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If n is positive integer, is 4^n+n^2+1 divisible by 2?

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by Max@Math Revolution » Sun Jan 27, 2019 11:46 pm

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[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6
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Source: — Data Sufficiency |
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by fskilnik@GMATH » Mon Jan 28, 2019 4:29 am
Max@Math Revolution wrote:[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6
$$n \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\left( * \right)$$
$$\underbrace {{4^n}}_{\left( * \right)\,\,{\text{even}}} + {n^2} + 1\,\,\,\,\mathop = \limits_{\left( * \right)}^? \,\,\,\,{\text{even}}\,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\boxed{\,n\,\,\mathop = \limits^? \,\,{\text{odd}}\,}$$

$$\left( 1 \right)\,\,{n \over 4} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle $$
$$\left( 2 \right)\,\,{n \over 6} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle $$

The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by kyuhunl » Tue Jan 29, 2019 11:46 pm
Max@Math Revolution wrote:[GMAT math practice question]

If n is positive integer, is 4^n+n^2+1 divisible by 2?

1) n is a multiple of 4
2) n is a multiple of 6
4^n is always even.
n^2 is even if n is even, odd if n is odd.
4^n+n^2+1 is even if n is odd, odd if n is even.
1) and 2) are both individually sufficient: D
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by Max@Math Revolution » Wed Jan 30, 2019 1:02 am
=>

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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Since 4^n is a multiple of 2, we only need to look at n^2+1.
If n is an odd number, 4^n+n^2+1 is divisible by 2.
If n is an even number, 4^n+n^2+1 is not divisible by 2.
The question asks if n is an odd number.
Thus, each of conditions is sufficient.

Therefore, D is the answer.
Answer: D
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