The "powers" that be...

Is there a power of 2 that contains every digit from 0-9 the same number of times?

Another way of saying it is....

You have an integer in which every digit from 0-9 appears, and each digit appears exactly the same number of times. Is this a power of 2?

This was a question I was given in Number Theory yesterday, and even though it's mildly trivial, for whatever reason, I really enjoyed it. The answer is no... but I'm not going to explain why until later.

I found out yesterday that next year at Gustavus I get to live with one of my very best friends in the newest, nicest dorm on campus. She was given the position of CF (the same thing as an "RA") in the dormitory called Southwest, and she got to choose one person to live with. We each get our own room, but we share a living area. I'm very proud of her for being given this job, because everyone wants to be the CF in Southwest!

In a couple of weeks, my mom and my sister are coming to visit me. They have an itinerary of three cities beginning in Prague. A neat feature of the BSM program is that they incorporate two four-day weekends into the schedule of classes. Unfortunately this means having class on two Saturdays to make up for it, but it is nevertheless really nice. During the first of these four day weekends, my mom and sister are arriving in Prague, and so therefore, I am able to come and meet them! Not only will it give me an opportunity to visit the Czech Republic, but I also get to spend more time with my family! A couple of days later, I will ride a night train back to Budapest with them, and then be able to show them around this fantastic city for about three days. They go on to Vienna, but I can't go with them since I will have class. I'm really excited!!

Either this weekend or next weekend I am going to go to Krakow. Apparently it's just gorgeous. Some students who were here last semester said that it was their favorite destination. I have a sneaking suspicion it will end up being next weekend because more people will be able to go with us. Also, this weekend there is a very reknowned scholar coming to speak to the mathematics students, and we all want to attend his presentation.

One of my favorite places to do homework is a shop called Kom Kom Czokolad. It's a Dutch chocolate shop where they sell truffles and hot chocolate. This hot chocolate is not the awful, watery American stuff. It is literally melted down chocolate. They give you a spoon to help you eat it since it is so thick. They also give you a small glass of water since it is so rich. It is absolutely beyond belief. However, they have huge tables and multiple levels, so it's ideal for spreading out and doing math either alone or with a group.

Now, here way to solve the problem I posed earlier. Take an integer with digits 0-9. It doesn't matter what order they appear because the trick here is to add them up. So you have 1+2+3+...+9+0. This equals 45. If each digit appears more than once, you know that each digit appears exactly the same number of times. Therefore, this summation of the digits can be simply multiplied by a constant, giving k(1+2+3+...+9+0). Regardless of what k is, we know that the number must be divisible by 3, since the sum of the digits is 45, and 45 is divisible by 3. However, by the fundamental theorem of arithmetic, 2^n will never be divisible by three. Basically, this is merely stating that 2 times 2 times 2 times 2.... etc will never be divisible by three since each positive integer has a unique prime factorization, and a power of 2 will therefore never contain a 3 within it's factorization. Thus, we have proved that an integer containing all digits the same number of times will never be a power of two. Like I said earlier, this is pretty trivial.... but I like it!

Happy (Ash) Wednesday!