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Re: x is a nonzero integer [#permalink]
01 May 2012, 10:28

İİ is not correct, since nowhere mentioned that even integers MUST BE positive I is true, since the avrg sum= (sum of even integers)/ 5 (odd integer)=even/odd=even

III is also out. u can check it , if u pick some integers _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]
01 May 2012, 13:41

4

This post received KUDOS

Expert's post

If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer. II. x is a nonzero integer. III. x is a multiple of 5.

(A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III

First of all notice that we are asked which of the following MUST be true, not COULD be true.

Also notice that the average of 5 consecutive even integers equals to the median, so it's just a middle term. So, I must always be true: {even, even, even, even, even}

Next: II. x is a nonzero integer. Not necessarily true, consider: {-4, -2, 0, 2, 4}; III. x is a multiple of 5. Not necessarily true, consider: {0, 2, 4, 6, 8}.

Answer: A.

Hope it's clear.

P.S. In your example if n=-4 then x=n+4=0, so II is not always true. _________________

Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]
17 Jul 2014, 21:00

Bunuel wrote:

If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer. II. x is a nonzero integer. III. x is a multiple of 5.

(A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III

First of all notice that we are asked which of the following MUST be true, not COULD be true.

Also notice that the average of 5 consecutive even integers equals to the median, so it's just a middle term. So, I must always be true: {even, even, even, even, even}

Next: II. x is a nonzero integer. Not necessarily true, consider: {-4, -2, 0, 2, 4}; III. x is a multiple of 5. Not necessarily true, consider: {0, 2, 4, 6, 8}.

Answer: A.

Hope it's clear.

P.S. In your example if n=-4 then x=n+4=0, so II is not always true.

Hello Bunuel

I have a dout in this statement that you made :the average of 5 consecutive even integers equals to the median.

Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink]
23 May 2015, 07:33

Expert's post

shallow9323 wrote:

Hi Bunuel,

First if all i'd like to Thank you for all the help. I am deeply appreciate your efforts.

Is zero an even integer ?

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

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