The mean of a set of n consecutive positive integers is m1. One of the

\(a_1+a_2+...+a_n= m_1*n\)
\(a_1+a_2+...+a_n= m_2*(n-1)\) (one of the integers removed from the set )

Is n even?

(Statement1): \(m_1=m_2\)
--> In order \(m_1=m_2\) to be true, the one integer removed from the set should be median and n should be always odd
(Always NO)--Sufficient

(2) The least element of the original set of integers is odd.
NO matter whether the least element of the set is odd. No info about what connection is there between \(m_1\) and \(m_2\)
insufficient

The answer is A.

We are given that the mean of n consecutive numbers is m1. One of integers is removed to create a new set of positive integers. So, we now have n-1 integers in the set. We are to determine if n is even.

First of all, let's form two sets of consecutive integers: one set with n even and the other with n odd.
Let n=5, A: 1,2,3,4,5
Let n=6, B: 1,2,3,4,5,6
We know that m1 for set A=3 while that for B=3.5

As a general rule, the mean of any odd set of consecutive integers is always an integer, while that for an even set of consecutive integers is always a fraction or non-integer.

Statement 1: m1=m2
Sufficient.
Let's take Set A {1,2,3,4,5} and remove 3 from the set, we have a new set Aa {1,2,4,5}. The average of the new set is (1+2+4+5)/4 = 3. If you take out the mean of a set of numbers and that mean happens to be a member of the set of numbers, the rest of the numbers will still form another set with a mean equal to the mean of the original set. This is always true since the mean is that one number that can be used to replace all elements of the set.

For an odd set of consecutive integers, since the mean is always equal to the median of the set, taking out the media will create another set that has a mean equal to the original set. Hence, we can conclude based on statement 1 that n is odd and not even.

n cannot be even because any number out of n even consecutive numbers will result in a mean which is different from the mean of the original set.

Statement 2: The least element of the original set of integers is odd.
Insufficient.
Just knowing that the least element in a consecutive set of integers is not enough to determine if that set is odd or even. As can be seen above, both sets A and B have the least element as 1, which is odd. Meanwhile set A is composed of odd number of elements while set B is composed of even number of elements.

The answer is option A.

#1
for set S= ( 1,2,3,4,5) and (1,2,4,5) we get m1=m2
only for odd set of integers this relation is possible , n is not even
sufficient

#2
The least element of the original set of integers is odd
no relation b/w m1 & m2 and we get m1 and m2 for n is odd and even
insufficient
IMO A
The mean of a data set of n consecutive positive integers is m1. One of the integers is removed from the set to create a new set of positive integers with a mean of m2. Is n even?

(1) m1=m2

(2) The least element of the original set of integers is odd.

>> Answer: A
I think the first statement alone is enough to arrive at the answer. If a number is removed from a set and the mean doesn't change. The removed number must equal mean (if it is less than mean, the new mean should be greater). So we know that mean of the original set is an integer.
If n is even, the mean of the original set should have .5 (for example the mean of 3, 4, 5, 6 is 3.5).
Therefore n is odd.