If x is even integer, which of the following must be an odd

One can spot right away that if \(x\) is any even number then \(x^2\) is a multiple of 4, which makes \(\frac{x^2}{2}\) an even number and therefore \(\frac{3x^2}{2}+1=3*even+1=even+1=odd\).

Answer: E.

If you don't notice this, then one also do in another way. Let \(x=2k\), for some integer k, then:

A. \(\frac{3x}{2}=\frac{3*2k}{2}=3k\) --> if \(k=odd\) then \(3k=odd\) but if \(k=even\) then \(3k=even\). Discard;

B. \(\frac{3x}{2}+1=\frac{3*2k}{2}+1=3k+1\) --> if \(k=odd\) then \(3k+1=odd+1=even\) but if \(k=even\) then \(3k+1=even+1=odd\). Discard;

C. \(3x^2\) --> easiest one as \(x=even\) then \(3x^2=even\), so this option is never odd. Discard;

D. \(\frac{3x^2}{2}=\frac{3*4k^2}{2}=6k^2=even\), so this option is never odd. Discard;

E. \(\frac{3x^2}{2}+1=\frac{3*4k^2}{2}=6k^2+1=even+1=odd\), thus this option is always odd.

Answer: E.

Similar question to practice: if-a-and-b-are-positive-integers-such-that-a-b-and-a-b-are-88108.html

Hope it helps.
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Hi,

Since, x is even we can assume x = 2 or x = 4,
such that x/2 is both odd/even as per the value of x.

Check each option with these values. We get the answer when both x=2, 4 gives an odd value.

Regards,

I narrowed this question down to B and E. Based on the rules alone why couldn't it be B?

(3x/2)+1

if x is even then we have an even + odd which would be odd?

thanks for the clarification.

Because the theory is important but also to reach the answer through the most efficient way.

\(x=2\)(as statement says) \(OR x=4\) (thanks this we know for instance that A is not always true)

\(\frac{6}{2}\)\(= 3+ 1 = 4\) \(OR 7\) (is not always true: one time even one time odd). That's it Same for the other answer choices. You can obtain E in 30 seconds
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thanks.

certainly i'm trying to answer "odd/even" questions in the most efficient manner possible.

what i would have done on the real CAT is narrowed it down to B and E, then like you let x = 2 or 4 and plugged in to see.

i kind of got tripped up. typically when we multiply a integer by an even we ALWAYS get an even but (3/2) is frac, therefore multiplying it by an even may or may not make it even?

That's not correct.

Given that \(x\) is even, thus \(x=2k\) for some integer \(k\). Substitute in option B:

\(\frac{3x}{2}+1=\frac{3*2k}{2}+1=3k+1\) --> if \(k=odd\) then \(3k+1=odd+1=even\) but if \(k=even\) then \(3k+1=even+1=odd\).

As you can see \(\frac{3x}{2}+1\) can be even (for example if x is 2, 6, 10, ...) as well as odd (for example if x is 4, 8, 12, ...).

Hope it's clear.
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thank you for the help.

Option E

x is an Even inetegr. Odd: O, Even: E & Fraction: F

A. \(\frac{3x}{2}\) = \(\frac{3*E}{2}\) = \(3*E\) \(or\) \(3*O\)= \(E\) \(or\) \(O\)
B. \(\frac{3x}{2}+1\) = \(\frac{3*E}{2} + 1\) = \(3*E + 1\) \(or\) \(3*O + 1\)= \(O\) \(or\) \(E\)
C. \(3x^2\) = \(3E^2\) = \(3*E\) = \(E\)
D. \(\frac{3x^2}{2}\) = \(\frac{3E^2}{2}\) = \(3*E\) = \(E\)
E. \(\frac{3x^2}{2}+1\) = \(\frac{3E^2}{2}+1\) = \(3*E + 1\) = \(E + 1\) = \(O\)

if I take x=2, first option A - would be 3*2/2 = 6/2 =3 and 3 is odd..why is this wrong?

The question asks which of the following MUST be odd, not COULD be odd. Option A could be odd but it's not always odd while E is odd for any even x.

Hope it's clear.
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.Hello experts!

I answered correctly this one but after that, I started to think the next doubt:

If 0 is considered an even number...

Then D is not going to become an integer because 0*2 will be 1 so:

(3(1) divided by 2) + 1

What am I doing wrong?

Thank you so much!

Plug in and try to negate -

A. \(\frac{3x}{2}\)

Or, \(\frac{3*4}{2} = 6\)

B. \(\frac{3x}{2} + 1\)

Or,\(\frac{3*6}{2} + 1 = 10\)

C. \(3x^2\)

Or, \(3*2^2 = 12\)

D. \(\frac{3x^2}{2}\)

Or, \(\frac{3*2^2}{2} = 6\)

(E) Plug in any number it will always be ODD, Answer must be (E)
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