Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S?
One way to go about approaching this question is to try adding up consecutive integers until you get 231. This is definitely a feasible solution path—just not a very efficient one. It would probably take you about 2 minutes to do this, and you could very well make a calculation error. Afterwards, you’d also have to remember the question and find the median—and not the greatest integer of Set S—all of which would take you closer to 3 minutes.
The GRE makes this solution path possible because—at least my reasoning goes—they want to tempt you with it. The question wouldn’t make the sum of Set S some really larger number that nobody in their right mind would try to calculate (with or without the handy calculator). But by falling for the trap, you may get the question correct, but you will also end up burning precious minutes that could be better spent elsewhere.
For a fast solution to the question above, we need to know a handy formula: (x)(x + 1)/2 = sum of series, in which x is the great integer in a consecutive series of positive integers starting with ‘1’. Therefore, we get the following:
At this point you might feel stymied. 462 is quite a large number and factoring it could take awhile. Notice, however, the (+x) in the middle of the equation above. That shows you that when you add up the two numbers whose product is negative 462, you get +1. In other words, one of the numbers is negative and the two numbers together sum to 1, so if you take the absolute value of the negative number, the two numbers are only one integer apart.
That’s good news because instead of trying to factor a bunch of ‘2’s out of 462, we know that we need two very similar numbers. The square root of 400 is 20, so you know that the two factors of 462 we are looking for have to be greater than 20. 21^2 = 441. And 462 ends in ‘2’. That should lead you to 22 x 21, which equals 462.
(x + 22)(x -21) = 0
x = -22 and 21
We can discount -22 because x has to be positive. Therefore the largest integer in S is 21. But be careful not to choose (E). The median of any set of consecutive integers is the sum of the lowest and greatest integer divided by 2, which gives us 11, answer (B).