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### FAQ

Where can I find a Form 990 for a non-profit?
GuideStar nonprofit reports and Forms 990 for donors, grantmakers, and businesses
Is yellowcircle.net a legitimate charity?
So, to answer your question, is it a "legitimate charity"...it would seem that the answer is Yes. it is registered with the IRS as a nonprofit (Exempt Organizations Select Check) so it would seem that it is at least legit in the sense that it is incorporated and set up as a nonprofit. Charity Navigator also has a profile on them here: Charity Navigator - Advanced Search though its not super helpful as it seems as of 2014 they were reporting less than \$50,000 in gross income so that means they are required to file an abbreviated set of financial disclosures. A 990-N rather than the full 990 as it were. It would seem from their website they have seen some growth in the last year so when they file their 2015 financials in a few months and post them that might give you a better look inside the organization. Not sure if that's the answer  you're looking for, I can add that if they're a "good" charity then they should have a decent customer service team to handle these sorts of inquires.
Is it worth pursuing a 501(c) registration? Is it difficult to file taxes? What about getting a general liability insurance?
Filing a Form 1023, 1023-EZ, 1024, or 1024-A to receive a determination of 501(c) status from the Internal Revenue Service can be relatively straight-forward. Having 501(c) status will make the organization exempt from federal corporate income taxes, which could potentially save the organization quite a lot. Donors of charitable contributions to most 501(c)(3) organizations can usually receive a tax-deduction, which can help motivate donors.Organizations with 501(c) status are typically required to file an information return to the IRS each year, either a Form 990, 990-EZ, or 990-N. Some organizations outsource this to a professional, and some do it themselves. There are also websites that can help make sense of completing the form. I like Form 990 Online a lot.General liability insurance may be wise regardless of whether the organization has 501(c) status. A good insurance agent or broker can help with that.There is a lot of good information here on Quora. Search it out.Good luck!
Where can I find a site that lists organizations in violation of their 501(c)(3) status?
Ben is right.  This list of revocations is here: http://www.irs.gov/charities/cha...Anyone coming to this question with a general interest in finding out...If a donation to a specific organization qualifies for a deduction?[1]Whether an organization is a 501(c)(3)?If a 501(c)(3) is it a public charity or a private foundation?.... should start here: http://apps.irs.gov/app/eos/ and click the Are eligible to receive tax-deductible contributions option.[2]  If you can't find an organization there, then check the revocation list, and if they don't appear there be ready to call the organization or dig deeper.In my experience, the IRS site works much better if you know an organization's EIN.  I go to the Foundation Center's site to find them because their search tool is better: http://foundationcenter.org/find...There are third party vendors that make this process much easier.  I don't know enough about the timeliness of their data, the differences between paid and free services, and what types of verification the IRS requires for tax purposes to comment on them.  [1] To be clear, the only thing you can do from the IRS's website is check to see if an organization's tax status allows for donations to be deducted.  Whether a specific tax payer making a specific donation to that organisation can deduct the donation is a separate question.  No tax advice here.  [2] Note that the Are eligible to receive tax-deductible contributions option on page http://apps.irs.gov/app/eos/ includes organizations that have filed the 990-N, which is the third option users can click on that page.   I'm not sure why the IRS lists them separately or, for that matter, why
Select n positive integers, such that for all pair combinations the sums of each pair generates the most unique primes. How would you solve it?
We can establish an upper bound through a simple parity argument. Two odd numbers cannot be added together to get a prime, because the result is even. Therefore, to generate a prime, the pair must consider of one odd number and one even number. Given E even numbers and N-E odd numbers in our set, the number of odd sums is$E(N-E) = EN-E^2$This is maximized at $N-2E = 0$, or $E = N/2.$ For example, with $n = 20$ the maximum number of primes appearing can be no more than $10 \cdot 20 - 10^2 = 100$. Our upper bound is thus $(N)(N/2)-(N/2)^2 = N^2/2 - N^2/4 = N^2/4$.What other number-theoretical constraints can we find? A standard result on prime constellations is that if $p+a_1, p+a_2, p+a_3, \ldots, p+a_k$ are all prime, then there can be no prime number that divides the product $(n+a_1)(n+a_2)(n+a_3)\ldots(n+a_k)$ for all $n$. Equivalently, for any prime $p$, the set of remainders $\{ a_i \bmod p \}$ must omit one of the possible values. See Admissible prime constellations (For $p k$, it is obvious that at least one of the congruence classes will be missing since there are more remainders than numbers in our constellation, so the check can be done mechanically over a finite set.)It is conjectured that whenever a prime constellation is admissible, there are infinitely many primes matching that constellation. So, if we’re only interested in finding some set of n, rather than the smallest possible, we can pick an admissible constellation starting with $a_1 = 0$ and search for its occurrences, each of which will give us an odd number to add to our set. The constellation need not be the smallest possible for $N/2$, but that seems more elegant. Unfortunately, this approach seems like it will lead to very large odd numbers!For example, with the constellation 0 2 6 8 12 18 20 26 30 32, the starting values are: 11, 33081664151, 83122625471, 294920291201, 573459229151, 663903555851, 688697679401, 730121110331, 1044815397161, and 1089869189021. (Solution from here: Prime) This is a solution to the N=20 case but is much less compact than the one provided in the comment above. That suggests we really need to make more use of our flexibility picking the even numbers. :) But this solution should work as long as we don’t mind searching a long time, and relying on the status of unproven number theory conjectures.What if we start from the other end? We can generate lots of primes, then look at what constellations exist within our sample. Greedy search might well give acceptable results. It seems this is the point at which some algorithmic cleverness is called for. The closest problem I can think of is maximum k-subset intersection, where we want to find exactly k subsets whose intersection size is maximized. (But we don’t actually care about maximization, we only care about satisfying that the intersection size is at least k.) See http://www.ic.unicamp.br/~eduard... where the full problem is shown to be NP-hard and hard to approximate.Our sets are actually far from arbitrary. Each of the sets of distances is a shifted version of the one before. I don’t see a ready way to use that other than to make building them more efficient.For each candidate distance $d$ for our constellation, build a set of prime numbers $p$ where $p + d$ is a prime. Then, perform greedy search of the distances, using the distance that preserves the largest intersecting set of prime numbers. (Any distance appearing in less than N/2 subsets can be excluded immediately.) Candidate solutions need to be checked that all the sums are distinct, but if there are enough primes in the set then we can trim some of them.If the starting set of primes is large enough, this method seems to work pretty well and find a solution more or less immediately. If the primes aren’t big enough we can burn a lot of CPU fruitlessly searching. It might make sense to terminate the search pretty quickly and adjust the starting set of primes, rather than trying to explore the tree more deeply.For N = 20 (primes up to 10000 ) only one backtrack:[0, 210, 330, 420, 990, 1200, 1890, 2520, 2730, 3060] [23, 127, 331, 499, 887, 1151, 1277, 2383, 2789, 3307](The primes 3491 and 3593 satisfy this constellation as well)For N = 22 (primes up to 12000 ), no backtracking:[0, 30, 210, 780, 990, 1200, 2310, 4440, 5070, 5610, 6786] [233, 647, 1151, 1567, 2341, 2857, 4241, 8543, 8941, 9467, 10837]For N = 24 (primes up to 100000), I hacked my backtracking routine to give up earlier on the leaves and found this with only a few dead-ends:[0, 204, 840, 2310, 3570, 4620, 5100, 5880, 7704, 8190, 9510, 13020] [457, 1229, 3253, 4219, 6217, 8297, 14827, 48337, 48799, 73379, 78989, 83987](I’m not happy about that solution, ideally we would start over with a different first choice in case that was the problem, but not enough to spend a lot of time on it. Or introduce some randomization, or something.)N=26: no success, primes up to 200000 are too big for my inefficient python implementation.
What are the fast reading techniques that you have used to read 990+ books?
$10^n - 3, n \in \mathbb{N}$is provably prime for $n = \{1, 2, 3, 17, 140, 990, 1887, 3530, 5996\}$; there don’t seem to be any other primes of this type$10^{10000}$. Such primes are evidently not common at all; I suppose then I have to say that the largest prime of this type I know of for a certainty would have to be $10^{5996} - 3$. Numbers of this type might be referred to as “near-repunit” primes; see A089675 in the OEIS (numbers $n$ such that $9R_n - 2$ is a prime number, where $R_n = 11\ldots1$ is the repunit (A002275) of length $n$).The last three of the above qualify as titanic primes per Samuel Yates’ original definition $\left(p \gt 10^{1000}\right).$I tested numbers of this type up to $n = 10000$, but found only the nine reproduced above to be prime. My results were as follows:$\begin{array}{ |l|l| } \hline \mathit{n} & \mathbf{Primality} \\ \hline \\ \hline 1 & \text{Pierpont prime (by inspection)} \\ \hline 2 & \text{Pierpont prime (trial division to 7)} \\ \hline 3 & \text{Prime (trial division to 31)} \\ \hline 17 & \text{Prime (Miller-Rabin-Selfridge)} \\ \hline 140 & \text{Prime (PARI/GP implementation of Atkin-Morain ECPP)} \\ \hline 990 & \text{Prime (PARI/GP implementation of BPSW compositeness test)} \\ \hline 1887 & \text{Prime (PARI/GP implementation of BPSW compositeness test)} \\ \hline 3530 & \text{Probably prime (Miller-Rabin-Selfridge)} \\ \hline 5996 & \text{Probably prime (Miller-Rabin-Selfridge)} \\ \hline \end{array}$If you like, you could try something like this in Mathematica™:Print["Primes of the form ", Superscript[10, "n"], " - 3:"]; Print[""]; m = 2000; For[n = 1, n m, If[PrimeQ[10^n - 3], Print[Superscript[10, n], " - 3"]]; n++] ently found a web page maintained by Makoto Kamada who, with other contributors such as Tyler Cadigan, Maksym Voznyy, Henri Lifchitz, and Markus Tervooren, proved the primality of the nine numbers listed above, and also discovered the probable primes corresponding to $n = \{13820, 21873, 26045, 87720, 232599\}$. The first four of those are gigantic primes per Chris Caldwell’s definition $\left(p \gt 10^{10000}\right).$ I suppose that $10^{232599} - 3$ technically is as well, as I don’t know of any specific naming convention for primes with between 100,000 and 1,000,000 digits.M. Tervooren has evidently searched for prime numbers of this type up to $n = 407197.$ It’s odd that there seem to be only fourteen primes of this type with fewer than 407,197 digits — that’s a density of only about 0.00344%, which seems far lower than I’d expect:$\{2, 3, 5\} \nmid 10^n - 3\\ 10^n - 3 \equiv 1 \pmod{2}\\ 10^n - 3 \equiv 1 \pmod{3}\\ 10^n - 3 \equiv 2 \pmod{5}$This would seem to suggest that since $\frac{11}{15}$ or $73.\bar{3}\%$ of integers $5$ are “5-smooth” (the odds against their being composite are 7:2), the probability of the primality of numbers such as $10^n - 3$ which cannot be multiples of 2, 3, or 5 should be much higher.