I bailed out.
Spoiler (it’s B5):
http://www.math.niu.edu/~rusin/problems-math/putnam.02
I can understand it, but it would have taken me a while to figure it out… I was heading down that road but I think I would’ve moved on to a different problem!
The problem statement implicitly ruled out a leading 0, by stating that 200 (00200) is not a palindrome in base 10.
I spent a few minutes last night but only guessed that 2002[sup]2[/sup] is a lower bound for the minimum base [(2002b)[sup]2[/sup] + 0 + 1 = (b[sup]3[/sup]-1) + (b[sup]2[/sup] -1 ) + (b-1)], but that doesn’t help solve the problem. Maybe I’ll just click the spoiler link for the answer.
Ok, back to the original problem, what method did you (who solved it) use to do it? I got the right answer, after starting over twice because of botched signs and what not.
The method I used was first to figure out the diagonal cross section of a cube - what it looks like viewing it from a corner:
It’s a rectangle with height s, and width s(sqrt(2)), where s is the length of a side, and the origin is at the midpoint of the bottom of the rectangle.
Once that was done, I just solved a pair of simultaneous equations:
The slope of a side of the cross-sectional triangle is -3, so its equation (assuming it straddles the origin and sits on the x-axis) is y=-3x+3. The line that passes through the origin and the Quadrant I corner of all similar rectangles of the proportion above is y=x(sqrt(2))
After that, it’s just plain old algebra.
So what other methods did you guys use?
rassa-frassa-fricken-fracken…
Sorry. That post was mine, not Jerseydiamond’s.
The Putnam is given on a Saturday, in two parts, each three hours long. Six problems in the morning, six in the afternoon, each problem worth ten points, partial credit is given.
I think the median score is usually zero, idnit?
Is it me, or are people be quite rude to me?
Anyway,
Quite simple. 1x1 is obviously gives an ever-increasing series of numbers that increases with the number of factors in N.
First, realize that 1x1 can be written N = bb + ab + 1. I prefer to use round numbers, so I make M = N-1 and say M = bb + ab.
Now, re-write it to say M/b = b + a.
So, if M is made up of the factors 2,3,4,5 then b can some factor, or combination of factors that make the remaining factors less than b (since a < b, as described above). For the above example, b can be equal to 10, allowing a to be 2. b can also be 8 allowing a to be 7. If you keep on increasing factors, then you will keep on having more variations of a and b.
Now, after that, I think I deserve some civility. Especially since the original problem statement didn’t mention that 0x0 was not allowed.
What’s this palindrome stuff? Did somebody hijack my thread?
RM Mentock Time magazine says the average score is 1.
This is not a contradiction. The statement that the median score is 0 just means that at least half of the scores were zero, without saying anything about what any of the other scores were. The average, however, does use information from all of the scores in it. So a few people who do really well would improve the average, but not the median.
As another example of this distinction, the average number of posts of an SDMB member is around 300 (last I calculated it), but the median number is 1, and the mode (the number common to the most people) is 0.
If you were offended by what I said, I didn’t intend to be rude; I was only going off iamthewalrus(:='s statement of the problem, as I’ve never heard of the Putnam until this thread. Unless you were taking issue with the implicit restriction on the leading 0. Anyway, I wasn’t even addressing you specifically, I was just trying to speak to that point.
(And also, when someone asks you to explain a problem here, they often mean exactly what they say. So I would take KarmaComa’s request at face value, which you fulfilled rather well.)