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Let say Z1=k; so Zn becomes k+n-1; Thus total of this sequence becomes = Z1+Z2+....+Zn=k+k+1+k+2+...+k+n-1= n.k + (n-1).n/2 = (n/2) . (2k+n-1) So the average of it is equal to = ((n/2) . (2k+n-1))/n= k + (n-1)/2 If this is an integer; n-1 must be even so n must be odd.

Thus the answer is A.

I was wrong in my previous post.

Thanks to all of them that makes me to see the truth.

Let say Z1=k; so Zn becomes k+n-1; Thus total of this sequence becomes = Z1+Z2+....+Zn=k+k+1+k+2+...+k+n-1= n.k + (n-1).n/2 = (n/2) . (2k+n-1) So the average of it is equal to = ((n/2) . (2k+n-1))/n= k + (n-1)/2 If this is an integer; n-1 must be even so n must be odd.

Thus the answer is A.

I was wrong in my previous post.

Thanks to all of them that makes me to see the truth.

Thanks, maliyeci. That's very good approach to solve this problem without picking numbers method.

First time I faced this question, I chose (C) because I didn't feel confident that if n was even,then the average wouldn't be a integer.

I don't intend to sound like a critic, but there is no such rule

Division results always depend upon the factors and their corresponding powers in both - the divident, and the divisor. Im just saying this so that you don't follow a wrong rule, nor do others who will read this post.

If the dividend is having greater powers of 2 (even prime factor) than the divisor [ex: 8; 4], the quotient will be even. If the dividend has the same powers of 2 as the divisor [ex: 30; 6], both the 2's will get cancelled and the quotient will be odd.

For the quotient to be a decimal, the divisor must have at least one prime factor at a greater power than the dividend [eg: 15; 75] OR the divisor should have one or more prime factors which are not present in the dividend [eg: 14; 10]

This is some kind of PhD which I cant comprehend, but looks logical.

Regarding this question ( a different view ):

1) Av is odd and integers are consecutive => there are equal number of terms on the right side and left side of this average(odd) => sum of integers on the right side of this av can be odd or even, depending on whether we start from an odd integer or even integer

Now, lets say sum of right side is ODD, then we have the same ODD sum on left side of Av, so we get Total = ODD+ Av (which is ODD) + ODD => sum is ODD

Now, lets say sum of right side is EVEN, then we have the same EVEN sum on left side of Av, so we get Total = EVEN + Av (which is ODD) + EVEN => sum is ODD

Hence, 1 is suff

2) number of integers is ODD. => sum depends on where we start from

Lets start from EVEN integer. If N is ODD, then we have one extra even integer ( rest is EVEN+ODD pairs ) => Total = ODD ( from even + odd pairs ) + EVEN ( the starting point ) => sum is ODD

Now lets start from ODD integer. If N is ODD, then we have one extra odd integer ( rest is EVEN+ODD pairs ) => Total = ODD ( from even + odd pairs ) + ODD ( the starting point ) => sum is EVEN

I don't intend to sound like a critic, but there is no such rule

Division results always depend upon the factors and their corresponding powers in both - the divident, and the divisor. Im just saying this so that you don't follow a wrong rule, nor do others who will read this post.

If the dividend is having greater powers of 2 (even prime factor) than the divisor [ex: 8; 4], the quotient will be even. If the dividend has the same powers of 2 as the divisor [ex: 30; 6], both the 2's will get cancelled and the quotient will be odd.

For the quotient to be a decimal, the divisor must have at least one prime factor at a greater power than the dividend [eg: 15; 75] OR the divisor should have one or more prime factors which are not present in the dividend [eg: 14; 10]

Please ignore my comment about ODD, EVEN division. Actually I attempted this question by determining ODD and EVEN criterion which worked fine. Then somehow (I don't know why) this logic came to my mind. I can not expect this from me. This is a blunder logic.

All I can say is sorry for the confusion. This is not a concept which I use, this is a wrong concept which I developed in couple of seconds after solving this question.

My biggest worry is how to prevent someone else using this if they had seen this post and never cared to come back. This is the worst feeling which I am going through because if a single person attempt this wrong due to me then partially I become responsible for it.

Guys,my suggestion is to open up a thread which can contain all those concepts which are proven wrong because it might happen that this might not have got caught now but say after 1 month or so if someone founds it then he should know where to report it. This will insure that no one is following a wrong thing.

May be I don't know what I am saying...........I am too disturbed. !!!!!

Please ignore my comment about ODD, EVEN division. Actually I attempted this question by determining ODD and EVEN criterion which worked fine. Then somehow (I don't know why) this logic came to my mind. I can not expect this from me. This is a blunder logic.

All I can say is sorry for the confusion. This is not a concept which I use, this is a wrong concept which I developed in couple of seconds after solving this question.

My biggest worry is how to prevent someone else using this if they had seen this post and never cared to come back. This is the worst feeling which I am going through because if a single person attempt this wrong due to me then partially I become responsible for it.

Guys,my suggestion is to open up a thread which can contain all those concepts which are proven wrong because it might happen that this might not have got caught now but say after 1 month or so if someone founds it then he should know where to report it. This will insure that no one is following a wrong thing.

May be I don't know what I am saying...........I am too disturbed. !!!!!

RELAX!!!!!

You didnt do anything sooooo bad to be so disturbed there would hardly be anyone who would learn ur "mistaken concept" so quickly and never care to come and take a look at the thread. In case it does happen, its his bad luck. Thats all

Just be a bit more careful next time before posting something you have derived yourself, post it as a theory-to-be-verified rather than just "a theory". That way people would use their own judgement before blindly believing.

If the average of a set of consecutive integers is an integer, then the total number of integers in the set will always be odd.

ie, if there are 5 consecutive integers, only then their average will be an integer, if there were 4 integers, the avg will be a decimal.

Now, the average of consecutive integers will always have equal number of integers below it as above it. Thus, say A is the avg, and A has x number of integers below it, and x number of integers above it.

so total number of integers in the set = A + 2x

Now Even + odd = odd So if A is odd, adding it to 2x will always give an odd integer, ie, the sum of all the integers will be odd.

If the average of a set of consecutive integers is an integer, then the total number of integers in the set will always be odd.

ie, if there are 5 consecutive integers, only then their average will be an integer, if there were 4 integers, the avg will be a decimal.

Now, the average of consecutive integers will always have equal number of integers below it as above it. Thus, say A is the avg, and A has x number of integers below it, and x number of integers above it.

so total number of integers in the set = A + 2x

Now Even + odd = odd So if A is odd, adding it to 2x will always give an odd integer, ie, the sum of all the integers will be odd.

If the average of a set of consecutive integers is an integer, then the total number of integers in the set will always be odd.

does this logic apply to sum of consecutive odd integers and consecutive even integers ?

The logic applies, not as a rule, but just the logic. set of consecutive even integers CASE 1: number of integers (N) is even Average will always be odd because it will lie between the 2 central even integers. Yet, the total sum will always be even.

CASE 2: N is odd Avg will always be even, the central even number, and the sum will also be even.

set of consecutive odd integers CASE 1: N is even Avg will always be even, ie, the odd integer between the 2 central odd numbers. Sum will also be even, because even number of odd integers always add up to an even integer.

CASE 2: N is odd Avg will always be odd, the central integer, and the sum will always be odd as well. Because the half of integers below avg will be even, and half of those above avg will also add up to even integer, and even+even+odd(avg) = ODD integer.

Try to get the logic, there are no fixed rules, but with logic you can derive rules for various scenarios. Just remember that O +E = O E+E = E and O + O = E

OR

Any number of even integers will always add up to even. Odd number of odds will add up to Odd even number of odd will add up to even.

Harvard asks you to write a post interview reflection (PIR) within 24 hours of your interview. Many have said that there is little you can do in this...