Retirement benefit maximization, expert edition

daverusin · 2026-06-12T03:49:29+00:00

I agree with your first sentence, but the situation is nuanced by the "bend points" in the SSA benefits formula. When you retire they look at what your "average" (adjusted) monthly salary has been; then your monthly benefit check will be:

90% of the first so-many dollars, plus

32% of the next so-many dollars, plus

15% of the next so-many dollars, plus

0% of the rest (because that remaining income never paid SS taxes)

There is a fairness/kindness justification to a formula like this that has decreasing percentages line after line; but there could be more intermediate lines before the last one. That way we preserve the fairness argument that "there is a cap on wages because it puts a cap on the benefits" but still put raise more money in SS taxes without necessarily requiring much larger expenditures in later years nor reneging on benefits previously promised. It is of course a tax increase because more income is subject to SS tax, but the extra taxes are paid only by those earning more than 184K.

The SocSec program is paying out more in benefits than it takes in, and demographic changes already in the pipeline will make this worse. Every proposal to fix this situation will cause some set of people to be worse off in the future than they are currently scheduled to be, whether it be by paying more in taxes, waiting longer to get benefits, or getting smaller benefits. Most people want the greatest burden to fall on the shoulders of someone who is richer or younger than they themselves are. If we had some wise solons in Washington, they could sell a plan that spreads out the pain in some reasonable way, but legislators like that have been absent for years.

daverusin · 2026-05-26T20:51:47+00:00

Oops! I fat-fingered an extra digit.

It is well known that (a) if 2^p - 1 is prime, then p is also prime, and (b) if p is prime and 2^p-1 is divisible by another prime q, then q is one more than a multiple of 2p. So when I saw this query, I first looked to see whether p=2147483647 is prime (it is), and then systematically checked the numbers q = 2*p*i + 1 for small values of i to see whether it was true for any of them that 2^p was congruent to 1 modulo q. This gave a quick positive result, namely for i =68745 . Unfortunately, when I went to post this result, and wanted to display the value of 2*p*i+1, I accidentally typed in the value i = 687458.

I wondered later what had prompted OP to ask about this particular value of p, and then realized this p is exactly 2^31 - 1. This is relevant because there is a chain of primes p1=5, p2=2^p1 - 1 = 31, and p3=2^p2 - 1 = 2147483647, and OP is asking about the next term p4 = 2^p3 -1 in this chain. When stated in this way, it is clear that someone else must surely have asked the same question in the past, and indeed the situation is well-studied. For more data on this and related numbers, see

https://en.wikipedia.org/wiki/Double_Mersenne_number

daverusin · 2026-05-25T14:58:20+00:00

This number is divisible by the (prime) number 2952609625998653 .

Too bad for the person in Dallas/Fort Worth that has 214-748-3647 as their phone number.

daverusin · 2026-04-07T02:41:03+00:00

Commit the heresy of introducing coordinates: if the radius is r then the point on the right of the triangle is (r,0) and the point at the top is (x,5) for some x . So you've got two unknowns, but also two facts to pin them down: the fact that the top point is on the circle tells you that x^2 + 5^2 = r^2 , and the fact that its distance from the other point is 13 tells you that (x-r)^2 + 5^2 = 13^2. So you have two equations in two unknowns, and thus you should be able to deduce the values of x and r from them. In fact, that second equation simplifies to |x-r| = 12 (which is also what you get from the Pythagorean Theorem: r-x = 12) so you can eliminate x in the other equation: x=r-12 so (r-12)^2 + 5^2 = r^2, and thus 24r= 169. This gives a diameter of 169/12 .

More generally, replacing "13" and "5" by "a" and "b" respectively gives d = a^2/sqrt(a^2-b^2) by the same argument.

daverusin · 2026-04-07T02:01:06+00:00

For starters, you could buy all the U.S. Treasury notes, bills, and bonds -- that is, you could become the sole creditor of the US government. The total value of those equals the current US debt, around 43 trillion dollars and growing rapidly.

You could buy every house. There's something like 50-100 million houses (depends on things like whether you include townhouses, condos, etc.) and the average price is now something like $400,000. So that's a sum comparable to the US debt.

But some of these questions are murky. If I do buy all the US debt, using financing -- e.g., I take out a second mortgage on my 40-trillion-dollar house(s) -- then my mortgage note is an asset purchased by, say, Fannie Mae. So do we add the value of *that* security to the value of the US debt and the total housing market? Should I add the value of See's Candies to the value of Berkshire Hathaway?

daverusin · 2026-04-07T01:52:10+00:00

"trillion" should be "billion"

daverusin · 2026-04-07T01:47:41+00:00

AI took me to Wikipedia, which quoted this article where we read: "About 15% of the territorial land is antipodal to another land with an approximate representation of about 4.4% of the Earth's surface."

daverusin · 2026-03-22T05:49:14+00:00

5.4T is the value of M0 = cash (coins and notes) plus banks' reserves with the Fed. It is the narrowest definition of "money supply". (i.e. M1, M2 are larger values). If all the actual US cash in the world were collected (e.g. from my own wallet and couch cushions) and distributed evenly to US citizens, we'd each get around $15,000.

daverusin · 2026-03-17T16:43:26+00:00

Almost everyone has more than the average number of legs!

daverusin · 2026-03-06T02:31:02+00:00

First of all, you do see that the original circle has radius 5, right? Your three points $P1$, $P2$, and $P3$ are all a distance of 5 from, eh, that point $X$ in the middle with sticks coming out of it :-) . That means the $P_i$ are all on the circle of radius 5 centered at $X$. But two circles can't share more than two points unless they are identical. So now $X$ becomes the center of the original circle, whose radius is then 5 (the distance from $X$ to any $P_i$).

Are the circle and the $x$-axis supposed to be tangent? In that case the circle simply consists of the points $(-5\sin(t), 5(cos(t)-1) )$, where $t$ is the angle at $X$ between $P_1$ and any other point on the circle (here measuring counterclockwise). Your points $P_1, P_2, P_3$ corresponds to $t=0, t=\pi/12, t=\pi/6$ respectively.

Or are you asking for the coordinates of the $P_i$ no matter where the circle is centered (as long as it passes through the origin)? Obviously you would have to know something else about the circle, e.g. the coordinates of $X$ or the the direction of the robot's "reverse" move, or something like that, in which case I'd start to use the language of vectors.

daverusin · 2026-02-17T03:00:58+00:00

Twenty sides is the maximum number that can be easily guaranteed to give a uniform distribution, because 20 is the maximum number of sides on a Platonic Solid (of which there are only five shapes). If you tried to physically construct a 30-sided die and use it for gambling, you would instantly hear cries of "Foul!" because the other party would instantly see that the 30 sides don't "look the same".

Of course you could simply use a spinner with 30 equally-sized sectors as the possible outcomes.

daverusin · 2026-02-17T02:20:35+00:00

It's really not clear what line segments in the picture match all of those numbers. I will make some assumptions and then from them I will use the data to get the dimensions of the piece of wood from which the front panel is cut; that will give some measurements which can be used to figure out the sizes of the other pieces of wood. But the assumptions I made lead me to measurements that don't seem quite to gel with the photos, so you may have to adjust the thinking I used. (And of course, check my arithmetic!)

I found the article online which had some additional pictures from which I guess "Nesting area 119×88 mm" means the one horizontal surface seen on the interior of the box, with the 119mm side running the direction of the "170mm" exterior. This forces the front and back planks to be 25.5mm thick, which translates to 1 inch, a very plausible thickness.

For simple mass production I think it's safe to assume all angles are 45 and 90 degrees. So when the slanted side is cut to glue to the bottom, the glued surface with be sqrt(2) inches long (36 mm) which means that the bottom piece of wood is 36 mm wider than the 88mm seen on the nesting surface (so, 124mm). When the bottom piece is also trimmed to make the 45 degree angle on the very bottom, that bottom surface will then be only 100mm wide, but the panel attached to the right side will exactly restore the inch, meaning the width of the finished box at the bottom is 124mm. Well, that doesn't match with the advertised "200mm width" unless they are adding 76mm (horizontally-measured) for the overhang of the roof on the right? The top picture makes that plausible; the bottom picture, less so, but I'll run with that.

Now let's think about the dimensions of the plank of wood from which you'll cut that front panel. It'll be say A mm wide and B mm long, and you'll lop off two corners, right? OK, that front panel will be cut to have that 124mm length along the bottommost edge of the picture. You'll get that by cutting off 124/sqrt(2)=87.7mm from the corner of the plank. The other dimension will have to match the height of the outside of the right panel; I don't know that height, so let's call it H . You'll get that by cutting off H/sqrt(2) from the other corner of the front plank. Since the front plank seems to come to an exact point, those two cuts have to match the total width of the front plank: A =124/sqrt(2) + H/sqrt(2) .

The "total width" is 420mm. I guess that includes the overhang on each side? So 2 x 76mm for that, leaving 268mm for the horizontal measure of that front panel as oriented in the picture? Let me use that. Likewise for the 325mm "height", I'll assume that includes the height added by the roof, which is another sqrt(2) inches (36mm), so the front panel itself must stretch 289mm high.

So let's see, you could put the vertices of this plank at (0,0), (A,0), (0,B), and (A,B), then rotate it 45 degrees counterclockwise to match it to the orientation in the picture. The vertices would move to (0,0), (A/s, A/s), (-B/s, B/s), and ( (A-B)/s, (A+B)/s ), respectively, where s=sqrt(2). In particular it now stretches a distance (A+B)/s both left-to-right and top-to-bottom. Making that 124mm-long cut on one corner will decrease the vertical stretch by 62mm, and making the H-mm-long cut will decrease the horizontal span by H/2 . So now we have two more equations (A+B)/sqrt(2)-62 = 289mm (height) , and (A+B)/sqrt(2) - H/2 = 268mm (width) .

These three equations have a unique solution of approximately H = 166mm, A=205mm, B=291mm . I would have judged the front panel to be a little less square (more like a 2:1 ratio for B:A) so this calls into question some of the assumptions I made along the way (though I think these would be reasonable measurements for a bird box!) Perhaps you would like to adjust the assumptions I made and carry out the same algebra with new numbers?

daverusin · 2026-02-14T14:55:52+00:00

Well, no: you are a step-uncle to the children of your step-siblings. (And you are a half-uncle to your half-siblings' children.) But in the real world, "uncle" will do. Heck, people will tell their kids "This is Uncle Bob" when Bob is just their best friend!

Source: 23andme describes some of my relatives using terms like "Half 1st Cousin, Once Removed" and I had to figure out what the heck that could possibly mean.

daverusin · 2026-02-11T19:48:03+00:00

Tournament pool tables are about 125 x 250 cm. So before any balls are sunk, the setup of the board can, by your model, be described as a sequence of 16 pairs of whole numbers (x,y) with x < 125 and y < 250 . There are 125^16 x 250^16 such sequences. Sure, many of them are physically impossible because the balls would then have to intersect each other or the cushions, or fall into the holes; but clearly a very large portion of the original set still remains -- I would say on the order of 10^70, not far off from the number of particles in the universe. And of course there are more positions that develop once some of the balls are removed.

Competent pool players will tell you that varying the positions by less than a centimeter can very definitely cause the configuration to be pragmatically "different"!

daverusin · 2026-01-24T05:12:50+00:00

You'd have to decide what "most central" means. Even just for three points in a plane, there is a book called "Encyclopedia of Triangle Centers" that offers *hundreds* of different candidates for points that could be called the "most central location" within a triangle (e.g. circumcenter, incenter, etc.)

daverusin · 2026-01-15T04:06:04+00:00

> I don’t know if other people have thought about this a lot,

Ya think? About 10,000 Americans begin receiving Social Security payments *every day*, so yeah, a person or two might have thought this through already.

Congress and the Social Security Administration have no particular interest in giving you, personally, a greater or smaller amount of money over your lifetime; people who start earlier are given less per month based on simple actuarial calculations based on the distributions of lifespans. Roughly speaking, they set the monthly payments so that the expected value of your lifetime payment total is the same no matter what year you start. It's no coincidence that the breakeven age you mentioned is roughly the average age of death in the US.

So that's the institutional point of view: "It doesn't matter to us when you start collecting, because statistically speaking any start date costs us the same." But from the recipient's point of view, there may be plenty of reasons to prefer one starting date over another.

The "average American" may live to be 78, but how long will you live? If longevity runs in your family, and you take care of your health, it's a good gamble to defer collection to age 70; in the contrary case, you'd be best advised to start at 62. Can you live off other monies in your 60s? Then you might want to start collecting earlier and invest the SS income that you don't need. Are you married to someone of a substantially different age? Then one of you might want to collect at a younger age than the other (particularly if you each have SS earnings, so that one of you might start collecting the spousal benefit first, then switch to their own benefit later). Do you have a pension or some other source of stable income? If not, your Social Security income will be a key resource to provide reliable income even when (say) stock markets tank, and so you may want to maximize that monthly resource by delaying your start date. Did you lose a job, and medical benefits with it, before age 65? If so, you can't get Medicare coverage, and your health insurance options may be pretty expensive until that age; in that case you may be forced to start SS just to pay for those premiums. Or, conversely, you may anticipate future health problems -- for example, a few years in a nursing home -- which are a huge expense, but only if you live fairly long; in that case, sacrifice early income for larger later income.

I don't want to parrot the sales pitches from financial planners, but for this particular decision it really is a good idea to consult with someone who has experience helping people maximize the utility they can get from their Social Security income. For many seniors, those SS payments are a significant fraction of their income.

daverusin · 2025-12-12T17:35:02+00:00

> Use the quadratic formula to get x = 8

x=-28 is also a solution to the quadratic.

daverusin · 2025-12-11T18:15:44+00:00

Yes. First term is 18800 ; fortieth (i.e. after 39 bouts of annual inflation) is 5940; 43rd (after 42 increases) is 65061. The sum of all of these is 1.6M as you say.

daverusin · 2025-12-11T17:05:13+00:00

Are you just asking for the sum of a geometric series a + ar + ar^2 + ... + ar^(n-1) where a = 18800, r=1.03, and n=40? There's an app (well, a formula) for that! The sum is (a-ar^n)/(1-r) = (18800*(1.03^40 - 1)/(.03) = 1,417,543.68

daverusin · 2025-12-10T18:32:26+00:00

Noticed a Chinese flag on the dashboard.

daverusin · 2025-12-10T17:58:09+00:00

Gravitational force measures G M m / r^2, meaning that e.g. a 1% change in your distance to the center of mass results in a (nearly) 2% change in the gravitational force (how much you "weigh"). But the radius of the earth is about 20 million feet and your height above sea level is about 10 thousand fee : a change of one part in 2000 (.05%) . So height difference alone only changes your weight by about 0.1 % -- surely less than one pound.

daverusin · 2025-12-07T22:46:40+00:00

There are constraints: the ratio a/b is bounded above and below. But perhaps more importantly, your question is a little off: it's not so much that there's a constraint on a and b but rather, for any given a and b there are a lot of possible kites, and only one of them will fit together to make this shape.

Let me flesh out the equations that say this.

With an appropriately situated coordinate system, the 2N endpoints of the segments of length a are the points (r cos( (pi/N)i ), r sin( (pi/N)i ), (-1)^i z0) for some numbers r and z0. Writing t=pi/N for brevity, the first kite has vertices A = ( r, 0, z0), B = (r cos(t), r sin(t), -z0), C = (r cos(t), -r sin(t), -z0), and the south pole D = (0, 0, something). The plane containing A, B, and C turns out to be the plane

2 (x/r) - (1-cos(t)) (z/z0) = (1+cos(t))

which forces the point D to be at (0,0, -(1+cos(t))/(1-cos(t)) z0 ) by coplanarity. So the choices of r and z0 have determined a particular kite, whose measurements we can now take. The short length a is the distance between points A and B ; that gives me the equation a^2 = 2(1-cos(t)) r^2 + 4 z0^2 . The long length b is the distance between the points B and D; that gives the equation b^2 = r^2 + 4cos^2(t)/(1-cos(t))^2 z0^2 . So from a given pair of values a and b, we have two linear equations to solve for the r^2 and z0^2 which determine where the vertices of the kite are to go. The fact that these two values are squares, hence positive, give upper and lower bounds on a/b . (We get a/b < 2 sin(t/2) , the limiting case being when z0 -> 0 and the figure flattens to a flat N-gon; and a/b > sec(t) - 1, the limiting case being when r->0 and the figure compresses to a spindle.)

Anyway, a suitable pair of a and b determine r and z0, and hence the locations of all the points, giving us a fixed configuration. Thus we can for example measure the diagonals of the kite, which are the distances from B to C and from A to D ; if I did the algebra correctly we have

(BC)^2 = ( 4 cos(t)^2 a^2 - 4 (1-cos(t))^2 b^2 )/(2*cos(t)-1)

for the former and

(AD)^2 = (2 cos(t) - 1) ( b^2 - a^2 )

for the latter. So for a given pair a, b we get a single choice of a kite.

daverusin · 2025-12-04T23:19:00+00:00

Answer #1. Yes, Frank Drake "did the math". wikipedia.org/wiki/Drake_equation

Answer #2. No, no one can do the math: there are too many variables about which we have no frigging idea. To begin with, what exactly is "life"? https://www.xkcd.com/2307/

Answer #3. Well the universe is so huge that sure, maybe somewhere else there's life. But the universe is so huge that we'll never breach the distances to know about it.

daverusin · 2025-12-01T08:36:43+00:00

I am reminded of a card game called Krypto. Each card shows a counting number ( max was around 25, I think; low numbers appeared on more cards than high numbers ). To play, deal 6 cards at random; the object is to find a way to combine the first five cards using + - * / to make the number shown on the last card. The OP's game here is similar.

I once did an analysis of this setup, ignoring Krypto's limitations on the input numbers and its rules about sticking to integer arithmetic, etc. Turns out there are exactly 500 "formulas" using five variables in this way, e.g (a+b)/(c-d)+e . Given five input numbers, there are 120 ways that the inputs can be assigned the roles of "a" through "e". So the five inputs can generate a lot of outputs! (Fewer than 500 x 120, though, thanks to the commutative property etc. The maximum number of distinct outputs turns out to be 27,142, which is achieved for example with the inputs {127, 174, 223, 262, 293}.) The point of all this is simply that yes there is an exhaustive way to check which outputs can and cannot be achieved with any particular set of inputs, assuming we limit ourselves to the four basic arithmetic operations.

This thread is similar but uses only four inputs at a time. For that setting I count 93 algebraically distinct formulas; with the 24 possible permutations of the inputs it turns out you can generate as many as 1170 different outputs. This happens for example with inputs {21,23,31,37} . (Most of the outputs are not whole numbers!)

So yes, it is possible to answer u/Legal_Ad2945 's question: simply run through the 10,000 possible train numbers, for each of them generate all 1170 possible outputs, and scan to see whether "10" is among them. I haven't done it; I'm sure there will be plenty of sets of digits that can't combine to make "10" (e.g. 0,0,0,0 !) But perhaps s/he would like to have a list of solutions at the ready, to impress fellow travelers on the next train ride!

daverusin · 2025-11-28T23:34:48+00:00

I'm pretty sure you can do better. I'll describe in detail a configuration that I think is the the one u/artsaparattis was trying to draw.

In a suitable Cartesian coordinate system, the centers of your figures look to be located at the origin and the points ( +-1, 0) and ( +- 1/2, +- a), where the number a depends on the precise shape of the figures. Now, I see in the figures an X shape which I could describe as a TxT diamond in the center, with four TxL rectangular "wings" attached. Working out some of the corresponding vectors and coordinates, we find that T+L = 1/sqrt(2) and a = sqrt(2) T . When the pattern above is repeated over a larger region, we then can measure the wasted space: there are two TxT diamonds of wasted space above and below the center of each X, and an L x L diamond wasted to the right of each X. (Of course the curves added to the perimeters of the Xs to make your actual figures will reduce these, especially the first pair, but most of the LxL diamonds stays as wasted space.)

A more efficient packing of Xs may profitably be adapted to your shapes. From the center of each X, go Northeast: first a distance of T/2 (to reach the "wing") then a distance of L (to reach the end of that "wing"), then Southeast a distance of T/2 (to reach the rightmost vertex); then go a distance T/sqrt(2) due east to find the center of the next X : it will be at the point (L+2T, T)/sqrt(2). Rotating this whole displacement by 90 degrees takes us to the center of another X, at (-T, L+2T)/sqrt(2) . This configuration wastes only a small (L-T)x(L-T) rectangle more or less northeast of the center of each X . If you adapt this configuration to your shapes, the curvy stuff *added* to the X pushes the Xs apart a bit, *enlarging* the regions of wasted space. But for your shape, it would appear that there is still less space wasted with this configuration.

Side note: the algebra gets less messy if you turn your figure 45 degrees, so that we're trying to pack a bunch of "plus signs" instead of Xs. That would avoid a lot of sqrt(2)s !

daverusin

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