12

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 ...


3

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: $$ 7, 17, 19, 23, 29, 47, 49, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, \dots $$


3

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.


3

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 $~7$th-grade expressions :$$15(x / 5) + 15(x - 1) / 3 = 15(1 / 5) $$


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