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Wednesday, 23 January 2008

AN INTERESTING MATHEMATICS PUZZLE

Well, this is the first time I've tried to put any maths into my blog, and I can tell you for sure I'll have to learn a lot more about how to do all the typesetting and stuff before trying again!!! Hats off to the clever folks who do it all the time.

I wanted to share a puzzle that I tackled some while ago now, when I was doing my first undergraduate year in mathematics. It was satisfying in one way, but remained always rather unsatisfying in another.

The course in question was a rather gentle introduction to some topics in number theory, and the lecturer wanted to encourage us to experiment and learn for ourselves a bit, so as well as the compulsory assignments we did each week, he set a weekly "challenge problem", which was optional. Whereas the assignment tested that we had understood the course and could apply what we had learned, the "challenge" was intended to go some way beyond the course itself. The lecturer gave the added incentive that he would keep a record of the challenge problem solutions he received, and at the end of the semester there would be a prize for the best student.

Initially, about eight or so students submitted solutions, but it tailed off quickly, and by about half way through the semester, the only students still submitting solutions were myself and another student. This carried on until the final lecture, when we both waited expectantly for the winner to be announced. Instead, however, he said that he was still unable to separate us, and so he was launching a final challenge for everyone to think about over the Christmas break. He said he would wipe the slate clean, and, regardless of how many, if any at all, of the previous challenges had been completed, the person who delivered the correct answer, with proof, soonest after Christmas would win the challenge prize. Should he receive more than one correct answer, the prize would go to the student whose solution was the most elegant.

What he then handed out read as follows:

CHRISTMAS CHALLENGE

The following Christmas teaser was set by Dr J Brownowski in the New Statesman and Nation, 24th December, 1949:

Find the smallest natural number which is such that if the digit on the extreme left is transferred to the extreme right, the new number formed is one and a half times the original number


He also gave a hint, without which I am sure that I would not have been able to solve it. I had a solution by the soonest date, and so did two others. One turned out to be wrong. The other was also correct, and had been submitted by the same student with whom I had been neck-and-neck all along. Since the lecturer still could not separate us, he finally announced us as joint winners.

I said at the beginning that the process was satisfying, and this was partly of course the pleasure of winning, but also because I had in doing so undertaken some puzzles, including this one, that I would never have thought I could solve previously. The dissatifaction, though, was entirely with my solution to this last problem. My initial approach, using the hint, was fine, and led to a most general solution. However, the second half, which was in order to find the lowest possible solution has always seemed inelegant, and I've wondered whether there was a better way of finding the answer. I hope that I've eluded to the part I mean enough, since my own solution, in the graphic is, I hope, too small to read, and those who want to try it themselves can avoid enlarging it.

The hint, was, in ROT13,
vasvavgr qrpvznyf


2 comments:

Jean-Noël said...

I actually arrived at the same answer without using the hint (I found the hint rather confusing to be honest).

I just wrote the number as n = a*10^m+b so that n' = 10*b+a and using that 2n' = 3n I also got 17b = (3*10^m-2)a. As a < 10 it follows that 17 | (3*10^m-2) which leads to the solution.

Stefan said...

What happens if you change the digits the other way around? Put the most right digit to the left and find a number that gets 1,5 times as big as before.
It will take you much less time to find it!
My solution starts with 28...

PS: Jean-Noels solution is very nice!