TWO NUMBER MACHINES
1. Leonard’s Number Machine
The 1980’s were bad times for arts education in New York City’s public schools. My career as a music teacher was disappearing. I was working at Taft H.S. and each year, I taught fewer and fewer music classes and more classes in other subjects. To keep myself employed, I took advantage of a program that offered free courses for teachers leading to recertification in mathematics. One of those courses was a calculus class at LaGuardia Community College. I only remember the professor’s first name – Leonard.
Leonard introduced the class to a little mathematical solitaire game to illustrate what the nature of mathematics is. The game begins with the selection by the player of any positive integer. If it’s odd, multiply it by 3 and add 1. If it’s even, divide by 2 as many times as necessary to obtain an odd number, and then repeat the procedure. The odd numbers you land on will sometimes be bigger than the previous odd number and sometimes smaller. For example the odd number following 7 in this game is 11. [(7)(3) + 1] / 2 = 11 The odd number following 17 is 13. [(17)(3) + 1] / 2 = 26 / 2 = 13.
Leonard stated that regardless of what number we started with, the game would eventually end with and endless loop of 1 – 4 – 2 – 1 – 4 – 2 – 1 …….. He said that a computer could test billions of numbers and no counterexample would be found. However that is insufficient to a mathematician. He challenged us to create a proof that all starting numbers eventually lead to 1.
I never met Leonard’s challenge, but I gave it a little bit of thought. To prove it, we must show that no odd number other than 1 will generate a series of numbers that loops back on itself. If we show that a repitition of an odd number other than 1 can never occur, then, although it may take a very long time, it is inevitable that you will get to an odd number that, when you triple it and add 1, will land you on a power of 2. Some examples of odd numbers that lead to a power of 2 are 5, 21, 85, 341……
2. Joey’s Number Machine
Years later, I made up a number solitaire game of my own. My game is played exclusively with odd numbers. As in Leonard’s game the player can choose any odd number to start the game. Each number will be divided by 3 and by 5 as many times as possible without leaving a remainder. If the number in play is NOT divisible by 3 or 5, then multiply it by 4 and add 1. If it still is not divisible by 3 or 5, then one more time you multiply by 4 and add 1. After that you WILL be able to divide by 3 at least once.
I observed interesting results in this game. The inevitable number that loops back on itself is NOT 1. It is 13, and no matter where you start you will always arrive at 13. It may take a long time.
Some examples:
Start with 1. Then you get the following series: 1, 5, 21, 7, 29, 117, 39, 13
Now let’s start with 13.
13, 53, 213, 71, 285, 19, 77, 103, 551, 2205, 49, 263, 1053, 117, 39, 13
The most remarkable result I observed was a very long series that is generated when you use 37 as your starting number. After a long time, the numbers grew in size until some numbers in the series reached more than a billion. That's remarkable, considering the small size of the number we started with.
Then there was a precipitous collapse and it crashed all the way down to 1, which, as we have seen, will soon lead us to the inevitable 13.
Here are the numbers in the crash landing of the series that began with 37.
1,165,188,743
1,553,584,991 (the maximum number in the series that is not divisible by 3 or 5)
414,289,331
22,095,431
1,178,423
1,571,231
9,311
2,483
3,311
883
4,711
3,769
6,701
1,787
2,383
4,237 (multiply by 4 and add 1 twice and you get a number divisible by
3 to the 7th power, crashing you all the way down to 31)
31 (becomes 125 or 5 to the 3rd power)
1
The 37 series is a model of some great civilizations in history. They flourish and grow for a long time and then crash rapidly into extinction.
Tuesday, January 31, 2012
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In order to play Joey's Number Solitaire game, you have to know the test for divisibility by 3.
ReplyDeleteConsider 1,553,584,991
step 1. eliminate all digits that are 0, 3, 6, or 9.
1555841
step 2. eliminate all groups of digits that add up to a multiple of 3. In this case, you can eliminate the three 5's because they add up to 15 and also eliminate the 8 and the 4 because they add up to 12.
step 3. Whatever is left tells you what the remainder will be when you divide the original number by 3. In this case, you're left with two 1's so if you divide the original number by 3 the remainder would be 2.
step 4. A remainder of 2 means you only need to multiply by 4 and add 1 once and then you'll be able to divide by 3 and repeat this procedure with the quotient. If the reaminder is 1, you'll have to multiply by 4 and add 1 twice before you can multiply by 3.
correction: the end of my comment should read "divide by 3"
ReplyDeletethis blog was really great, never seen a great blog like this before. i think im gonna share this to my friends.. solitaire card games
ReplyDelete