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# If a and b are positive integers, which of the following can

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If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 02:34
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If a and b are positive integers, which of the following cannot be odd?

A. (2+4a)/(4+4b)
B. (4a)/(b)
C. a/b
D. (4+a)/(2+4b)
E. (4+a)/(1+4b)

P.S: Bunuel/VeritasPrepKarishma- guys I would request you to provide an explanation on this question.Is there any generic approach for this kind of questions?

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Re: If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 06:08
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debayan222 wrote:
If a and b are positive integers, which of the following cannot be odd?

A. (2+4a)/(4+4b)
B. (4a)/(b)
C. a/b
D. (4+a)/(2+4b)
E. (4+a)/(1+4b)

P.S: Bunuel/VeritasPrepKarishma- guys I would request you to provide an explanation on this question.Is there any generic approach for this kind of questions?

$$\frac{2+4a}{4+4b}=\frac{2a+1}{2b+2}=\frac{odd}{even}\neq{integer}$$.

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Re: If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 02:48
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In my view, Best approach to such type of questions is to prove that the constraint is possible. Let's see how it works- 1 is odd integer & a, b can take any positive integer value. So if we can prove that any fraction can yield 1, we can reject that choice

A. (2+4a)/(4+4b) - Under any condition this can't be odd. Answer
B. (4a)/(b) = 4.1/4 = 1 = Odd
C. a/b = 4/4 = Odd
D.(4+a)/(2+4b) = (4+2)/(2+4.1) = 1 = Odd
E.(4+a)/(1+4b) = (4+1)/(1+4.1) = 1 = Odd

Hope this helps.
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Re: If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 03:16
fameatop wrote:
In my view, Best approach to such type of questions is to prove that the constraint is possible. Let's see how it works- 1 is odd integer & a, b can take any positive integer value. So if we can prove that any fraction can yield 1, we can reject that choice

A. (2+4a)/(4+4b) - Under any condition this can't be odd. Answer
B. (4a)/(b) = 4.1/4 = 1 = Odd
C. a/b = 4/4 = Odd
D.(4+a)/(2+4b) = (4+2)/(2+4.1) = 1 = Odd
E.(4+a)/(1+4b) = (4+1)/(1+4.1) = 1 = Odd

Hope this helps.

Hi fameatop,
Other than plug-in, how one can be sure that A is the answer...? and plug-in takes time in this type of questions i guess...Your thoughts please!
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Re: If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 03:30
fameatop wrote:
In my view, Best approach to such type of questions is to prove that the constraint is possible. Let's see how it works- 1 is odd integer & a, b can take any positive integer value. So if we can prove that any fraction can yield 1, we can reject that choice
A. (2+4a)/(4+4b) - Under any condition this can't be odd. Answer
B. (4a)/(b) = 4.1/4 = 1 = Odd
C. a/b = 4/4 = Odd
D.(4+a)/(2+4b) = (4+2)/(2+4.1) = 1 = Odd
E.(4+a)/(1+4b) = (4+1)/(1+4.1) = 1 = Odd
Hope this helps.

Like Flametop has mentioned, I'd like to rephrase the question in a form that 'which of the following can yeild integers'. The expressions are such that one of them under no circumstances can yeild an integer. If one of them always stays a fraction, it can neither be considered odd or even.

Choice 1 can be reduced to $$(1 + 2a)/2*(1+b).$$Now for all values of a, the nuemerator will be odd, but the denominator is even (product of 2 and (1+b) which can be even or odd). Hence, the nuemerator and the denom are odd and even respectively, and do not depend on the values of a and b. Hence it will always be a fraction. For the rest of the choices, the value of the expression can be reduced to an integer for some values of a and b. Hence, according to me, the trick is to find the expression whose value is independent of the underlying variables and yeilds a constant value for all cases as above, where the value is a fraction no matter what.

Hope it helps!

Regards,
A
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Re: If a and b are positive integers, which of the following can  [#permalink]

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11 Jun 2013, 06:33
arpanpatnaik wrote:
fameatop wrote:
In my view, Best approach to such type of questions is to prove that the constraint is possible. Let's see how it works- 1 is odd integer & a, b can take any positive integer value. So if we can prove that any fraction can yield 1, we can reject that choice
A. (2+4a)/(4+4b) - Under any condition this can't be odd. Answer
B. (4a)/(b) = 4.1/4 = 1 = Odd
C. a/b = 4/4 = Odd
D.(4+a)/(2+4b) = (4+2)/(2+4.1) = 1 = Odd
E.(4+a)/(1+4b) = (4+1)/(1+4.1) = 1 = Odd
Hope this helps.

Like Flametop has mentioned, I'd like to rephrase the question in a form that 'which of the following can yeild integers'. The expressions are such that one of them under no circumstances can yeild an integer. If one of them always stays a fraction, it can neither be considered odd or even.

Choice 1 can be reduced to $$(1 + 2a)/2*(1+b).$$Now for all values of a, the nuemerator will be odd, but the denominator is even (product of 2 and (1+b) which can be even or odd). Hence, the nuemerator and the denom are odd and even respectively, and do not depend on the values of a and b. Hence it will always be a fraction. For the rest of the choices, the value of the expression can be reduced to an integer for some values of a and b. Hence, according to me, the trick is to find the expression whose value is independent of the underlying variables and yeilds a constant value for all cases as above, where the value is a fraction no matter what.

Hope it helps!

Regards,
A

Thank you arpanpatnaik- I like this strategy....was looking for something generic like this!
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Re: If a and b are positive integers, which of the following can  [#permalink]

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14 Sep 2018, 17:44
bagdbmba wrote:
If a and b are positive integers, which of the following cannot be odd?

A. (2+4a)/(4+4b)
B. (4a)/(b)
C. a/b
D. (4+a)/(2+4b)
E. (4+a)/(1+4b)

In answer choice A, notice that 2 + 4a = 2(1 + 2a) and 4 + 4b = 4(1 + b). So

(2+4a)/(4+4b)

2(1+2a)/[4(1+b)]

(1+2a)/[2(1+b)]

We see that (1 + 2a) is an odd number and 2(1 + b) is an even number, regardless what a and b are, so the quotient can’t be an odd number since it’s not even an integer.

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Re: If a and b are positive integers, which of the following can   [#permalink] 14 Sep 2018, 17:44
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