No examples needed. 2C - 10% of 2C + 32 = 0.9 x 2C + 32 = 1.8C + 32 = (9/5)C + 32.
It also works for 5, 12, 13; but it’s clumsier.
Not a formula, but I found out Poncelet’s Porism on my own. Although I must admit that I only found it out for polygons inside circles. I didn’t extend it to all conic sections, exterior as well as interior.
I discovered the “shuffle idempotent” that could, in principle, tell you without fail whether a given permutation is a shuffle. The only trouble is that applied to a deck of 52 cards it would have 52! (that’s 52 factorial) terms, a bit unfeasible. But it has theoretical interest. It took only 6 years of coming back to the question from time to time and gnawing away at it to do this.
I figured out that x[sup]n[/sup] - x is exactly divisible by n if n is prime, and I quite like my proof of it, but I don’t know whether this is at all noteworthy or just really trivial. 
In undergrad physics, someone asked if formulas would be provided on the midterm (some professors did). He told us not to bother with memorization. When he passed out the exam and there was no formula page, people began raising their hands, thinking their test was missing a page.
The professor turned to the blackboard and wrote in giant letters F=ma. “From this you can derive anything else you need”. I loved the class and got an A, but there were many angry people.
- That the ‘9’ in any base (that is, the digit that is the largest single digit number, one less than the base itself) has the rule that, given a number, the ‘9’ divides that number if and only if it divides the sum of the digits of that number.
Most people know that in base 10, you can add the digits of a number to see if it’s divisible by 9. It also works for 7 in base 8 (octal), for 15 in base 16 (hex), etc.
It also appears to be true of the divisors of the ‘9’ - 3 and 5, divisors of 15, follow the rule in base 16. 3 follows the rule in our familiar number system. I’ve never bothered to figure out why that is, though.
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A formula for generating Pythagorean triples.
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That is a an element of a group is a generator of that group, then its inverse is also a generator of the group (fairly obvious, but it also leads to the conclusion that groups have even numbers of generators except for the primitive case of a one-element group).
Oh, yeah, that. It’s a good one to show a middle-school kid, and becomes easy to understand if one draws the squares as squares:
o
oo
oo
ooo
ooo
ooo
oooo
oooo
oooo
oooo
…
Write an n-digit number as ABCD…N and it equals 10[sup]n-1[/sup]A + 10[sup]n-2[/sup]B + 10[sup]n-3[/sup]C + 10[sup]n-4[/sup]D + … N. That’s equal to A times a string of nines, plus B times a string of nines, plus… plus A+B+C+D+…N. Plainly the whole of the “string of nines” part is divisible by 9 (and hence 3) no matter the specific A, B, C etc. Hence if the other bit, A+B+C+D+…N, is divisible by 3 then the whole shooting match is, and more rigorously if it’s divisible by 9. Of course, if N is even then the number is divisible by 6. Generalise as you see fit for any number base, an application of the above logic shows that 3 and 5 must indeed work for base 16 and a whole bunch of numbers would work in base 61. 
This is Fermat’s little theorem.
In that case, I feel quite proud for figuring out both this and the binomial proof of it completely unaided. 
footnote: I don’t use modular arithmetic in the proof, but it’s a proof by induction using the given lemma about binomials.
Wait. I’m almost there.
A[sup]n-1[/sup] is equal to A00000000…000, which is equal to 999999999..999 + 1, though - not 99999999…999 + A.
So doesn’t “That’s equal to A times a string of nines, plus B times a string of nines, plus… plus A+B+C+D+…N.” become
That’s equal to A times a string of nines, plus B times a string of nines, plus… plus 1+1+1+1+1…+1"?
Never mind - I got it.
a[sup]n-1[/sup] is equal to (as you said) Ax999999999… + A.
Then everything falls into place. I never saw that because I did a sort of proof (to myself) by induction, working from the ones digit left. The ‘9’ part became obvious, but not the divisor part.
Thanks.
I worked out a proof of this pattern, but this post is too small to contain it.
No, actually I’ve forgotten some of the details and would have to go over it again before I can offer it here. I do remember that one part of it is to show that x[sup]n[/sup] -1 is divisible by x -1 for any integral power of n.
That’s easy: x[sup]n[/sup] - 1 = (x - 1)(1 + x + … + x[sup]n - 1[/sup]).
Once in high school I was thinking about hypercubes. I don’t remember why; maybe I had run across Flatland in the library or something. (I’ve never read it though.) I was wondering how they they know how many faces a tesseract has, or why a cube has 12 edges, or something. I realized that if you took a line segment, a square, and a cube, all with edges 1 unit long and put them on a graph, with one vertex at 0,0,0, then it was all just a counting problem, counting 1’s and 0’s. I figured out the formula in this section: Hypercube - Wikipedia of the Wikipedia articale on hypercubes.
I found it easier to verify it for small values of n, and then recognize that
(x[sup]n+1[/sup] - 1) can be rewritten as x(x[sup]n[/sup] - 1) + (x - 1)
so the divisibility (x[sup]n[/sup] - 1) by (x -1) holds for all n, from which it follows that kx[sup]n[/sup]/x - 1 yields a remainder of k. From there it’s not hard to show that, for example, 3126[sub]DEC[/SUB] is divisible by 9 because the sum of its digits is.
Easier still: x[sup]n[/sup] - 1 = (x - 1)P(x) + R where P(x) is some polynomial in x and R is the remainder. When x = 1, x[sup]n[/sup] - 1 = 0(P(x)) + R = R = 0. Hence x[sup]n[/sup] - 1 = (x - 1)P(x), as required.
Formulas are a dime a dozen. I come up with them, use them, and generally forget them all the time. They occupy me when I’m bored with something tedious, and tend to be too specific to what I’m doing to be worth remembering. I can’t think of any that really stand out–maybe the one I came I came up with to calculate time saved by increasing average driving speed; I’d have to reconstruct it now, but its purpose was to talk myself out of speeding more on a really long and boring drive.
Not a formula, but I unwittingly reinvented Newton’s Method approximation when I was a lad, and made small forays into limit theory and some related basic concepts of calculus before I found out what calculus was. I was a little disappointed when I reached my high school Intro to Calc class and found that someone had already done the interesting bits.
Well played sir. Well played.
I recall feeling quite pleased with myself for coming up with a proof that 0.99 repeating was = 1, only to realize shortly after that it was hardly original…