Roulette's Frets

Copyright 2003 Sophisticated: The Magazine. All rights reserved.


y son's question, "What sum results from adding the numbers on a Roulette wheel?" reminds me of a question put to the greatest mathematicians of all time, Carl Friedrich Gauss (1777-1855).
As a young child, he was asked by his teacher: "What is the sum of the first 100 numbers?" The teacher's intention, as the story goes, was to keep young Carl busy for a while and out of mischief.  In but a few minutes, however, he announced the solution, "5,050."

Astounded, obviously, his teacher asked, "How did you do that so fast?"

"Oh, I just mentally lined up the numbers in a row and took the first number, 1, and added it to the last number, 100.  Then, I took the second numbers from both ends, 2 and 99, and got 101, again.  It was obvious that those sums would all be the same, so I just multiplied 101 by the total number of pairs, which is 50."

Gauss was probably kidding his teacher.  As we shall see, the sophisticated solver will know how to use "mathematical induction," which is an invention of the late 16th century, 200 years before Gauss.

Meanwhile, back to the question about the Roulette wheel. The solution, of course, is assured by a trivial but lengthy summing up of 36 numbers, a procedure made slightly less tedious with the use of a calculator.

Nota bene, the "0" and "00" add nothing to the sum, only to the frets.
The mythical young Gauss might have lined up the 36 numbers in symmetrical pairs, each summing to 37, then, using an 18th-century calculator, he would merely multiply 37 by the total number of pairs, which is 18, and get the solution straight away...

"Mark of the Beast," some will shudder, probably with an exclamatory punctuation.  "There's no denying Roulette's Frets, just add up those numbers right there on the wheel."

We can stop right here, then.  Naah.

et us instead take the opportunity to explore a powerful invention by a certain 16th-century Italian scientist named Francesco Maurico. The Maurico invention was put to good use in the 17th century by our old friend Pierre de Fermat and also by our new friend Blaise Pascal -- hey, isn't he the guy regarded by some as the inventor of the Roulette wheel?

In the late 19th century, along came Augustus De Morgan -- yeah, that De Morgan, the English mathemetician who, along with George Boole, gave us symbolic logic and stuff, resulting, ultimately, in the computer and the Internet and... Where was I?  Oh right, 300 years after Maurico, De Morgan finally gave his invention its modern name: Mathematical Induction.

Here's how it works.  There are two steps...

Step 1. Find a formula that works for some specific value of an integer.
Step 2. Show that if the formula indeed works for any given value of that integer, then it must work for the next value.
That's it.  That's mathematical induction, albeit expressed informally. You see right away where it leads, don't you. Inevitably, integer after integer, you have proven that one formula fits all sizes, so to speak.

Let us try our hand with mathematical induction by first applying the Gauss solution to the Roulette wheel puzzle, putting it this way...

666 = (36 / 2)(36 + 1)
Then, doing what comes natural to a child (generalizing from the particular), we can make self-evident substitutions...
S(n) = (n / 2)(n + 1)
...which you see rearranged here to fit a conventional formula for the sum of the first n integers...
S(n) = n(n + 1)/2.
t is difficult to overstate the importance of that formula, for we have all benefitted mightily from it throughout the 20th and into the 21st century, in a galaxy of practical applications, such as for planning communications networks and transportation systems and... Sorry, I have gotten away from the subject of mathematical induction.
That formula S(n) = n(n + 1)/2 seems to work all right for n = 36, but can it be proved to work for any value of n?
Step 1 is done already.  Step 2 starts with substituting n + 1 "the next value" for n.
Nota bene, nobody is saying that n = n + 1.
We have then, that...
S(n+1) = (n + 1)(n + 1 + 1)/2
...which can be rearranged algebraically to read...
S(n+1) = [n(n + 1) + 2(n + 1)]/2
S(n+1) = n(n + 1)/2 + (n + 1)
Look what we have here: That lefthand side can be obtained independently by taking the formula for S(n), the sum of the first n integers, and merely adding on the (n + 1)st integer and writing...
S(n+1) = S(n) + (n+1)
...which indeed comports with the results of our substitution, bringing us right back where we hoped to be, QED...
S(n) + (n + 1) = n(n + 1)/2 + (n + 1),
S(n) = n(n + 1)/2.
Of course, we did not need to start with n = 36; however, inspired by my son's question, we sure did.  Accordingly, our proof by mathematical induction does not apply to any values of n less than 36.  Better to have started with n = 1, so you will have full coverage of all integers.

ave you noticed that the sum increases according as the square of the value of n?  Double n and you quadruple the sum.  Not quite, but almost.  With 72 frets, a super-Roulette wheel would have -- here, I'll do the math for you -- 2,628 as its sum.  No more Mark of the Beast, but nobody in Las Vegas will have any reason to care about that, I suppose.

On the other hand, in the late 20th century, the management of a huge enterprise faced a strategic decision that called for an understanding of what might be called the "square-law of benefit" enjoyed by communications networks.  For networks, by the way, it is necessary to limit potential routings by excluding connectivity of a node to itself, which explains the formula shown in "Track Record."

Whereas the product of the first n integers has a name ("factorial"), the sum of the first n integers does not.  One coinage has been proposed in Puzzles with a Purpose (see Factorial Factoids), "totorial," with the notation & being analogous to the ! used in factorials. 

Thus, n& = S(n) and is pronounced "n totorial."  Cool, huh?

or our purposes here, let us conclude with a reversal in the formula.  Suppose we know the sum, S(n), and want to find out the value of n.  By rearranging and algebraically manipulating our formula, we have...

2S(n) = n(n+1),
2S(n) = n2 + n,
n2 + n - 2S(n) = 0
...which can be solved by the quadratic formula so that...
n = {[8S(n) + 1]1/2 - 1} / 2
You are invited to create a puzzle that applies this formula and submit it to Puzzles with a Purpose.

Meanwhile, that 8 there in the radicand reminds me of the 5 that shows up in the formula first credited to Robert Simson at the University of Glasgow...

= (51/2 - 1) / 2
...and makes me fret about what number that little white ball will land on next.

Epilog: Totorials at Work

Another problem sent to me by my son reads...

The sum of 3 consecutive integers is 417.  What are they?
...which can be generalized to read...
The sum of k consecutive integers is T(k).  What are they?
The derivation ought to be a slam dunk using totorials, for we see that for some unknown value of n...
T(k) = (n + k)& - n& [see definition of "&" above]
T(k) = S(n + k) - S(n)
T(k) = (n + k)(n + k + 1)/2 - n(n + 1)/2
2T(k) = n2 + 2nk + n + k2 + k - n2 - n
n = T/k - (k + 1)/2
For the specific problem...
T(3) = 417
n = 417/3 - (3 + 1)/2 = 137.
So the three consecutive integers are 138 + 139 + 140 = 417
What happens if T(k) is not divisible by k?
What happens if k is an even number?

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