b)*g(a) = 0\): Every positive integer has at least one odd factor: 1. Therefore, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, in turn, means that \(b\) is an odd integer (odd integers do not have even factors).

\(f(a) = 1\): \(a\) has one even factor. This means that \(a\) must be 2 (the only positive integer with only one 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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braddouglas10
O great Bunuel, Math Genius birthed by Harvard and MIT scholars who found the cure to the quadratic equation, do you have any more similar problems?
Bunuel
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


\(f(b)*g(a) = 0\): Every positive integer has at least one odd factor: 1. Therefore, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, in turn, means that \(b\) is an odd integer (odd integers do not have even factors).

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

Check questions on functions here.

Hope it helps.
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Thank you, great Bunuel, Math Genius birthed by Harvard and MIT scholars who found the cure to the quadratic equation.
Bunuel
braddouglas10
O great Bunuel, Math Genius birthed by Harvard and MIT scholars who found the cure to the quadratic equation, do you have any more similar problems?
Bunuel
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


\(f(b)*g(a) = 0\): Every positive integer has at least one odd factor: 1. Therefore, \(g(a)\) cannot be 0, which implies that \(f(b) = 0\). This, in turn, means that \(b\) is an odd integer (odd integers do not have even factors).

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

Check questions on functions here.

Hope it helps.
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Great Ques!!!!
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Great Ques!!!!
______________
Thank you!
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