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For those curious, below are some kinds of primary math questions we model after to test teacher applicants. In my experience, many kids are much more enthusia

For those curious, below are some kinds of primary math questions we model after to test teacher applicants. In my experience, many kids are much more enthusiastic about this kind of "challenging" problems than the drills in many standard textbooks, as long as the problems are chosen to match their level. They definitely learn a lot more as well. Note that although they do require a little arithmetic to solve, the challenging part is not arithmetic. Some problems focus on geometry, logic, patterns, or other kinds of puzzles. More examples can be found at the source below. "- The edge of a cube is 8 cm. All the faces are painted orange. It is then cut into small cubes of edge 1 cm. How many small cubes have exactly two faces painted? - What is the greatest possible number one can get by discarding 100 digits, in any order, from the number 1234567891011121314151617…57585960? - Eleven consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2012, what number did Maria erase? - You must color each square in the figure below in red, green or blue. Any two squares with adjacent sides must be of a different color. In how many different ways can this coloring be done? Figure at question 11 here: https://gato-docs.its.txstate.edu/jcr:450cce10-3b6a-4ddd-a195-5c26efe39bed/2012_PWMC_individual.pdf https://gato-docs.its.txstate.edu/jcr:450cce10-3b6a-4ddd-a19... " Source: the Primary Math World Contest (usually held in Hong Kong) https://www.txstate.edu/mathworks/PMWC/previous-pmwc-tests.html https://www.txstate.edu/mathworks/PMWC/previous-pmwc-tests.h...

esmi 7y ago
Is the answer to the second question allowed to be in engineering form? :) What grade level is question 3 considered?
- nopinsight 7y ago
  What do you mean by 'engineering form'? Using variables in the answer? It would be suitable for students between grades 4 and 6 (the highest level we currently teach), depending on how talented the kid is. To be clear to other readers, these are hard questions that most people, including older kids and adults, probably can't answer without proper learning and practice.
  - esmi 7y ago
    In the US we call this engineering form. https://en.m.wikipedia.org/wiki/Scientific_notation#E-notation https://en.m.wikipedia.org/wiki/Scientific_notation#E-notati... I misread the question though and now realize it’s not needed.
    - monktastic1 7y ago
      I'm also from the US but I've never heard the term "engineering form." FWIW, Wikipedia thinks it's something a little different: https://en.wikipedia.org/wiki/Engineering_notation https://en.wikipedia.org/wiki/Engineering_notation.
    - Armisael16 7y ago
      I've never heard that (or anything else) called 'engineering form'.
Shivetya 7y ago
and here I sit totally flummoxed by the questions yet later on you state these were suitable to grades up to sixth? So for those who have let their math skills lapse or need help to get back in the right frame of mind to learn, where online is the best resource? Khan Academy?
- nopinsight 7y ago
  To be clear, these are hard contest questions that most people, including older kids and adults, probably can't answer them without proper learning and practice. I have not checked them out in detail, but it appears that https://artofproblemsolving.com/ https://artofproblemsolving.com/ is a good resource. They also have interesting online and physical book series: https://artofproblemsolving.com/store https://artofproblemsolving.com/store (I have no affiliation.)
cecilpl2 7y ago
72? 99999785960? 199?
- monktastic1 7y ago
  FWIW, I got the same answers. 1. 12 edges * 6 non-corner pieces per edge. 2. There are 111 digits there. You can't keep six nines and have enough digits left. So you keep five (from 9, 19, 29, 39, 49), and now have six digits to keep from 5051525354555657585960. It can't start with 8 (or you still won't have enough digits left), so you choose 785960. 3. First notice that 0+...+10 = 55. To make it end in 2, you have to discard the 3. In general if you have n+...+(n+10) you discard n+3. Now n+...+(n+10) = 5.5*(n+(n+10)) by Gauss' trick, and subtracting (n+3) you have 10n + 52 = 2012. Very much doubt I could have solved them in elementary school(!) though.
  - asdffdsa 7y ago
    Could you explain why you can't just do 99999988888? 100 digits in any order, what am I missing?
    - dan-robertson 7y ago
      You’re allowed to delete whichever digits you like. But you aren’t allowed to reorder the digits which are there. I.e. if you had to delete one digit you could go from 189 to 89 but not to 98.
chongli 7y ago
Eleven consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2012, what number did Maria erase This is a very challenging problem. Without a calculator (or computer), it took me a system of an equation with an inequality in 2 variables, using the fact that the solutions are integers, to solve it. I would be very surprised (and excited) to meet an 8th grader (let alone a 6th) who could solve this problem without any external help the way I did. I would not expect an elementary school student to have this kind of algebra knowledge. Most elementary school students around here don’t even know what algebra is.
- malaya_zemlya 7y ago
  try performing all calculations mod 10.
  - chongli 7y ago
    I’m not sure how to make use of your hint. I solved the problem for n, the largest number in the sequence of eleven, and a, the number Maria erases, by the following: n - 11 < a < n + 1 11(n + n - 10)/2 = 2012 + a I’m interested to know what other methods may be used to solve it (apart from brute force).
- username90 7y ago
  You don't need algebra. You know that 10 numbers are 2012, so on average 201, so the middle of the numbers needs to be around there. After that you just try a few sequences around 201 and you'll find it pretty quickly, most can add 11 numbers pretty fast in 6'th grade. Forcing yourself to only rely on the solution strategies taught by school makes you very bad at problem solving.
  - chongli 7y ago
    I specifically set out to find a solution that doesn’t involve any brute force (trial and error). Trial and error can waste all of your time and get you nowhere on nontrivial problems.
deleted 7y ago
[deleted]
tacomonstrous 7y ago
I'm a working mathematician and would need a non-trivial amount of thought to get any of these problems. I was also pretty poor at this kind of thing in high school and college, but fortunately the skills involved are almost but not quite disjoint from what is needed to do research on math.
- novok 7y ago
  What are the working skills needed for research in math?
- nopinsight 7y ago
  I know some students who excelled at this kind of problems when they were in elementary school and now possibly on path to be good researchers in math and theoretical CS. One is doing math PhD at Harvard; another received a PhD in theoretical CS from Berkeley with stellar track record (over 20 papers when he graduated after just 4 years in PhD--many/most in top venues, as well as perhaps the most prestigious best paper award for CS Theory PhD student when he was in 2nd year). Not all went on the research path though; some are working in applied software engineering. A few who did math undergrad degrees at places like MIT but I'm not sure where they are going next. Another conspicuous example: Terrence Tao was also the youngest ever International Math Olympiad gold medalist (he won it when he was 12-13!). There appears to be a significant correlation between subsets of people who are good at certain types of math and theoretical CS research and at solving this kind of problems when younger. I read that there are mathematicians who focus on solving problems and those who build frameworks. Perhaps the skills for solving these problems are closer to the former?

esmi 7y ago

Is the answer to the second question allowed to be in engineering form? :) What grade level is question 3 considered?

nopinsight 7y ago
What do you mean by 'engineering form'? Using variables in the answer? It would be suitable for students between grades 4 and 6 (the highest level we currently teach), depending on how talented the kid is. To be clear to other readers, these are hard questions that most people, including older kids and adults, probably can't answer without proper learning and practice.
- esmi 7y ago
  In the US we call this engineering form. https://en.m.wikipedia.org/wiki/Scientific_notation#E-notation https://en.m.wikipedia.org/wiki/Scientific_notation#E-notati... I misread the question though and now realize it’s not needed.
  - monktastic1 7y ago
    I'm also from the US but I've never heard the term "engineering form." FWIW, Wikipedia thinks it's something a little different: https://en.wikipedia.org/wiki/Engineering_notation https://en.wikipedia.org/wiki/Engineering_notation.
  - Armisael16 7y ago
    I've never heard that (or anything else) called 'engineering form'.

Shivetya 7y ago

and here I sit totally flummoxed by the questions yet later on you state these were suitable to grades up to sixth? So for those who have let their math skills lapse or need help to get back in the right frame of mind to learn, where online is the best resource? Khan Academy?

nopinsight 7y ago
To be clear, these are hard contest questions that most people, including older kids and adults, probably can't answer them without proper learning and practice. I have not checked them out in detail, but it appears that https://artofproblemsolving.com/ https://artofproblemsolving.com/ is a good resource. They also have interesting online and physical book series: https://artofproblemsolving.com/store https://artofproblemsolving.com/store (I have no affiliation.)

cecilpl2 7y ago

72? 99999785960? 199?

monktastic1 7y ago
FWIW, I got the same answers. 1. 12 edges * 6 non-corner pieces per edge. 2. There are 111 digits there. You can't keep six nines and have enough digits left. So you keep five (from 9, 19, 29, 39, 49), and now have six digits to keep from 5051525354555657585960. It can't start with 8 (or you still won't have enough digits left), so you choose 785960. 3. First notice that 0+...+10 = 55. To make it end in 2, you have to discard the 3. In general if you have n+...+(n+10) you discard n+3. Now n+...+(n+10) = 5.5*(n+(n+10)) by Gauss' trick, and subtracting (n+3) you have 10n + 52 = 2012. Very much doubt I could have solved them in elementary school(!) though.
- asdffdsa 7y ago
  Could you explain why you can't just do 99999988888? 100 digits in any order, what am I missing?
  - dan-robertson 7y ago
    You’re allowed to delete whichever digits you like. But you aren’t allowed to reorder the digits which are there. I.e. if you had to delete one digit you could go from 189 to 89 but not to 98.

chongli 7y ago

Eleven consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2012, what number did Maria erase This is a very challenging problem. Without a calculator (or computer), it took me a system of an equation with an inequality in 2 variables, using the fact that the solutions are integers, to solve it. I would be very surprised (and excited) to meet an 8th grader (let alone a 6th) who could solve this problem without any external help the way I did. I would not expect an elementary school student to have this kind of algebra knowledge. Most elementary school students around here don’t even know what algebra is.

malaya_zemlya 7y ago
try performing all calculations mod 10.
- chongli 7y ago
  I’m not sure how to make use of your hint. I solved the problem for n, the largest number in the sequence of eleven, and a, the number Maria erases, by the following: n - 11 < a < n + 1 11(n + n - 10)/2 = 2012 + a I’m interested to know what other methods may be used to solve it (apart from brute force).
username90 7y ago
You don't need algebra. You know that 10 numbers are 2012, so on average 201, so the middle of the numbers needs to be around there. After that you just try a few sequences around 201 and you'll find it pretty quickly, most can add 11 numbers pretty fast in 6'th grade. Forcing yourself to only rely on the solution strategies taught by school makes you very bad at problem solving.
- chongli 7y ago
  I specifically set out to find a solution that doesn’t involve any brute force (trial and error). Trial and error can waste all of your time and get you nowhere on nontrivial problems.

deleted 7y ago

[deleted]

tacomonstrous 7y ago

I'm a working mathematician and would need a non-trivial amount of thought to get any of these problems. I was also pretty poor at this kind of thing in high school and college, but fortunately the skills involved are almost but not quite disjoint from what is needed to do research on math.

novok 7y ago
What are the working skills needed for research in math?
nopinsight 7y ago
I know some students who excelled at this kind of problems when they were in elementary school and now possibly on path to be good researchers in math and theoretical CS. One is doing math PhD at Harvard; another received a PhD in theoretical CS from Berkeley with stellar track record (over 20 papers when he graduated after just 4 years in PhD--many/most in top venues, as well as perhaps the most prestigious best paper award for CS Theory PhD student when he was in 2nd year). Not all went on the research path though; some are working in applied software engineering. A few who did math undergrad degrees at places like MIT but I'm not sure where they are going next. Another conspicuous example: Terrence Tao was also the youngest ever International Math Olympiad gold medalist (he won it when he was 12-13!). There appears to be a significant correlation between subsets of people who are good at certain types of math and theoretical CS research and at solving this kind of problems when younger. I read that there are mathematicians who focus on solving problems and those who build frameworks. Perhaps the skills for solving these problems are closer to the former?