chikki420
GMATGuruNY
aalriy
If the sum of n consecutive positive integers is 33, what of the following could be the value of n?
I. 3
II. 6
III. 11
A. I only
B. II only
C. III only
D. I and II
E. I, II and III
For any set of consecutive integers:
\(median = \frac{sum}{count}\)
I: 3 integers
\(median = \frac{33}{3} = 11\) --> 10, 11, 12
II: 6 integers
\(median = \frac{33}{6} = 5.5\) --> 3, 4, 5, 6, 7, 8
III: 11 integers
\(median = \frac{33}{11} = 3\) --> not viable, since some of the integers below the median will be negative
Why to use median concept for the question which belongs to Arithmetic progression, besides the term MEDIAN isnt even mentioned in the question.
The median of consecutive integers must be either an integer or a value halfway between two integers (-1.5, 0.5, 10.5, etc.).
Here, calculating the median is an efficient way to determine whether a viable set of consecutive integers can be yielded by Statements I, II and III.
A valid statement must satisfy two conditions:
1. The median must be either an integer or a value halfway between two integers
2. The integers surrounding the median must all be positive
As shown in my earlier solution:
Statements I, II and III all satisfy the first condition.
Only Statements I and II satisfy the second condition.
Thus, the correct answer is D.