The average (arithmetic mean) of 17 consecutive integers is an odd num

Sum of first n numbers is \(1+ 2+ ... + n = \frac{n(n+1) }{ 2}\)

Now, sum of first 17 numbers will be \(\frac{17*18}{2} = 153\) and its average will be \(\frac{153}{17} = 9\) (Odd)

1. Largest number is Odd (17)
2. Sum of all integers is Odd (153)
3. Difference between largest and smallest integer is even. ( 17 - 1 = 16)

Hence, only (E) holds true, Answer must be (E)

a9 is odd
For this a1 is odd and a17 is odd
so largest integer is odd
For consecutive integers
Sum of the terms= AM*no. of terms
and AM= (First +Last)/2=odd (given)
Sum of the terms= odd* odd=odd
a1 is odd and a17 is odd
Difference=a17-a1= Even
E:)

Since the mean (or average) of a set of consecutive integers is equal to the median, we see that the median is also odd. Now let’s analyze the Roman numeral statements.

I. Largest integer is even.

The largest number in terms of the median (or the middle number) is (17 - 1)/2 = 8 more than the median. Since the median is odd and odd + 8 = odd + even = odd, we see that the largest number is odd also. Statement I is not ture.

II. Sum of all integers is odd.

Since sum = average x quantity and here we have average = odd and quantity = 17, we see that Sum = odd x 17 = odd x odd = odd. Statement II is true.

III. Difference between the largest and smallest integers is even.

In I, we see that the largest number is 8 more than the median, i.e., largest number = median + 8. Therefore, the smallest number should be 8 less than the median, i.e., smallest number = median - 8. So we have:

largest number - smallest number = (median + 8) - (median - 8) = 8 + 8 = 16

Statement III is true.

Answer: E

Question:
The average (arithmetic mean) of 17 consecutive integers is an odd number. Which of the following must be true?
I. Largest integer is even.
II. Sum of all integers is odd.
III. Difference between largest and smallest integer is even.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


Solution:
We know that: The average (arithmetic mean) of 17 consecutive integers is an odd number.

Consecutive integers refer to integers having a common difference of 1, for example: 11, 12, 13, 14, ..
In such a scenario, the arithmetic mean (average) is always equal to the median
[In short: in an arithmetic progression: mean equals the median]

Thus, for 17 integers, the median is the \([(17+1)/2] th\) term i.e. the 9th term
Since the above median (or mean) is an odd number, we have:

The 9th number is odd (hence: 8th is even, 7th is odd ... and the 1st number is odd)
Similarly, the 17th term is 8 more than the 9th term i.e. 8 more than an odd number; hence is odd

Let us look at the statements:
I) The largest number is the 17th term, which is odd
II) Sum of all numbers = 17 x (Mean) = 17 x (odd number), which is odd
III) Difference between the largest and smallest numbers = Odd - Odd = Even

Thus, statements II and III are correct

Answer E

The average (arithmetic mean) of 17 consecutive integers is an odd number - This means the middle number (9th) is odd.
So the 17 numbers are made up of 8 numbers before and after the middle number. e.g. if the middle number (average) is 9, the numbers would be 1 to 17 - 1 to 8 before 9 and 10 to 17 after 9.

I. Largest integer is even.
Largest integer here is 17. Not even.

II. Sum of all integers is odd.
We have 8 even integers and 9 odd integers. Since we have an odd number of odd integers, the sum will be odd.

III. Difference between largest and smallest integer is even.
Smallest and the largest integers both will be odd. So their difference will be even.

Answer (E)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.