The only decent CHMMC problem I saw this year

by v_Enhance, Nov 18, 2012, 7:45 PM

Compute the sum of the quadratic residues modulo 4096.

Palmer you should un-graduate

This post has been edited 1 time. Last edited by v_Enhance, Nov 18, 2012, 7:45 PM

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hmph that's not a good problem imo because it is trivial by slightly (but not very) obscure results

EDIT : The the quadratic residues modulo $2^n$ are the numbers of the form $4^a(8k+1)$

And hmph I did see some of the other problems such as $3^x + 4^x = 5^x$

This post has been edited 2 times. Last edited by dinoboy, Nov 18, 2012, 8:01 PM

by dinoboy, Nov 18, 2012, 7:56 PM

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Define "slightly obscure".

(I'm not sure what you're referring to.)

Also if you saw the rest of the problems...

by v_Enhance, Nov 18, 2012, 7:58 PM

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Oh darn, I feel like I should probably know that.

Also the individuals test was all algebra

by v_Enhance, Nov 18, 2012, 8:05 PM

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Hm, I thought tiebreaker #3 was pretty nice (although pretty easy, and a bit annoying to compute):

Let $\sigma(n)$ denote the sum of the positive factors of $n$ . For example, $\sigma(10) = 1 + 2 + 5 + 10 = 18$ . For how many $n \in \{1, 2, 3, \dots, 100\}$ is $\sigma(n) < n + \sqrt{n}$ ?

by AIME15, Nov 18, 2012, 8:08 PM

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Hmm so apparently I-15 and T-10 were also nice (I didn't look at them whoops).

darn I wish I had a copy of the team test now.

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by v_Enhance, Nov 18, 2012, 8:10 PM

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wait that's pretty silly $n$ always has a factor bigger than its >= square root unless its prime and its well known there are 25 primes under 100 (darn mathcounts memorization of silly stuff)

edit : darn

This post has been edited 1 time. Last edited by dinoboy, Nov 18, 2012, 8:48 PM

by dinoboy, Nov 18, 2012, 8:26 PM

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by v_Enhance, Nov 18, 2012, 8:37 PM

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Hm I think T10 went something like this:

(random variables chosen)
Let $P=\binom{2^{2012}}{1} * \binom{2^{2012}}{2}....\binom{2^{2012}}{2^{2012}}$ . Let $M$ be the number of zeroes $P$ has when written in binary. Let $N$ be the number of zeroes $M$ has when written in binary. Compute $N$ .

and why was the individual test all algebra

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by giratina150, Nov 18, 2012, 8:39 PM

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I think that tiebreaker #4 was pretty good, although it used similar logic to Individual #14.

by r31415, Nov 19, 2012, 1:34 AM

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@dinoboy: I never said the problem was hard. I said it's relatively nice.

by AIME15, Nov 19, 2012, 3:30 AM

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By "silly", I meant that it wasn't a good problem.

This post has been edited 1 time. Last edited by dinoboy, Nov 19, 2012, 5:52 AM

by dinoboy, Nov 19, 2012, 5:52 AM

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Define it.

I "should have" got a 15. But I misbubbled a lot.
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I got a 6 too. and an index of 190.5

I should've gotten a 10.... but no one likes me.
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The only decent CHMMC problem I saw this year

by v_Enhance, Nov 18, 2012, 7:45 PM

11 Comments