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If a and b are integers, is a^2+b^3 an odd number?

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If a and b are integers, is a^2+b^3 an odd number? [#permalink]

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New post 02 Jan 2018, 00:10
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[GMAT math practice question]

If \(a\) and \(b\) are integers, is \(a^2+b^3\) an odd number?

1) \(3a+4b\) is an odd number
2) \(a\) and \(b\) are consecutive integers
[Reveal] Spoiler: OA

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If a and b are integers, is a^2+b^3 an odd number? [#permalink]

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New post 02 Jan 2018, 07:56
MathRevolution wrote:
[GMAT math practice question]

If \(a\) and \(b\) are integers, is \(a^2+b^3\) an odd number?

1) \(3a+4b\) is an odd number
2) \(a\) and \(b\) are consecutive integers


We need to know whether \(a\) & \(b\) are odd or even

Statement 1: \(3a+4b=Odd => 3a=Odd-Even\) (\(4b\) will always be even irrespective of \(b\))

Hence \(a=\frac{ODD}{3}=ODD\). But we do not have information about \(b\). Insufficient

Statement 2: for two consecutive integers, one will be even and the other will be odd. Odd and Even number raised to any positive integer power will be ODD and EVEN respectively.

Also \(Odd+Even=Odd\)

Hence \(a^2+b^3=Odd\). Sufficient

Option B
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Re: If a and b are integers, is a^2+b^3 an odd number? [#permalink]

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New post 03 Jan 2018, 23:28
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

We can modify the original condition and question as follows.
There are two different ways in which \(a^2+b^3\) can be odd:

i) \(a\) is even and \(b\) is odd.
ii) \(a\) is odd and \(b\) is even.

Since condition 2) tells us that a and b are consecutive integers, one of them must be odd, and the other must be even. In both cases, the answer is ‘yes’.
Therefore, condition 2) is sufficient.

Condition 1) implies that a is odd, but tells us nothing about b. Therefore, it is not sufficient.

Therefore, the answer is B.

Normally, in problems which require 2 equations such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Answer: B
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Re: If a and b are integers, is a^2+b^3 an odd number? [#permalink]

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New post 12 Jan 2018, 16:53
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(a\) and \(b\) are integers, is \(a^2+b^3\) an odd number?

1) \(3a+4b\) is an odd number
2) \(a\) and \(b\) are consecutive integers


We need to know whether \(a\) & \(b\) are odd or even

Statement 1: \(3a+4b=Odd => 3a=Odd-Even\) (\(4b\) will always be even irrespective of \(b\))

Hence \(a=\frac{ODD}{3}=ODD\). But we do not have information about \(b\). Insufficient

Statement 2: for two consecutive integers, one will be even and the other will be odd. Odd and Even number raised to any positive integer power will be ODD and EVEN respectively.

Also \(Odd+Even=Odd\)

Hence \(a^2+b^3=Odd\). Sufficient

Option B


I'm confused as to why (1) isn't sufficient. if 3(a)+4(b) is odd and 4(b) is always even, then 3(a) will always be odd.. therefore a will always be odd.. therefore shouldn't odd^2+even^3=odd?
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If a and b are integers, is a^2+b^3 an odd number? [#permalink]

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New post 12 Jan 2018, 18:25
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YYZ wrote:
niks18 wrote:
MathRevolution wrote:
[GMAT math practice question]

If \(a\) and \(b\) are integers, is \(a^2+b^3\) an odd number?

1) \(3a+4b\) is an odd number
2) \(a\) and \(b\) are consecutive integers


We need to know whether \(a\) & \(b\) are odd or even

Statement 1: \(3a+4b=Odd => 3a=Odd-Even\) (\(4b\) will always be even irrespective of \(b\))

Hence \(a=\frac{ODD}{3}=ODD\). But we do not have information about \(b\). Insufficient

Statement 2: for two consecutive integers, one will be even and the other will be odd. Odd and Even number raised to any positive integer power will be ODD and EVEN respectively.

Also \(Odd+Even=Odd\)

Hence \(a^2+b^3=Odd\). Sufficient

Option B


I'm confused as to why (1) isn't sufficient. if 3(a)+4(b) is odd and 4(b) is always even, then 3(a) will always be odd.. therefore a will always be odd.. therefore shouldn't odd^2+even^3=odd?


Hi YYZ

As you rightly mentioned that \(4b\) is even but from this we do not know the value of \(b\) per say

for eg. if \(4b=12\), then \(b=3=>b^3=27=Odd\)

but if \(4b=8\), then \(b=2=>b^3=8=Even\)

In both these scenarios our question stem will yield different result.
If a and b are integers, is a^2+b^3 an odd number?   [#permalink] 12 Jan 2018, 18:25
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