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1d |
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How to prove these two ways give the same numbers? formatting / grammar |
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1d |
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How to prove these two ways give the same numbers? This seems a perfectly clear question to me, with no reason to close or downvote. Two sequences of operations are being done: in the first sequence, start with a pair of numbers $(73, n)$, and at each step replace it by the pair $(73, [\text{odd part of }73 + n])$. In the second, start with a fraction $\frac{n}{73}$, and at each step replace it by the new fraction $2^x \frac{n}{73} - 1$, where $x$ is the smallest number such that $2^x \frac{n}{73} > 1$. The question is to prove that the sequence of $n$s is the same. |
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2d |
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Conjecture regarding trapping rational numbers in some special intervals Ooh, then I expect this is as hard as the Lonely Runner Conjecture, which many people (including me) have spent enough time on to understand that it's probably a hard problem. :-) |
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2d |
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Conjecture regarding trapping rational numbers in some special intervals What is the motivation for this problem / how did it arise? |
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2d |
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Show that the only solution to $\phi(n) =n-2$ is $n=4$ Actually, a simpler argument: if $n$ is composite, say $n = ab$ with $a, b > 1$, then $a, 2a$ are two numbers not relatively prime to $n$, so $b = 2$, similarly $a = 2$, and therefore $n = 4$. This argument can be extended to prove that $\phi(n) = n - k$ has only finitely many solutions, for any $k$. If $n = ab$, then considering the numbers $a, 2a, \dots, ba$ gives $b \le k$, similarly $a \le k$, so $n \le k^2$. |
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May 16 |
answered | Show that the only solution to $\phi(n) =n-2$ is $n=4$ |
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May 16 |
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Show that the only solution to $\phi(n) =n-2$ is $n=4$ This idea is a bit backwards: you have listed two integers which are counted in $\phi(n)$, and suggesting finding one more, which will prove that $\phi(n) > 2$, rather than $\phi(n) < n - 2$. |
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May 15 |
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What is the mathematical foundation of Control Theory? I think this question is too broad. As I understand it, control theory is a branch of mathematics in itself (falls under arXiv category math.OC). See the other questions tagged control-theory here: math.stackexchange.com/questions/tagged/control-theory |
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May 15 |
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Prime numbers stretch to infinity, but what about the distance between them? I just found this survey article that has a lot more details: ams.org/journals/bull/2007-44-01/S0273-0979-06-01142-6/… ("Small gaps between prime numbers: The work of Goldston-Pintz-Yildirim") |
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May 15 |
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Partial Solution to the Twin Primes Conjecture — What does it imply? The twin-prime conjecture is already assumed to be true by many. A weaker version (with $\le 16$ in place of $2$) has been proved assuming the Elliott–Halberstam conjecture, in 2004 by Daniel Goldston, János Pintz and Cem Yıldırım. Their result is much stronger than this one. The interesting thing about this one is that it doesn't require any unproved conjectures, so it's a big breakthrough and its methods would likely lead to new things, but it's premature to ask for its applications to engineering. |
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May 14 |
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Interesting and irritating problem. The first part is not quite right: according to the wording of the question, the pattern repeats every $98$ days, not $99$ days ($98 = 1 + 2 + \dots + 10 + \dots + 2$, no $1$ at the end because it's part of the next period.) Also, in the second part, one of the cycles following the 5-day cycles is a 6-day cycle, but the second one is a 4-day cycle. This doesn't affect the answer of $8/10$, though. (You don't even have to mention these: you can just say, any day after one of the $10$ days; so there are exactly $10$ of them.) |
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May 14 |
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Deriving the 37-percent rule for dating math.stackexchange.com/questions/45266/… |
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May 12 |
awarded | combinatorics |
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May 11 |
comment |
Can the order of 2 mod p be arbitrarily small (relative to $p - 1$)? Meanwhile, I'm also convinced that there cannot be a much more elementary proof than this. Writing this proof backwards, for my own understanding: Let $r_p(2)=M$, and $\operatorname{ord}_p(2)=N$, so that $p-1=MN$. Considering some primitive root $a$ of $p$, we have $2\equiv a^k$ for some $k$, and as $1\equiv2^N\equiv a^{kN}$, we need $(p-1)|kN$, so $M|k$ and $2\equiv a^k=(a^{k/M})^M$. So we do necessarily need $p$ to satisfy both that $M|p-1$, and that $2$ is $x^M$ for some $x$ (i.e., $x^M-2$ has a root mod $p$). Something like Chebotarev's density theorem is probably needed to guarantee this. |
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May 11 |
accepted | Can the order of 2 mod p be arbitrarily small (relative to $p - 1$)? |
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May 11 |
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Can the order of 2 mod p be arbitrarily small (relative to $p - 1$)? Sorry for the delay in marking this as accepted. I was holding off because I hoped to learn and understand the proof of this very useful theorem first, but I don't seem to be getting the time for it right now. Meanwhile I was wondering if there might be a more elementary proof that doesn't require such heavy machinery, but it seems unlikely, now that I think of it more. (Next comment.) |
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May 11 |
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How often does it happen that the oldest person alive dies? This "answer" is based on a misunderstanding of the question, which is stated sufficiently clearly. Even assuming you read only the title of the question and not its body (which deserves an automatic -1), the question asks about how frequently the actual event of dying happens. It's not "When does it happen that the oldest person alive eventually dies?". For each person, the person dies only once; they're not dying every day, let alone "always". (Except to the extent that we're all dying a little every day, every minute.) |
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May 11 |
revised |
How to find the smallest number with just $0$ and $1$ which is divided by a given number? precomputation |
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May 11 |
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How to find the smallest number with just $0$ and $1$ which is divided by a given number? another example |
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May 11 |
answered | How to find the smallest number with just $0$ and $1$ which is divided by a given number? |
