- #71
- 19,568
- 25,557
The missing rest of the month is added now.
The answer is correct, the reasoning is not: ##32\equiv 2 \,(3)\wedge 32\equiv 2\,(5)## and ##32## does not end on a ##7.##TeethWhitener said:Edit: ignore what I wrote. I didn’t consider that the number could be prime all by itself. The first two conditions (##2\mod3## and ##3\mod4##) tell us that the solution will be an integer of the form ##12n-1##. The first and third (##2\mod5##) conditions tell us that it ends in a 7. The smallest number that satisfies this is 47.
Pretty sure a little piece of code could do this much quicker, but here’s my calculator-free solution:
The smallest prime factor will be at least 7. Looking at the modular behavior of multiples of 7, we find that a family of solutions consisting of 7 multiplied by a number satisfying the following expressions simultaneously:
$$3n+2$$
$$4n+1$$
$$5n+1$$
The smallest number of this form is 41 (this is easy to find by noting that ##4n+1## and ##5n+1## together imply ##20n+1##), which gives a solution of ##7\times41=287##.
To check whether this is the smallest integer satisfying the equations, we note that ##287<\sqrt{17}##, so the smallest prime factor of the solution will either be 7, 11, or 13. The requirements for the multipliers of 11 are:
$$3n+1$$
$$4n+1$$
$$5n+2$$
The smallest number satisfying these expressions is 37 (again, easy to find noting that ##3n+1## and ##4n+1## together imply ##12n+1##), which gives the solution ##11\times37=407>287##.
The requirements for multipliers of 13 are:
$$3n+2$$
$$4n+3$$
$$5n+4$$
The smallest number satisfying these expressions is 59 (the conditions are equivalent to ##3n-1##, ##4n-1##, and ##5n-1## together, or ##60n-1##), giving ##13\times59=767>287##, so our answer is in fact 287.
Almost correct! In case ##|X|=1## we have ##\mathcal{P}(X)=\{\emptyset\, , \,\{x\}\}## but ##\emptyset \not\in f(X)## hence ##f## isn't surjective in that case either: ##\emptyset \notin X.##fishturtle1 said:We consider the first map first. Suppose ##f(a) = f(b)##. Then ##\lbrace a \rbrace =\lbrace b \rbrace##. This implies ##a = b##. So, ##f## is injective. If ##\vert X \vert = 1##, then ##f## is surjective. Suppose ##\vert X \vert \ge 2##. Then we can write ##X = \lbrace a, b, \dots \rbrace##. Then ##\lbrace a, b \rbrace \in \mathcal{P}(X)## and ##f(x) \neq \lbrace a, b \rbrace## because ##\vert f(x) \vert = 1## for all ##x \in X##. So, ##f## is surjective iff ##\vert X \vert = 1##. Lastly, we see ##f^{-1}(\emptyset) = \lbrace x \in X : f(x) = \emptyset \rbrace = \emptyset## because ##\vert f(x) \vert = 1## for all ##x \in X##.
Next, we consider the second map. Let ##X =\lbrace a ,\dots \rbrace## and define ##A = \lbrace a \rbrace## and ##B = \emptyset##. Then $$g((A,B)) = A \cup B = \lbrace a \rbrace \cup \emptyset = \lbrace a \rbrace = \emptyset \cup \lbrace a \rbrace = B \cup A = g((B,A))$$
and ##(A,B) \neq (B,A)##. This shows ##g## is not injective. Let ##C \in \mathcal{P}(X)##. Then ##g((C,\emptyset)) = C \cup \emptyset = C##. This shows ##g## is surjective. Lastly, observe for arbitrary ##A, B \in \mathcal{P}(X)##, if ##A \cup B = \emptyset##, then ##A \subseteq \emptyset## and ##B \subseteq \emptyset##. This implies ##A = \emptyset## and ##B = \emptyset##. So, ##g^{-1}(\emptyset) = \lbrace (\emptyset, \emptyset) \rbrace##. []
fresh_42 said:The answer is correct, the reasoning is not: ##32\equiv 2 \,(3)\wedge 32\equiv 2\,(5)## and ##32## does not end on a ##7.##
Another idea how to prove it?
Sorry, but you wroteTeethWhitener said:The condition ##2\mod5## means the number ends in a 2 or a 7, and ##3\mod4## means the number has to be odd.
and not second and third!The first and third (##2\mod5##) conditions tell us that it ends in a 7.
Cauchy is correct, but it can be used in either of the two directions, hence it is necessary to at least sketch which proof is meant (a proof where the chain rule is seen would be preferred), rather than saying Cauchy.mathwonk said:for #9, I believe that a loop is null homologous iff every holomorphic one - form integrates to zero over it. does that do it? (just throwing out an approach for complex analysts, since a topologist presumably knows that homology is a continuous invariant; or does 0 - homologue have some other meaning?)
Oh, oops…fresh_42 said:Sorry, but you wrote
and not second and third!
Here is the proof by the Chinese remainder theorem:There is a solution ##x## since ##3,4,5## are pairwise coprime by the Chinese remainder theorem, and all solutions are congruent modulo ##M=3\cdot4\cdot5=60.## The calculation isTeethWhitener said:Oh, oops…
I guess it will do no harm to elaborate on these two solutions for those who do not "see" it at once.ergospherical said:For 30 I observe that ##\nabla \cdot F = 2z##, so by theorem of Gauss ##\int_A \langle F, n \rangle dS = \int 2z dV +\cancel{123\int_{\mathrm{cover}} dS} - \cancel{123\int_{\mathrm{base}} dS} = \pi R^2 h^2##.
(Directly, one could parameterise ##A## by ##\Phi(\theta, z) = (R\cos{\theta}, R\sin{\theta}, z)## to obtain a surface element ##n dS = R(\cos{\theta}, \sin{\theta}, 0)d\theta dz## and then an integral ##\int_A \langle F, n \rangle dS = R^2 \int z dz \int d\theta = \pi R^2 h^2##).
does this mean no three points in ##\mathcal{P}## are collinear? Or only a specific ##p_1, p_2, p_3##?fresh_42 said:Let P be a finite set of points in a plane, that are not all collinear.
fresh_42 said:28. Consider the circle segment above ##A = (-1, 0)## and ##B = (1, 0)## of $$
x^2 + \left(y + \dfrac{1} {\sqrt{3}}\right)^2 = \dfrac {4}{3}
$$
The point ##P := \left(\dfrac{1} {\sqrt{3}}, 1 - \dfrac{1} {\sqrt{3}} \right)## lies on this segment. Calculate the height ##h## of the circle segment, and ##|AP| + |PB|##.
Thank you! My bad. I have corrected this. I'm afraid it won't be the last sloppy formula.julian said:Hi @fresh_42. I notice you have given the solutions for December's math challenges. For problem 2 you've written
\begin{align*}
I = \frac{1}{2} \log^2 \frac{a}{a+1} = \log^2 \sqrt{\frac{a}{a+1}}
\end{align*}
The final expression (the RHS) seems wrong because isn't ##\log^2 \sqrt{\frac{a}{a+1}} = \dfrac{1}{4} \log^2 \frac{a}{a+1}##?
Thanks for the flowers. I wrote many of these late at night, and who wants to proofread 500+ pages?julian said:I was just thinking to myself, it is not like @fresh_42 to be sloppery.
fresh_42 said:31. Let ##\mathcal{P}## be a finite set of points in a plane, that are not all collinear. Then there is a straight, that contains exactly two points.
You do not need an algorithm or any iteration. Consider pairs ##(g,P)## of a straight line in the plane and a point ##P\not\in g.## Then choose a pair such that ##\operatorname{dist}(g,P)## is minimal.Not anonymous said:Sorry, please ignore my previous post (post #90). I think I found a mistake in that solution for question #31. I will attempt to find a correct solution later and make a new post.
I didn't expect a reply to my withdrawn incorrect solution. ** thanks for that hint. Below is a solution, hopefully correct this time.fresh_42 said:You do not need an algorithm or any iteration. Consider pairs ##(g,P)## of a straight line in the plane and a point ##P\not\in g.## Then choose a pair such that ##\operatorname{dist}(g,P)## is minimal.