the 332-232 problem:

really, really, trivial.  just a repeated application of (a2-b2) = (a-b)(a+b)

from that procedure, however, you get only 3 prime numbers under 100: 5, 13 and 97.  now there's the less trivial part of finding a 4th prime.  unless you count 1 as a prime, which you really shouldn't do.

if i were in a high school math competition, due to time pressure, i would take advantage of the question saying 4 prime numbers under 100.  the easiest thing would be to calculate the next factor, which is 38+28.  this is reasonably fast to calculate even without a calculator, and you get 6561+256=6817.  this number happens to be easily factorizable, by inspection--clearly, 6817=17*401.  (both 68 and 17 are multiples of 17).  and i would move on to the next question, having gotten what the question asks for: 5, 13, 17 and 97.

there might be more sub-100 factors left in the 216+316 part, but in a competition you would just move on.  i verified that this number has factors 3041 and 14177, i.e. no sub-100 primes.  there might be an elegant way of showing that 216+316 does not have a sub-100 prime.

Spot on! This is exactly my same procedure! Congratulations on finding the answer!

Basically you have 3³²-2³², which means (3¹⁶+2¹⁶)(3¹⁶-2¹⁶). Then you grab (3¹⁶-2¹⁶) and factor it until you get the following:

36 medals, 6 days.

i originally wrote down the equations for the series... but noticed that it gets too complex too quickly, with the recursive equations hardly easy to solve. after staring at it for a while, i realized it's best to work backwards--starting from the last day. once you realize the last day has to be on a day that's a multiple of 6 (because of the 1/7 condition), the rest is basically tedious work and showing that this answer is the unique solution.

i've got a few questions interviewers like to ask. (yeah, i'm probably quite a bit older than you guys and am working). these aren't overly complicated, but you're under tremendous pressure to solve it within minutes since, well, there's somebody sitting right across you. here's one that's pretty good. i first heard about it from somebody who interviewed for goldman sachs, but it's very well-known--well-known enough that i was asked the same question by a hedge fund i interviewed with later on.

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you have two identical plates. you're standing next to a 100-floor building and your job is to figure out at what height (i.e. floor) these plates will break by releasing the plates from various floors. what is your optimal strategy? (to be exact, minimize the maximum number of trials, but i guess that was implicitly assumed since in my interview the guy didn't specify.)

so, for example, one strategy could be you do something like a binary search. first, you release the plate from the 50th floor. if it doesn't break, you move up to the 75th floor and try again. now if it does, you know the plates break somewhere between the 50th and 75th floor. of course, you're now only left with 1 plate, so to figure out that exact floor, you'll need to test it floor by floor starting from the 51th to the 74th floor. so if the floor is the 75th floor, you'll need like 2+24=26 tests.
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have fun! 

one word: don't fret if you don't get it within a few minutes.  usually, when the interviewers sees that you're getting stuck, some hints would be coming your way.  (obviously it's better not to receive any hints, but some hints is okay).

I didn't know you could approach infinity from the right.

 You know, it's funny. I had a different answer that I thought was right and hell, it is, if you allow decimals to become the natural number before them (i.e. 7,99 is 7). Redoing the exercise through my very own procedure, I arrive at your same answer, it seems there's something I didn't consider (or considered wrong) last time.

However the other answer is also right if what I said above is true. Funny huh? My answers were 57 and 7. Do the try, it fits extremely well, and arguably it is a better answer since the number of medals given per day doesn't match the number of days except for the last one (and maybe some days, I don't recall).

Also, the equations are:

For the first day: (m+6)/7=k and for the second day: (6m+78)/7=k; where k is the number of medals given per day. If we say k is the same for every day, then (m+6)/7=(6m+78)/7. Solving this gives us the answer m=36 and then k=6, and then I proceed to show that it's right.

Your answer is right, but I don't get your argument for the day being a multiple of 6 if there's the 1/7 condition.

My other answer ended up being the same first day but a different second day, where I messed up the part of what remained) and k would give me a fractioned result. That's when I made the addition of the first day and the second day, to find a new k. I equaled this one with the first day and k surprisingly gave me a 12. Then I got m=57 and n=7. Doing the procedure, I found that in certain divisions for the number 7, I got decimals, and I decided to lower these decimals to the highest natural number before them. In the end, everything fit like a puzzle.

So who's right? I'd say you and my new answer are right because I seriously believe I messed up the equation for the second day in my first procedure. Congratulations!

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For your interview problem, I'm going to say this:

I begin with the 10th floor, if it doesn't break, I go up to the 20th if it doesn't break, the 30th, and so on.
At one of those ten floors it must break, then I just go down to the one I tried before +1 (so for example, if it broke on the 40th floor, I go to the 31st) and start going from there and up.

This would take a maximum of 20 trials, supposing it breaks at the 100th floor for the first plate, then I lower to the 91st and it doesn't break until I reach the 100th floor again.

That's what I'd do anyway. Can you really get a better strategy than that?

EDIT: So I was thinking about this problem some more and I came with similar solutions that also give me a maximum of 20 trials...it's if you jump in intervals of 8, 9, 11, 12 and 13. Any lower or higher than that and the maximum turns into 21.

i didn't check if 57 and 7 would work if you round down any fractional number... but i'd say that if the original problem didn't say you can round up or down, you probably would be better off sticking with natural numbers. especially since there is a solution that works! but 57 & 7 would be a cool extension to the problem.

with regards to your equation, i don't think the problem implied that the same number of medals were given out each day, so i don't think you can quite make that assumption.

here's why the last day has to be a multiple of 6 because of the 1/7 condition--it's really obvious once you see it. since on the previous day 1/7 of the the rest of the medals were given out, that means there are 6/7 of medals left for the next day. now, on the last day, you give away all the remaining medals, which has to be 6/7 of some integer. now you see why that last day has to be a multiple of 6!

once you realize that the number of days n = 6*i where i is some integer, of course you proceed to test if i=1 works. turns out it does, and gives a very nice, symmetric solution (6 medals a day for 6 days), so you can bet your money that it's the right answer. however, you still need to show that it's the only solution. this part isn't that hard although it's a little tricky, but essentially you show that only when i=1 you can ensure you get integer value of medals awarded on every single day.

for the interview problem, yes, you can do better than your solution :) i must admit that if i were in an interview and haven't heard about it before, i probably would not have come up with the right solution. at least not without hints :)

you're getting close with jumping 8, 9, 10, 11 floors. the "official" solution is: first you let go from the 14th floor. then 14+13=27th floor. then 27+12=39th floor. and so on.

so the "official" solution has a better worst-case scenario performance than your solution. however, if we go beyond the "worst-case scenario performance", say best "average performance" (or "amortized" performance), then i have no idea what solution is optimal. i would guess that the official solution would do better, but it's not obvious. maybe there's a theorem that links the two but i'm not aware of it. i doubt the existence of such a theorem anyway (because i know in general worst-case performance says nothing about amortized performance), but it's not my field of expertise so i could easily be wrong.

It doesn't, in fact, in the beginning I considered k to be different each time (k₁, k₂, etc.), where all k are integer numbers, and the only thing that matters is the absolute value (for example, my answer was not 12 in the other try, it was -12; but I only had to consider the absolute value for it to work).

Then I thought that maybe k₁= k₂ is one solution and proceeded to do the exercise that way, and it worked, giving answers 36 and 6.

For when k₁/=/k₂, the answer would be 57 and 7, but of course the value of the 1/7 part of the addition is not an integer all times. Then you have to lower to the highest natural number before it for it to work. If you're not allowed to do that, k won't be an integer, and since it must be an integer,  then the only answer is 36 and 6.

I understand what you mean about 6 being a multiple of the answer, but keep in mind that you're not giving away 1/7 each day, you're giving away 1/7 +1, 1/7+2, 1/7+3.

In any case, we got the answer right congrats!

I'll post a problem later, though it'll probably be a little easier. Well, if you know geometry that is. Umm...this is an embarrasing question...what's the button combination for printing the screen?

Hey guys!

I plan on reviving continuing this topic, starting with some awesome news, I was chosen for the Selection phase of the Ecuadorian Mathematic Olympiads! This means I'm among the 20 best young mathematicians of my country! I feel really great about it, even though I was not in the first 3 places.

Here's the link if you want to check (so you can see I'm not lying haha):  http://www.icm.espol.edu.ec/intercolegial/fase_seleccion_oem.html

I'm Bruno Giuseppe Poggi Cevallos and I study in the Arco Iris school. For the record I'm not from Guayaquil, but due to the fact that I had to go to Guayaquil to do the test, my city says Guayaquil hehe

And to kick up this topic again, here's a problem from the last olympiads. Please do not use calculators on this one.

Consider the algebraic expression:

(1+(2008+Π)⁴+(2009+Π)⁴)  -(2008+Π)²-Π-1
(1+(2008+Π)²+(2009+Π)²)

Determine its value.

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By the way, the part (1+(2008+Π)⁴+(2009+Π)⁴) is dividing (1+(2008+Π)²+(2009+Π)²), and that as a whole is being algebraically added to -(2008+Π)²-Π-1

****

Now if you want to see all the problems of my tests and their solutions, go here:

http://www.icm.espol.edu.ec/intercolegial/resolucion.html

The first link that says "Resolución" is my last test, and this problem, among others, can be found there. Good luck trying to solve them on your own!