interesting math problem

[gumpzilla]

I realize that these math problems aren't everybody's cup of tea, but I think there are a few OOTiots who might enjoy this one. I grabbed it from the USSR Olympiad Problem book.

Suppose you have thirteen gears, each of which has integer weight. You know that these gears have the property that if you choose any twelve, you can separate that group of twelve into two groups of six which will be of equal weight. Prove that all of the gears must have the same weight.

Feel free to post reasoning as to how to go about this, and if you have seen it before, hold back for a bit, I guess. (I haven't looked up the solution yet, nor fleshed out my argument totally rigorously, but I'm pretty sure that I now understand one mechanism for proving this)

[gumpzilla]

In retrospect, this post may be a little intimidating to the OOT masses, so allow me to comfort you with the soothing balm of familiarity.

Stick it in her pooper.

[Jazza]

cool problem, i appreciate posts like these

the first thing you have to realize is that either all the numbers are odd, or all the numbers are even, if this were not the case you could pick 12 numbers whose total sum is odd, and therefore when you split them into two equal groups, one odd, one even, and therefor not equal

if they are all even, you could divide them all by 2 and it shouldn't change whether or not all the numbers worked to begin with

if they are all odd, you could add 1 to all of the numbers, and then divide by two, and it should not change whether or not the numbers work

if at any time you end up with a mixture of odd and even numbers you know these numbers and hence the original numbers will not work

now you could repeat both of these processes, and as long as all of the numbers were finite to begin with, you will figure out the numbers don't work, or you will evenutally end up with all 1's

and the only way to end up with all 1's is to have all the numbers the same to start with

[bighomage]

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I grabbed it from the USSR Olympiad Problem book.

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Where do you get this stuff? I used to be a stud on math contests but I've kind of fallen off and it would be cool to look at some old problems.

[chaas4747]

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[ QUOTE ]
I grabbed it from the USSR Olympiad Problem book.

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Where do you get this stuff? I used to be a stud on math contests but I've kind of fallen off and it would be cool to look at some old problems.

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I don't think you can use the word stud to describe someone in a math contest.

[bighomage]

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I don't think you can use the word stud to describe someone in a math contest.

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Ya, you can. Try it. Pussy comes flocking like you were an NBA player.

[gumpzilla]

The USSR Olympiad Problem Book is published by Dover Publications. I love Dover. I'm sure you can get it off Amazon. It has solutions to every problem in the book, as well. If you don't want to spend $15 or so on that, try Googling for USAMO, Soviet Olympiad or Putnam problems. Putnam problems are generally beastly, so they wouldn't be my starting point.

[GFunk911]

they need to be odd or even, once u solve odd you +1 and solve for even. let's say they are even.

They all need to be a multiple of A apart, where A is some constant. A must be even for all numbers to be even or all odd. If they are not multiples of 2 apart, they can't all be odd or all be even, so they must be at least multiples of 2 apart.

Any 12-sum will have a sum of 12C + xA, where C is the smallest gear, A is the above constant and x is some number. In addition, each gear has the weight C + xA. The half-sum will be 6C + (x * .5A). The C is irrelevant. You can't make a sum of 6C + .5Ax with weights of C + Ax, unless A=0.

[gumpzilla]

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Any 12-sum will have a sum of 12C + xA, where C is the smallest gear, A is the above constant and x is some number. In addition, each gear has the weight C + xA.

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You're using the same x for both the sum and for an individual gear, as far as I can tell, which is problematic.

[GFunk911]

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Any 12-sum will have a sum of 12C + xA, where C is the smallest gear, A is the above constant and x is some number. In addition, each gear has the weight C + xA.

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You're using the same x for both the sum and for an individual gear, as far as I can tell, which is problematic.

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Sorry, they are different x's. Probably should have used x and y to signify.

Addendum: I didn't prove that oyu can't do it with different weights. Here's more

Say you have 12 of the weights in groups of 6 so they equal. The 13th weight is A more than a weight in the 12. You swap them. You not cannot make the 2 6-groups equal, since this would require moving .5A from the swapped group to the unswapped group, and all your weights are multiples of A. The weight you swap in needs to be 2Ax more (or less) than the weight swapped out, meaning that your weights need to be in multiples of 2A.

Repeat problem with 2A in space of A, 4A in space of 2A, etc.

Better?

[gumpzilla]

Much.

[GFunk911]

i don't need to conform to your standards, you fascist opressor.

i mean ... awesome.