Hardest Question at MathCounts 2012?

February 1, 2013 at 1:01 am 5 comments

Rumor has that some big professional football competition will occur on Sunday, but I’ll be setting my sights on a more intellectual competition that takes place on Saturday. Middle school students across the country will be participating in the MathCounts Chapter Competition. Good luck to all mathletes!


As a member of the MathCounts Question Writing Committee, one of my responsibilities is to ensure that the questions on the competition are not too hard, not too easy, but just right. Consequently, the committee reviews a lot of data regarding the percentage of students that answer each question correctly.

Rather accidentally, I wrote the most difficult question that appeared on the 2012 Chapter Competition.

The chart below shows the number of students from a random sample (n = 1200) who answered each of the 30 questions correctly on the Sprint Round:

MathCounts 2012 Chapter Sprint

Although Problem 29 was answered correctly by just 42 students, it was not the most difficult. Students are given 40 minutes to complete all 30 problems in the Sprint Round. Consequently, many students never get to the problems at the end of the round, and the last several questions are typically answered correctly by only a handful of students.

On the other hand, Problem 14 was answered correctly by only 50 students, and it occurs in the middle of the round. By design, the questions in the round proceed from least difficult to most difficult, so problem 14 was intended to be a medium question. Questions on either side of Problem 14 were answered correctly by 500 to 700 students, and the members of the question writing committee suspected that Problem 14 would be answered correctly by a similar number. Boy, did we miss the mark!

(You’ll notice that we also incorrectly guaged the difficulty of Problem 9, though not quite as dramatically.)

So, what was this doozie of a question? Try it for yourself:

If Friday the 13th occurs in a month, then the sum of the calendar dates for the Fridays in that month is 6 + 13 + 20 + 27 = 66. What is the largest sum of calendar dates for seven consecutive Fridays in any given year?

What do you think? Are you as smart as the 4.167% of mathletes who got this problem correct? To find out, take a look at the calendar for May and June 2013.

Entry filed under: Uncategorized. Tags: , , , , , .

Results of a Wonderlic-SAT Comparison One Joke Per Cent

5 Comments Add your own

  • 1. Dumb DUmbson  |  March 12, 2015 at 12:06 pm

    SO EZ i got answer in 5 seconds.

  • 2. Stephen Hawkings  |  February 8, 2016 at 3:07 pm

    mathcounts is a joke competition. it doesn’t test how good you are at math but how good of a human calculator you are. Solving this problem isn’t too difficult, its just that it takes more than 15 seconds to solve so everyone skips it.

  • 3. Brian Lin  |  July 5, 2016 at 3:33 pm

    @above, it’s Stephen Hawking, fyi 😛
    Also, mathcounts DOES focus on speed. However, to be proficient in MathCounts, one must also have a proficient background in math. After all, there is a reason why EVERYONE in the past 5 years who have made the top 12 in MathCounts Nationals made USA(J)MO or higher (top 500 in the nation for high-schoolers).
    Mathcounts also isn’t a joke: it is the most popular and most competitive middle school math competition.
    Trust me, as I made the national competition in my 8th grade year. Being a human calculator isn’t guaranteeing success. It is a merely tool which is more useful in MathCounts than in other competitions.
    How does a person a geometry problem by being a human calculator? It doesn’t happen. Plus, you are given 80 seconds on average to solve the problem. Thus, it makes sense that one would spend at least 80 seconds on #14.

  • 4. Ptown_Beast  |  January 12, 2017 at 8:04 pm

    so what is the solution to get to 142???

    • 5. venneblock  |  January 12, 2017 at 10:28 pm

      If you click on the link for the calendar of May and June 2013, the dates highlighted in red give a sum of 17 + 24 + 31 + 7 + 14 + 21 + 28 = 142.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Trackback this post  |  Subscribe to the comments via RSS Feed

About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.

Past Posts

February 2013
« Jan   Mar »

Enter your email address to subscribe to the MJ4MF blog and receive new posts via email.

Join 245 other followers

Visitor Locations

free counters

%d bloggers like this: