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Math Challenge Problem – A “Mean” Series

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?

  1. 10.5
  2. 11
  3. 12
  4. 13
  5. 22



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:

(x)(x + 1)/2 = 231

x^2 + x = 462

x^2 + x - 462 = 0

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).


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8 Responses to Math Challenge Problem – A “Mean” Series

  1. Vikas September 30, 2013 at 10:13 pm #

    I solved it with other method.

    Chris please correct me if my logic is correct –

    If there are n consecutive integer – if n is odd then sum of these numbers must be divisible by n – if n is even then it will be never divisible by n.

    if n Is even – its sum 231 will not divisible by 12 and 22. so options C and E are gone.

    n is odd – its sum 231 must be divisible by 11 and 13…and 13 is not factor of 231…so option D is gone.

    we left with option 10.5 and 11.

    sum= (f+l) * N/2 ( for AP series)

    when we plug in the value (1+21)*21/2 = 231

    Hence correct answer – B

    • Chris Lele
      Chris Lele October 3, 2013 at 1:56 pm #


      I like the logic :). You could even take it one step further, so you don’t have to rely on plugging anything into the formula. If the median is 10.5, then n = 20. 231 clearly isn’t divisible by 20, so the answer has to be (B).

      • Bilal January 13, 2014 at 12:59 pm #

        “If the median is 10.5, then n = 20. 231 clearly isn’t divisible by 20” can you explain it a little more, how’d you come up with 20.231….. i am new to quant stuff so difficulty understanding some stuff posted here.

        • Chris Lele
          Chris Lele January 15, 2014 at 11:01 am #

          Happy to help!

          Let’s take the first statement: If the median is 10.5, then n = 20.”

          The logic is whenever there is an even number of consecutive integers the median is going to be some number followed by .5. The reason is it is halfway between two integers. The quick way to calculate the median of a list of consecutive numbers is to add the first and last, then divide by 2. In this case, 1 + 20 = 21/2 = 10.5.

          I came up with 20 integers, by looking at the answer choices, specifically the middle ones. If they are in the “middle” (since they are the median), then that list of consecutive would have to be somewhere in the 20’s.

          The number 231 is provided by the question :).

          Hope that helps 🙂

  2. Kate Boyd September 30, 2013 at 1:49 pm #

    You might want to proofread this post. There is at least one spelling error, and two incomplete sentences which make the post difficult to understand:

    “…we know that we need to (two?) very similar numbers.”

    “Notice, however, that the (+x) in the middle of the equation above (???).”

    “In other words, one of the numbers is negative and the two numbers (???), so if you take the absolute value of the negative number…”

    Well, I’m a math dolt, so I perhaps wouldn’t understand it anyway. 🙂

    Best, Kate

    • Chris Swimmer
      Chris Swimmer September 30, 2013 at 5:59 pm #

      Hey Kate! Good catch! Sometimes Chris writes faster than he thinks ;). I’ve fixed the errors though.

      Chris S

  3. Ashutosh September 30, 2013 at 11:56 am #

    Hi Chris, you’re amazing, apart from the solution you gave for solving this math problem. I’m quite impressed by the fact that you’ve used few GRE vocab words in the math solution. Which actually reinforces the words in our memory.

    Thank you!!

    • Chris Lele
      Chris Lele October 3, 2013 at 1:58 pm #

      You are welcome!

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