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Hint: $$a+b+c=abc$$ and $a<b<c$ so we have $$3c>abc\implies ab<3$$ Soince $ab >1$ we have $ab=2$ so $a=1$ and $b=2$ and ...

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More generally, this happens for the fraction $1/n$ exactly when $10$ is a primitive root mod $n$. Those $n$ are the ones in A167797: $$7, 17, 19, 23, 29, 47, 49, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, \dots$$

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Working with integers $$n(n+1)(n+2)=3n+3=3(n+1)$$ With $n=-1$, we have $$-1,0,1$$ as a solution Otherwise $$n(n+2)=3$$ $$n^2+2n-3=0$$ $$(n+3)(n-1)=0$$ $$n=3,n=1$$ Thus we have $$-3,-2,-1$$ or $$1,2,3$$ as solutions.

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If one exists, is there one that doesn't involve a perturbation off the real axis? Why perturb at all? It's easier to analyse if you stay on the real axis. By consideration of the roots of $z^2 - z + c$, the easy construction is $c = \tfrac14 + \varepsilon$. Without the $\varepsilon$ it would converge to $z \to \tfrac12$; $$(\tfrac12 - \delta + \alpha\... 3 Notice that the inductive step, like any inductive proof, assumes that 'The statement holds for k' .... i.e. (in this case) that with k blue-eyed islanders, none of them leaves before day k, but they do all leave on day k. So, at this point we indeed don't know that it is true, but rather we just assume that it is true, and see what follows. Well, ... 2 To answer this question, I'm going to use some ideas that will also be present in a (hopefully) forthcoming paper of mine on somewhat related questions on harmonic sums. I will make repeated (ab)use of the prime number theorem and little-o notation, p always denotes a prime number, and if anything is unclear, please feel free to ask. The idea is that we ... 2 If we know, that A=1,B=2,C=3 is a solution we can look for another solution with larger numbers by$$(A+a)+(B+b)+(C+c) = (A+a)(B+b)(C+c) \\ -----------------------------\\ (1+a)+(2+b)+(3+c) = (1+a)(2+b)(3+c)\\ 6+a+b+c = 6+ 2c+3b+6a+bc+3ab+2ac+abc\\ a+b+c = 2c+3b+6a+bc+3ab+2ac+abc\\ 0 = c+2b+5a+bc+3ab+2ac+abc\\ $$If no number a,b,c is negative, all must ... 2 "And generalised pattern of such identities would be interesting and appreciated" Well, while there is nothing particular about the triplet 59,60 and 61, we do have tan59+tan60+tan61=(tan59)(tan60)(tan61), the triplet being in degrees. This particular case comes from the identity tanA+tanB+tanC=(tanA)(tanB)(tanC) where A+B+C=180. 2 I think you did a very good job here. I don't think more sophisticated math tools would make the problem easier to solve. One of the things that will make your proofs shorter is practice -- you'll become more confident of knowing when you've made a solid point so that you don't have to repeat yourself. But you definitely deserve to be a tertiary student ... 2 Because \dfrac1{7} =.142857142857...  and all (and there is a lot) that follows from that. 1 There is no point equally distant to all points of a square. Indeed, if P were such a point, then the triangle PAB is isoscele, as PA=PB. Therefore, the orthogonal projection of P on the line (A,B) (sorry if the notation is unusual) is the middle point of the segment [A,B], call it I. Such a point belongs to the square, thus PI=PA=PB, however,... 1 If the volume of one cuboid is V, the three orientations give$$nV = abc \\ (n-1)V = (a-1)(b-1)(c-1) \\ (n-2)V = (a-2)(b-2)(c-2)$$Expanding the second and third, and substituting in the first, we get$$V = ab + ac + bc - a - b - c + 1 \\ V = ab + ac + bc - 2a - 2b - 2c + 4$$so$$a + b + c = 3$$and the (a-2,b-2,c-2) box is going to have some negative ... 1 Are there another triples (or not necessary triples) such that their multiple equal to their sum? Easy. Just take some random numbers, say 3 and 4. We have 3 \cdot 4 = 12, but 3+4=7. So, just pad it with 12-7=5 more 1's, and we have: 1+1+1+1+1+3+4=1\cdot1\cdot1\cdot1\cdot1\cdot3\cdot4=12 1 Let {x_n} be a sequence of strictly positive numbers with the following propriety:$$ \sum_{k=1}^n k x_k = \prod_{k=1}^n x_k^k $$for every positive integer n. It turns out that$$ \lim_{n \to \infty} x_n = 1 $$1 \lim_{n\to\infty}n\log(1+n^{-1})=1 1 I like using ~7th-grade expressions :$$15(x / 5) + 15(x - 1) / 3 = 15(1 / 5) 

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