April 9

What is the smallest positive integer which is known to not divide any (even or odd) perfect number? Also, how many positive integers <= 400 are known to not divide any (even or odd) perfect number? 49.217.123.95 (talk) 08:05, 9 April 2024 (UTC)[reply]

All positive integers divide a perfect number. Writing in which is odd, it divides a perfect number of the form for  --Lambiam 17:03, 9 April 2024 (UTC)[reply]

Are you sure about that, Lambiam? We know of only 51 perfect numbers, and we do not know whether there are infinitely many of them. Assuming there is a highest one, there must be infinitely many numbers greater than that, which, by definition, do not divide any lower numbers. Or, is your formula somehow proof that there are indeed infinitely many? -- Jack of Oz ^{[pleasantries]} 19:22, 9 April 2024 (UTC)[reply]

No, thank you for correcting me. All positive integers divide a number (in fact, infinitely many) of the form but these need not be perfect numbers.  --Lambiam 00:13, 10 April 2024 (UTC)[reply]

Check. :) -- Jack of Oz ^{[pleasantries]} 02:27, 10 April 2024 (UTC)[reply]

As far as I can tell, from our article on perfect numbers (in particular, the section on odd perfect numbers; even perfect numbers are precisely divisible by all odd primes of the form , and assuming that there are infinitely many such primes, all powers of 2 as well), it would seem that the answer is 105. GalacticShoe (talk) 02:22, 10 April 2024 (UTC)[reply]

Obviously this gives a trivial lower bound of 3 on the number of positive integers at most 400 known to not divide any perfect number. GalacticShoe (talk) 02:23, 10 April 2024 (UTC)[reply]

Even numbers cannot be factors of odd perfect numbers, thus there should be a larger lower bound. 220.132.230.56 (talk) 21:11, 12 April 2024 (UTC)[reply]

Good point. The sequence of even numbers that divide no perfect numbers is , basically even numbers not of the form where is prime and . GalacticShoe (talk) 23:47, 12 April 2024 (UTC)[reply]

So how many positive integers <= 400 are known to not divide any (even or odd) perfect number? 220.132.230.56 (talk) 13:35, 13 April 2024 (UTC)[reply]

Is it possible to use all n-cubes with n <= 5 (reflections counted as distinct), with totally 186 cubes, assembled into a 2*3*31 rectangular cuboid?

Also, for which positive integer N, it is possible to use all n-cubes with n <= N (reflections counted as distinct), assembled into a rectangular cuboid? 2402:7500:916:7991:1CB3:D7F2:BD28:3F3A (talk) 08:49, 9 April 2024 (UTC)[reply]

There are 11-polycubes with holes in the center, and it's impossible to fill this hole with anything but a single cube. There are several such shapes, and you only get one single cube, so N≥11 is impossible. You run into similar problems with holes when you try to tile with heptominos. You could stipulate that the pieces with holes not be included, but that would be a different problem. For N=5, it seems to me that best (or at least the most fun) approach would be to whip up a set with a 3-D printer and experiment. Sometimes a parity argument can give a simple impossibility proof, but that seems unlikely in this case. --RDBury (talk) 12:56, 9 April 2024 (UTC)[reply]

April 13

Hi!

There is a line integral ∫_L yds, where L is y²=2x and from (1/2, -1) to (2, 2). My answer is (5√5−2√2)/3, but L is a quadratic curve (instead of a line) and the answer seems odd, therefore I am not sure my calculation is correct or not.

How can I calculate a line integral result by computer (by Wolfram Alpha, python etc.)? James King 2009 (talk) 11:49, 13 April 2024 (UTC)[reply]

Swapping and , this is the arc length of the segment of the parabola given by the equation from to . It can be turned into an ordinary integral of the form

In view of the square root in this formula, the appearance in the result should not be odd, but I find a different result, involving not only square roots but also logarithms (or arcsines). Obviously, the length of the arc should exceed that of the straight line segment connecting the end points, which is but your result evaluates numerically to less than To check your result: its numerical value should be close to  --Lambiam 14:19, 13 April 2024 (UTC)[reply]

The question calls for the integral of yds, so the first moment wrt the x-axis, not arc length. So basically you're integrating . --RDBury (talk) 15:37, 13 April 2024 (UTC)[reply]

PS. if you just want a ballpark estimate to check your result, break the curve up and approximate by line segments. In this case the section from y=-1 to y=1 cancels itself out, so you really just have to do the segment from (1/2, 1) to (2, 2). If you replace the curve with a line segment, its center of mass is the midpoint, so (5/4, 3/2). The length is √13/2, so the moment is (3/2)(√13/2) ≈ 2.7. That's off from the original result by about .1 so I'd count it as verified. --RDBury (talk) 15:54, 13 April 2024 (UTC)[reply]

Symbolic integration by Maxima confirms it as well.  --Lambiam 20:54, 13 April 2024 (UTC)[reply]

Actually I should try that out, so thanks. There are limits to what you can do with pencil & paper (and maybe a calculator). I've found you can do wonders with an off-the-shelf spreadsheet program, and if you already know how to use one that would work as well. As with many things, the trade-off is the time it takes to learn how to use the tool in question. You can write a Python program when the task is too complex for a spreadsheet, but if you don't know Python already that's a lot of effort for solving one problem. The OP mentioned Wolfram Alpha as well, and I assume that would tell you the answer too if you know the right input string. I use Alpha occasionally, but mostly for things where the right command is obvious, "solve x^3-x-1=0" for example. --RDBury (talk) 02:17, 14 April 2024 (UTC)[reply]

For this specific integral it isn't too hard to do the integration purely mentally by change of variables so and thus  --Lambiam 10:09, 14 April 2024 (UTC)[reply]

April 15

I'm trying to read this article translated from a 1934 article in Swedish by B. Berggren. The word "catheter" is mentioned several times, and I gather it should be "cathetus" meaning a leg, or side other than the hypotenuse in a right triangle. I don't think I've heard of a cathetus before, nor a catheter with reference to a triangle. Is this term still used by anyone now? --RDBury (talk) 20:25, 15 April 2024 (UTC)[reply]

It makes it into wiktionary: cathetus, but there are no quotations. This says it's fairly rare, but some prefer it over the term "leg". -- Jack of Oz ^{[pleasantries]} 20:47, 15 April 2024 (UTC)[reply]

Thanks. I've mostly heard and used "side", but that's wrong if you think about it; triangles have three sides. I didn't think to look at NGram Viewer before, but I did just now and it seems that "cathetus" had about a 10 year window of popularity between the mid-1920's and mid-1930's, but still only about 3% as common as "hypotenuse". Perhaps it's more common in Swedish. --RDBury (talk) 06:44, 16 April 2024 (UTC)[reply]

The Swedish Wikipedia uses it in its article on the Pythagorean theorem. It did not help the translator that in Swedish the form kateter is both the word for "catheter" and the plural of katet, which means "cathetus".  --Lambiam 06:14, 17 April 2024 (UTC)[reply]

There's a room with 1,000 people, each with a computer. Their task is to access Wikipedia, and click Random article, editing whatever they see needs fixing each time, before repeating the same process endlessly. They're well paid. They do this for 8 hours a day, 5 days a week, until they either quit out of sheer boredom, kill themselves, or ascend to some higher plane of consciousness.

What are the chances that 2 or more of the people could be editing the same article simultaneously, causing an edit conflict when the second or later ones press Publish? And how long might it take for the first such conflict to occur? -- Jack of Oz ^{[pleasantries]} 23:20, 15 April 2024 (UTC)[reply]

This is essentially the same as the birthday problem for a set of randomly chosen people on a planet in which a year takes days, where is equal to the number of Wikipedia articles (currently 6,815,709). Assuming all are in the same time zone, working 9 to 5, the probability that any article is being edited simultaneously by multiple persons, using the square approximation, equals approximately

A slightly better approximation is given by the formula

How long it will take for a conflict to arise (assuming that all simultaneous edits are conflictual) depends on how long editors stay within the same article, which may be characterized by individual-dependent probability distributions and hard to impossible to tackle analytically. If they are synchronized, each editing one article per time slot in equal time slots, the expected number of time slots until conflict is approximately

 --Lambiam 03:31, 16 April 2024 (UTC)[reply]

Thanks. The first answer is about what I expected. But the time till the first conflict is rather counter-intuitive. If the time slots were 5 minutes, that means a conflict would occur after only a bit over an hour. I don't question your maths, but that is amazingly short for such a huge database. Thanks again.

But then, I guess the number of editors does make a rather significant difference (* cough), since the number of possible conflicts with 1,000 editors in each time slot is half factorial 999, a number with c. 2,565 digits, and about 10²⁴⁸⁵ times the number of atoms in the universe. Compared to this, a mere 6-odd million articles is infinitesimally small, so having to wait an hour for a conflict does seems a bit tardy, on reflection. If it were only 4 editors, there are only 6 possible conflicts and the expected time would increase massively, to about 500 days. Very nice. -- Jack of Oz ^{[pleasantries]} 23:56, 16 April 2024 (UTC)[reply]

Yes, at first glance it is counterintuitive, which is why the problem in its birthday suit setting is also known as the "birthday paradox".  --Lambiam 05:41, 17 April 2024 (UTC)[reply]

April 18

Interest compounds daily. The effective annual interest rate is 5.15%. Is there some formula to calculate what is the actual daily interest rate? Using a lot of trial and error on spreadsheets, I estimated that the daily interest rate is 5.02235% (or so). But, that was through a lot of trial and error … and it only comes “close to” an estimate of the actual daily interest rate. Is there some mathematical formula that will give me the exact figure? Thanks. 32.209.69.24 (talk) 02:04, 18 April 2024 (UTC)[reply]

To clarify, when I estimated the daily interest rate, I came up with 5.02235% (divided by 365) = 0.01375986301370%. Thanks. 32.209.69.24 (talk) 02:14, 18 April 2024 (UTC)[reply]

If I'm interpreting this correctly, then if you are given the annual interest rate , then you want a daily interest rate such that , where is the number of years. Since both sides are exponentials and the only way they can always match is for the bases to match, we can just remove the to get . Taking the root on each side, you get . Rearranging yields . For your value of , this yields , which is close to but slightly off from your value. GalacticShoe (talk) 02:15, 18 April 2024 (UTC)[reply]

Thank you so much ... for such a quick reply and for such a detailed explanation. Much appreciated. A follow-up question, if I may. That final formula that you cite contains a "root index" of 365. Is there a comparable formula that can be entered into an Excel spreadsheet? (I have no idea of how to -- or even if -- one can express a root index of 365 in Excel.) Thanks! 32.209.69.24 (talk) 02:29, 18 April 2024 (UTC)[reply]

Some preliminary searching indicates that you can just do "value^(1/365)", so for example "365*((1+A)^(1/365)-1)" GalacticShoe (talk) 02:44, 18 April 2024 (UTC)[reply]

Thanks a million! Very much appreciated! 32.209.69.24 (talk) 05:21, 18 April 2024 (UTC)[reply]

...and not pentillion? Someone who's wrong on the internet (talk) 05:43, 18 April 2024 (UTC)[reply]

All those number names are based on Latin (in this case quinque) rather than Greek (πέντε). --Wrongfilter (talk) 06:14, 18 April 2024 (UTC)[reply]

Then why is it pentagon and not quintagon? Someone who's wrong on the internet (talk) 09:07, 18 April 2024 (UTC)[reply]

Because -gon is Greek (γωνία = angle, from γόνυ = knee), hence the prefix is Greek, too. Some terms are Greek, others are Latin, mixtures are to be avoided. --Wrongfilter (talk) 09:21, 18 April 2024 (UTC)[reply]

German makes a bit more sense when it comes polygons, Dreieck=Triangle, Viereck=Quadrilateral, Fünfeck=Pentagon, ... . Literally (the number in German)+"Eck" = corner or angle. I imagine other languages not so heavily influenced by Latin and Greek are similar, none of this "If it's Tuesday we must be using Greek roots." The word for "polygon" is "Polygon" in German, breaking the pattern, though "Vieleck"="many"+"angle" is used as well. As for the numbers, people still don't seem to agree on what these names even mean (see Long and short scales). When in doubt, use 10¹⁸. --RDBury (talk) 14:21, 18 April 2024 (UTC)[reply]

My favourite shape is the octangular quadragon. -- Jack of Oz ^{[pleasantries]} 21:14, 18 April 2024 (UTC) [reply]

April 20

Imagine one tinder gender has a max 100 right swipes per day and the other have X max right swipes per day. Would it be possible to find the value of X that would be needed to make sure they have equal amount of matches per day at average with just that information I am presenting here or you would need internal data to solve this mathematical problem?

75% of tinder is male, 25% is female.

Woman swipe right 7% of time while man swipe 40%

Woman match with 33% of man they swiped right while men match with 2.5% of person they swiped right.

Woman vote at 200 profiles per day while man do with 137. 179.134.97.227 (talk) 17:34, 20 April 2024 (UTC)[reply]

I'm not familiar with tinder, but it seems that the question is about a selection process for strictly binary and straight users, in which each of the two genders is only presented candidates of the other gender. Each candidate presented is a match for female users with probability 0.07 × 0.33 = 0.0231 and for male users with probability 0.40 × 0.025 = 0.0100 .

Then the expected number of matches for a female user when presented 200 candidates equals 200 × 0.0231 = 4.62, while that for a male user when presented 137 candidates equals 137 × 0.0100 = 1.37 .

So far so good, but where does the maximum of 100 come in if a woman can do 200 swipes per day?

Given match rates p_f and p_m for the two genders, the expected numbers of matches for users of these genders equal n_f × p_f and n_m × p_m, in which n_f and n_m stand for the numbers of candidates presented to the respectively gendered users. To make these expected numbers equal requires achieving a ratio between these numbers of candidates such that

n_f : n_m = p_m : p_f.

For the data supplied, this means,

n_f : n_m = 0.0100 : 0.0231 = 100 : 231 .

So if female users are presented 100 candidates, male users need to get presented 231 candidates to achieve the same number of matches, to wit:

100 × 0.0231 = 231 × 0.0100 = 2.31 .

 --Lambiam 20:33, 20 April 2024 (UTC)[reply]

At tinder, people are presented with a "random" (not exactly), person, they can swipe left (not like) or swipe right (like), if male A like female B and female B like male A they match and can start to talk. The 200 swipes means at average, the woman rates at this an day (like OR dislike) 200 people. Max amount of right swipes would be the amount of likes they would be able to give before not being able to give likes for that day. Woman receive way more matches than man at this apps for various reasons, the question here is if that information presented at the question, would be possible to find what would need to be the limit at the amount of likes a man can do (assuming woman can do 100 per day) to make sure the amount of matches is the same per day at average.179.134.97.227 (talk) 21:09, 20 April 2024 (UTC)[reply]

A question about your terminology. If male A likes female B and female B likes male A, which of the two "receives" the match? It seems to me that if every match is between two users of different genders, then each gender will always have the same number of matches as the other gender. A difference, if any, can only be in who initiated the process that led to the match being established.  --Lambiam 21:18, 20 April 2024 (UTC)[reply]

"If male A likes female B and female B likes male A, which of the two "receives" the match? It seems to me that if every match is between two users of different genders, then each gender will always have the same number of matches as the other gender."

Man A and female B receive the match. A gender can have different numbers of matches at average, one example, if man 1 likes woman 1 and 2, man 2 likes woman 1 and 2, man 3 likes no one and woman 1 likes man 1 and woman 2 likes man 1. Thats man 1 having 2 matches and other man having no matches, thats 0.66 matches at male side and 1 match at average at the woman side.179.134.97.227 (talk) 21:30, 20 April 2024 (UTC)[reply]

Let M stand for the number of matches under some procedure. In your example M = 2, since we have the two matches man 1 ⇆ woman 1 and man 1 ⇆ woman 2. Let, furthermore, u_f and u_m denote the number of female and male users. In the example, u_f = 2 and u_m = 3. Then the average number of matches for the two genders are M⁄u_f and M⁄u_m. In the example, we get 2⁄2 and 2⁄3. One way to get the two averages equal is to make sure that M = 0, for example by not allowing any right swaps at all. The only other way does not involve the process for indicating preferences: make sure that u_f = u_m.  --Lambiam 18:22, 21 April 2024 (UTC)[reply]

April 22

With spheres there's only 1 right answer, there's also formulas that can be used to find points 0.5π radians from another point, is there a formula for the line with half the ellipsoid surface area on each side? Is there a formula for the line where an infinitely far Star of Bethlehem and an anti-Star of Bethlehem at the other end of the ellipsoid normal line would have equal zenith distances ignoring refraction and geoids? Are these lines the same? How far apart can they be? Sagittarian Milky Way (talk) 18:30, 22 April 2024 (UTC)[reply]

Unless an ellipsoid is a sphere (which the WGS 84 reference ellipsoid is not), no portion of it is a mathematical hemisphere. Any plane through its centre divides it though into two equal (congruent) parts. Usually the plane will be a meridional or the equatorial plane. In more general geodetic systems the equator and meridians, although not ellipses, also lie in a plane and can be used for a fairly fair cutting into two parts, which however will normally not be congruent. Calling the two parts "hemispheres", although not correct in a strictly mathematical sense, is nevertheless conventional.  --Lambiam 19:11, 22 April 2024 (UTC)[reply]

Yes but that's the easy way out, hemispheres centered on the equator or pole are exactly zero percent of all possible centers. Sagittarian Milky Way (talk) 21:56, 22 April 2024 (UTC)[reply]

Is division by any plane through the centre not general enough?

The sight lines to a point on the celestial sphere and to its celestial opposite are parallel. So are the directions to the respective zeniths from a given place on the ellipsoid and its antipodal place. Therefore the angular distances are the same.  --Lambiam 11:03, 23 April 2024 (UTC)[reply]

I must admit to being intrigued by the idea that the Star of Bethlehem might have been arranged by angels using WGS 84! However as far as I can see there's just ocean at the antipodal point for the birth of Damien Thorn. Pehaps we're safe for a while yet ;) NadVolum (talk) 19:42, 22 April 2024 (UTC)[reply]

The normal on the ellipsoidal through Bethlehem won't go through the center of the Earth and so won't go though the antipodal point. NadVolum (talk) 20:04, 22 April 2024 (UTC)[reply]

It doesn't really matter though cause the idealized celestial sphere/astronomical coordinate system is infinitely far, the lines to the star from anywhere on Earth would be parallel. It would matter for the "ranking all points by distance and picking the nearer half" way as an extremely flattened ellipse could have the (geographic, not geocentric) latitude minus 90 be only a few miles away (plus 90 in the southern hemisphere) Sagittarian Milky Way (talk) 22:15, 22 April 2024 (UTC)[reply]

April 23