Comment by nopinsight | Hacker News Reader

Comment by nopinsight | Hacker News Reader

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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 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-a19... "

Source: the Primary Math World Contest (usually held in Hong Kong) https://www.txstate.edu/mathworks/PMWC/previous-pmwc-tests.h...

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.

What are the working skills needed for research in math?

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?

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?

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/ is a good resource. They also have interesting online and physical book series: https://artofproblemsolving.com/store (I have no affiliation.)

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.

try performing all calculations mod 10.

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).

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.

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.

OTOH, puzzling something out through trial and error fosters curiosity and teaches a "fail fast" mindset. For a sixth grader, this "play" may be a more beneficial lesson than only being able to tackle problems for which you already know the optimal strategy (in this case, algebra)

I disagree. The hard problems in math, science, and engineering are not amenable to a "fail fast" mindset. They take years, decades, or sometimes centuries of study to solve.

This sort of puzzling through play is good for solving math contest problems that have been specifically designed for the purpose, and little else.

Simple examples may be useful. Exploring them, proving stuff about them it is how one may carve a foothold to solve a harder problem.

Example: The Josephus Problem - Numberphile https://youtube.com/watch?v=uCsD3ZGzMgE&t=86

You wondered how a sixth grader could solve it, I showed you how. Being able to do it using formal methods is good, but my point is that sixth graders are still pretty smart and can use more creative methods which still works very well as I showed.

Also ny solution wasn't brute force or trial and error, there are very few possible sequences of 11 numbers around 201, so you are guaranteed to find the solution quickly.

I'm a pure math student and when I first found the problem I sent it to my friend who is a stats student. She solved it very quickly, but admitted to using a calculator. None of my exams permit calculators, so I've gotten used to never using them.

Perhaps the elementary school math contest permits the use of a calculator? That would make it very quick for sixth graders to solve it using your method.

You could do it without a calculator like this:

Assume the middle of the sequence is 201 since 2012 / 10 ~= 201. Then the sum of the 11 numbers between 195 and 206 would be 11 * 201 = 2211. Subtract the sum of the values we didn't remove to get 2211 - 2012 = 199. 199 is between 195 and 206 proving that it is a valid answer.

Edit: Also I did my masters in pure math, doesn't mean that I cannot use common logic without the formalism.

199 is between 195 and 206 proving that it is a valid answer.

Try the same trick with this modified problem:

Fifty consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is 2009, what number did Maria erase?

You'll find that it does not work. The trick you did relies on the specific problem as stated. It is not true in general.

2009 / 49 = 41. 41 * 50 = 2050.

2050 is not the sum of the numbers 17..66, 2075 is, and subtracting 2009 from 2050 just gets you 41 again, which is not the correct answer.

I solved it like this: The numbers are x, x+1..., x+10 = 11x + 55. One is erased therefore 10x + 55 - y = 2012. 10x - y = 1957. We know X is whole so we need a number between 1957 and 1967 that divides 10 and that is X. 1960 so X = 196 and therefore y = 3. Answer is X+Y = 199.

I think a school student can do this math manually. The hardest part is the sum of the first N(in this case 10) numbers and then finding the divisor. Yes there might be a better solution.

But I think a kid would think up a brute force algorithm like this and it is easier for small N. The computation is only in O(N) complexity haha. And if an answer exists it will always be found - so it is a real solution. Might be the first algorithm you write if you are tasked to do this with software.

Due to a parity issue (50 here is even) we need to adjust username90's solution. In particular, a sequence of consecutive integers of even length has a half-integer in its middle.

In this case, let's consider the sequence of 50 consecutive integers 16, 17, ..., 40, 41, ..., 65 where 40.5 is in the middle and the sum is 40.5 * 50 = 2025. Subtracting 2009 we get a supposed erased value of 2025 - 2009 = 16 and this indeed correspond to a valid solution:

16, 17, ..., 65 => 17, ..., 65

We also could've considered the sequence 17, ..., 41, 42, ..., 65, 66 centered at 41.5 and this would've led to an erased value of 41.5 * 50 - 2009 = 66 which is another solution:

17, ..., 65, 66 => 17, ..., 65

Right, but we are talking about sixth-graders here. If a sixth-grader saw username90's solution and tried to apply it to the adjusted problem, they would not know that the result is incorrect without actually summing up all the numbers to check, an undesirable chore on a math contest with other problems to worry about.

The act of figuring out how to solve the problem is itself an act of trial and error. And in many problems, the clever part is figuring out how to target your trial and error.

For instance, if you want to find the global minimum of some function on the integers, you might realize you can find the roots of the derivative and test only the neighboring integers. If you want to divide 144359305 by 241, you're going to start by guessing how many times 241 goes into 1443, and then multiply to check whether that's right. If you want to compute floor(29489238428^(1/3)) by hand, you're better off smartly guessing and checking than trying to figure out a digit-by-digit cube root algorithm on the spot.

The eleven numbers are n,n+1,..., n+10. Their sum is 11n + 55. Let the number removed be n + k, then the sum of the remaining is just 10n + 55 -k = 2012. Or 10n-k = 1957. Then it is obvious that k=3 so that n is an integer. Hence, n=196

The eleven numbers are 196,197,..., 206, out of which 199 was removed.

That’s a very good solution. Thanks.

I began with n being the largest of the sum so that 11n - 55 = 2012 + k. I solved for k in terms of n and then used an inequality to find n which gave me k.

Now I see that the method I used relies on math beyond elementary school, which makes the problem a bit more complicated than it ought to be. On the other hand, the elementary school only method relies on the fact that you have 10 numbers remaining and so you can easily see the remainder. It wouldn’t generalize as well, especially in cases where there may be multiple solutions and you’re asked to find all of them.

Given 11 numbers: n, n+1,...,n+10 the sum of 10 of them:

    10*n+sum_conseq=2012

where max(sum_conseq)=0+1+...+10=(0+10)+(1+9)+(2+8)..=55

therefore erased number is n+3 because (2012-55-3) divisible by 10 -> n=196.

Is the answer to the second question allowed to be in engineering form? :)

What grade level is question 3 considered?

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.

In the US we call this engineering form.

https://en.m.wikipedia.org/wiki/Scientific_notation#E-notati...

I misread the question though and now realize it’s not needed.

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.

I've never heard that (or anything else) called 'engineering form'.

72?

99999785960?

199?

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.

Could you explain why you can't just do 99999988888? 100 digits in any order, what am I missing?

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.

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