Tuesday, February 19, 2008

More funny answers on math olympiad

(I had a prior post on FUNNY ANSWERS TO MATH OLYMPIAD QUESTIONS: here. This one is a different problem.)

In 2000 I made up and graded the following problem from the Maryland Math Olympiad from 2007 (for High School Students)
There are 2000 cans of paint. Show that at least one of the following two statements is true:
  1. There are at least 45 cans of the same color.
  2. There are at least 45 cans of all different colors.
It was problem 1 so it was supposed to be easy 95% of the students got it right and I suspect everyone reading this blog can do it. Note that the students taking this exam, Part II of a math olympiad, did well on part I, so they are good students. (Part I is 25 TOUGH multiple choice questions, point off if you are wrong, and Part II is 5 problems that require proofs.)

I got two funny answers:
ANSWER ONE: Paint cans are grey. Hence there are all the same color. Therefore there are 2000 cans that are the same color.
ANSWER TWO: If you look at a paint color really really carefully there will be differences. Hence, even if two cans seem to both be (say) RED, they are really different. Therefore there are 2000 cans of different colors.
Were they serious?

The first one points to a problem with the phrasing of the question- I clearly did not mean the cans themselves, and all of the other students knew that, but looking at the problem it could be interepreted that way. This person might have been serious.

The second one I can't imagine was serious.


  1. In an exam, ones mind becomes foggy to see the obvious such as (44)^2 < 2000
    Any tinge of sarcasm is injurious to young minds (IMHO)
    Please look at this years AMC 12 last question and see whether the readers of this blog can solve it in less than 5 minutes. a_{n+1} = \sqrt(3) * a_n - b_n and b_{n+1} = \sqrt(3) * b_n + a_n. Given that (a_{100},b_{100})=(2,4), find a_1+b_1

  2. What's the point of requiring proofs in these contests?

    If a high school student is good enough to solve difficult multiple choice questions, doesn't it follow that his/her proof skills would probably be pretty good -- or at least easily developed later in university?

  3. I'm not so sure they will develop good proving skills later in university or that you can make that assumption from the multiple choice. I know I was much better at the multiple choice sections of math contests in high school than at proofs. Also, it probably depends on the school and major as to whether they will develop any proving skills at all.

  4. I know I was much better at the multiple choice sections of math contests in high school than at proofs.

    What do you mean by this? Were your proofs correct (at least in your head) but not well written?

    I think what matters most of all here is insight -- not well written proofs.

    Getting difficult multiple choice questions correct is an indication that the student understands the major steps in a well written proof.

  5. It could be that both of them were serious. In my high school there were people who wrote these type of answers in exams and they were serious. Just because you are way too intelligent doesnt bring any right to you to mock at these students. Anyone can mock at you as well in certain circumstances.

  6. The difference between multiple choice type questions and proof type questions is the depth of the material.

    Traditionally, multiple choice questions are small nuggets which do not require very deep thought, but one needs to be fast. There is no point writing down proofs as they are very short anyway.

    Proof-type questions test depth in understanding and perseverance of the student. Further, tough questions should be given as proof so that partial credit can be awarded to students with an approach, or idea.

  7. Further, tough questions should be given as proof so that partial credit can be awarded to students with an approach, or idea.

    Why do you care about giving partial credit?

  8. I don't see how you can answer the question without the number of colors available... Why could you not have 2000 different colors?

  9. Then: ``2. There are at least 45 cans of all different colors.''

  10. The first one is a joke. Or, at least, I would have been very tempted to write that down as a joke.

  11. Sorry pal, you must admit that this is not a math problem. It is a ill defined problem, there must be better specified scientific wariables, like what shop, how many colours out of the pantone scale are different (20 000 ?!), wether it is the can or not.

    Say I was a 6 y old genius who has been with my father only to the wood pile shop. There is only linoleum there and certainly not even 45 brands (less than 45 outside colours) and the inside colour is - colourless - is that a colour? House colours are in so many nuances so they are ordered from a catalouge and arrive later.
    So my answer could be zero as well!

    OR, I certainly could prove at least 45 nuances/hues by comparing by the pantone scale, even if they are the same colour all, except if they are linoleum all.

    You can really spoil a career for a genious!

  12. Anonymous #1, that's a cute problem and I did get it in five minutes -- if I haven't screwed up the arithmetic the answer is a_1 = -2^{-98} and b_1 = 2^{-99}, so a total of -2^{-99}.

    I suppose if you don't see the trick, you could get to the right answer by cranking through three steps of the recurrence, _if_ you don't mess up all the sqrt{3}'s.

  13. There's no way either is serious. Both are desperate attempts at humor. Neither uses math, just extraneous real world 'facts'.

  14. Is it still provable in intuitionistic logic? "At least one of these statements is true" suggests intuitionistic disjunction to my mind.

  15. If you interpret "are the same color" as either "look the same color" or "are within a given tolerance of each other", then the relation becomes intransitive (i.e. if A and B are the same color, and so are B and C, then A and C are not necessarily the same color). So then the given statement becomes: any graph on n nodes has either a clique or an independent set of size sqrt(n). Is this true, by the way?

  16. you need to know how many colours are avalable first.

  17. If the n choices provided are either
    1) not mutually exclusive, or
    2) not exhaustive
    then it is not possible to prove either choice is correct or incorrect.
    In the case of 1) multiple choices will apply unless there is another qualifier.
    In the case of 2) [which is the case in this question] the missing choices may be correct.
    It is not possible show that either of these two statements is correct or incorrect.

    Poor questions must result in poor answers.

  18. [a_(n+1); b_(n+1)]=[sqrt(3) -1;1 sqrt(3)]*[a_n ; b_n]
    Solve for [a_n ; b_n]=inv[sqrt(3) -1 ; 1 sqrt(3)]*[a_(n+1);b_(n+1)]
    [a_99;b_99]=inv[sqrt(3) -1 ; 1 sqrt(3)]*[a_100;b_100]
    [a_98;b_98]=(inv[sqrt(3) -1 ; 1 sqrt(3)])^2*[a_100; b_100]
    [a_1;b_1]=(inv[sqrt(3) -1 ; 1 sqrt(3)])^99*[a_100; b_100]
    [a_1;b_1]=(1/2^100)*[0 2;-2 0]*[2; 4]=[1/2^97 ; -1/2^98]
    Therefore a_1+b_1=1/2^97-1/2^98=1/2^98