987654321 / 123456789

Posted by ColinWright 10/26/2025

987654321 / 123456789(www.johndcook.com)

637 points | 107 comments

Joker_vD 10/30/2025|

Somewhat interesting, 123456789 * 8 is 987654312 (the last two digits are swapped). This holds for other bases as well: 0x123456789ABCDEF * 14 is 0xFEDCBA987654312.

Also, adding 123456789 to itself eight times on an abacus is a nice exercise, and it's easy to visually control the end result.

andyjansson 10/30/2025||

Another interesting thing is that these seem to work:

base 16: 123456789ABCDEF~16 * (16-2) + 16 - 1 = FEDCBA987654321~16

base 10: 123456789~10 * (10-2) + 10 - 1 = 987654321~10

base 9: 12345678~9 * (9-2) + 9 - 1 = 87654321~9

base 8: 1234567~8 * (8-2) + 8 - 1 = 7654321~8

base 7: 123456~7 * (7-2) + 7 - 1 = 654321~7

base 6: 12345~6 * (6-2) + 6 - 1 = 54321~6

and so on..

or more generally:

base n: sequence * (n - 2) + n - 1

madcaptenor 10/30/2025||

This is in the original post, in the form

  num(b)/denom(b) = b - 2 + (b-1)/denom(b)

so you just need to clear the denominator.

monkpit 10/30/2025|||

> the last 2 digits are swapped

They are also +9 away from being in order.

And then 12345678 * 8 is 98765424 which is +9 away from also being in order.

mechanicalpulse 10/31/2025|||

I also went about looking at the difference rather than the order. In the hexadecimal case, the difference is 15 (0xEF vs 0x12). I thought, then, that for any base B with ascending digits A and descending digits D, (D-(B-1))/A=B-2.

For binary, it looks like (1-(b-1))/1=b-10 or (1-(2-1))/1=2-2=0 in decimal.

For trinary, it looks like (21-(b-1))/12=b-2 or (7-(3-1))/5=5/5=1 in decimal.

For quaternary, it looks like (321-(b-1))/123=b-2 or (57-(4-1))/27=54/27=2 in decimal.

Essentially and perhaps unsurprisingly, the size of the slices in the number pie get smaller the bigger the pie gets. In binary, the slice is the pie, which is why the division comes out to zero there.

monkpit 10/31/2025|||

Oops - the second one was supposed to say +8

debtta 10/31/2025||

On an 8 digit calculator the common variant of this was

    12345679 * 8 = 98765432

jedberg 10/30/2025||

This was by far the most interesting part to me. I've never considered that code and proofs can be so complementary. It would be great if someone did this for all math proofs!

"Why include a script rather than a proof? One reason is that the proof is straight-forward but tedious and the script is compact.

A more general reason that I give computational demonstrations of theorems is that programs are complementary to proofs. Programs and proofs are both subject to bugs, but they’re not likely to have the same bugs. And because programs made details explicit by necessity, a program might fill in gaps that aren’t sufficiently spelled out in a proof."

TheTon 10/30/2025||

As a kid, I was marginally decent at competitive math. Not good like you think of kids who dominate those type of competitions at a high level, but like I could qualify for the state competition type good.

What I was actually good, or at least fast at, was TI-Basic, which was allowed in a lot of cases (though not all). Usually the problems were set up so you couldn’t find the solution using just the calculator, but if you had a couple of ideas and needed to choose between them you could sometimes cross off the wrong ones with a program.

The script the author gives isn’t a proof itself, unless the proposition is false, in which case a counter example always makes a great proof :p

sockaddr 10/31/2025||

I used to do the same thing. I'd scan for problems on the test amenable to computational approaches and either pull up one of my custom made programs or write one on the spot and let it churn in the background for a bit while I worked on other stuff without the calculator.

layer8 10/30/2025|||

This is misleading in that the (Curry–Howard) correspondence is between proofs and the static typing of programs. A bug in a proof therefore corresponds to a bug in the static typing of a program (or to the type system of the programming language being unsound), not to any other program bug.

(Also: complementary != complimentary.)

Sharlin 10/30/2025|||

I don't think the author is referring to the C–H correspondence. Just the fact that

a) it can be actually helpful to check that some property holds up to one zillion, even though it's not a proof that it holds for all numbers; and

b) if a proof has a bug, a program checking the relevant property up to one zillion is not unlikely to produce a counterexample.

jedberg 10/30/2025||||

> Also: complementary != complimentary

I'm gonna blame autocorrect for that one, but appreciate you catching it. Fixed! :)

travisjungroth 10/30/2025||||

The point is to not be so tight, leaning on the correspondence. The fact that you’re coming at the problem differently (even that it’s a different problem, “for some” versus “for all”) is actually helpful. You’re less likely to make the same mistake in both.

There’s a technique for unit testing where you write the code in two languages. If you just used a compiler and were more confident about correspondence, that would miss the point. The point is to be of a different mind and using different tools.

nh23423fefe 10/30/2025|||

i think this is wrong. code is proofs, types are propositions

CamperBob2 10/30/2025|||

Code is proof that the operation embodied by the code works. I don't understand how it proves anything more generally than that, apart from code using exotic languages or techniques intended for just that purpose.

sigbottle 10/30/2025||

Well, in theory (and I guess more generally philosophy) land, sure, you can't really "prove absoluteness" outside of your axioms and assumptions. You need to have a notion of true and false, and then implications, for example, to do logic, then whatever the leap from there it takes to do set theory, then go up from there etc. it's turtles all the way down.

In practice land (real theorem provers), I guess the idea is that, it theoretically should be a perfect logic engine. Two issues:

1. What if there's a compiler bug?

2. How do I "know" that I actually compiled "what I meant" to this logic engine?

(which are re-statements of what I said in theory land). You are given, that supposedly, within your internal logic engine, you have a proof, and you want to translate it to a "universal" one.

I guess the idea is, in practice, you just hope that slight perturbations to either your mental model, the translation, or even the compiler itself, just "hard fail". Just hope it's a very not-continuous space and violating boundaries fail the self-consistency check.

(As opposed to, for example, physical engineering, which generally doesn't allow hard failure and has a bunch of controls and guards in mind, and it's very much a continuuum).

A trivial example is how easy it is to just typo a constant or a variable name in a normal programming language, and the program still compiles fine (this is why we have tests!). The idea is, that, down from trivial errors like that, all the way up to fundamental misconceptions and such, you can catch preturbations to the ideal, I guess, be they small or large. I think what makes one of these theorem provers minimally good, is that you can't easily, accidentally encode a concept wrong (from high level model A to low level theorem proving model B), for a variety of reasons. Then of course, runtime efficiency, ergonomics etc. come later.

Of course, this brings into notion just how "powerful" certain models bring - my friend is doing a research project with these, something as simple as "proving a dfs works to solve a problem" is apparently horrible.

layer8 10/30/2025|||

The types are the propositions proved by the proof. The proof is correct <=> the program is soundly typed.

travisjungroth 10/30/2025|||

I came here to quote that entire section as well I’m glad I checked the comments first.

I’ve never seen a more succinct explanation of the value of coding up scripts to demonstrate proofs.

I think I’ll tighten it up to “proofs have bugs” in the future.

Me001 10/30/2025||

[dead]

gus_massa 10/30/2025||

The other replies are good, but let's add another one anyway.

0.987654321/0.123456789 = (1.11111111-x)/x = 1.11111111/x - 1 where x = 0.123456789

You can aproxímate 1.11111111 by 10/9 and aproxímate x = 0.123456789 using y = 0.123456789ABCD... = 0.123456789(10)(11)(12)(13)... that is a number in base 10 that is not written correctly and has digits that are greater than 9. I.E. y = sum_i>0 i/10^i

Now you can consider the function f(t) = t + 2 t^2 + 3 t^3 + 4 t^4 + ... = sum_i>0 i*t^i and y is just y=f(0.1).

And also consider an auxiliary function g(t) = t + t^2 + t^3 + t^4 + ... = sum_i>0 1*t^i . A nice property is that g(t)= 1/(1-t) when -1<t<1.

The problem with g is that it lacks the coefficients, but that can be solved taking the derivative. g'(t) = 1 + 2 t + 3 t^2 + 4 t^3 + ... Now the coefficients are shifted but it can be solved multiplying by t. So f(t)=t*g'(t).

So f(t) = t * (1/(1-t))' = t * (1/(1-t)^2) = t/(1-t)^2

and y = f(0.1) = .1/.9^2 = 10/81

then 0.987654321/0.123456789 ~= (10/9-y)/y = 10/(9y)-1 = 9 - 1 = 8

Now add some error bounds using the Taylor method to get the difference between x and y, and also a bound for the difference between 1.11111111 an 10/9. It shoud take like 15 minutes to get all the details right, but I'm too lazy.

(As I said in another comment, all these series have a good convergence for |z|<1, so by standards methods of complex analysis all the series tricks are correct.)

debtta 10/31/2025|

An easier way to evaluate sum i/10^i is by squaring sum 1/10^i

If you multiply term by term every term has coefficient 1 of course. There are n terms with exponent n+1, made from the n sums of the first exponent and the second exponent.

Eg 1+5, 2+4, 3+3, 4+2, 5+1.

So (1/9)^2 = (sum 1/10^i)^2 = 1/10 sum i/10^i

The derivative trick is more useful generally, but this method gets you the solution to 0.12345678.. in an quick way that's also easier to justify that it works.

bobbylarrybobby 10/30/2025||

Let's prove it.

In general, sum(x^k, k=1…n) = x(1-x^n)/(1-x).

Then sum(kx^(k-1), k=1…n) = d/dx sum(x^k, k=1…n) = d/dx (x(1-x^n))/(1-x) = (nx^(n+1) - (n+1)x^n + 1)/(1-x)^2

With x=b, n=b-1, the numerator as defined in TFA is n = sum(kb^(k-1), k=1…b-1) = ((b-2)b^b + 1)/(1-b)^2 = ((b-2)b^b + 1)/(1-b)^2.

And the denominator is:

d = sum((b-k)b^(k-1), k=1..b-1) = sum(b^k, k=1..b-1) - sum(kb^(k-1), k=1..b-1) = (b-b^b)/(1-b) - n = (b^b - b^2 + b - 1)/(1-b)^2.

Then, n-(b-1) = (b^(b+1) - 2b^b - b^3 + 3b^2 - 3b +2)/(1-b)^2.

And d(b-2) = the same thing.

So n = d(b-2) + b - 1, whence n/d = b-2 + (b-1)/d.

We also see that the dominant term in d will be b^b/(1-b)^2 which grows like b^(b-2), which is why the fractional part of n/d is 1 over that.

I disagree with the author that a script works as well as a proof. Scripts are neither constructive nor exhaustive.

jph00 10/30/2025||

The author does not say a script works as well as a proof.

vatsachakrvthy 10/30/2025||

If you want to be lazier, after finding the generating functions one can plug into sympy to skip the algebra.

tetris11 10/30/2025||

I like calculator quirks like this. I remember as a kid playing with the number pad and noticing a geometric center of mass in number sequences

    ┌───┬───┬───┐
    │ 7 │ 8 │ 9 │
    ├───┼───┼───┤
    │ 4 │ 5 │ 6 │
    ├───┼───┼───┤
    │ 1 │ 2 │ 3 │
    ├───┼───┼───┤
    │ 0 │ . │   │
    └───┴───┴───┘

I remember seeing that (14787 + 36989) / 2 would produce 25888, in that the mean of geometric shape traced by the two sequences would average out in the middle like that

nonethewiser 10/30/2025||

The even simpler example is more striking imo.

(147 + 369) / 2 = 258

and

(741 + 963) / 2 = 852

anvuong 10/30/2025|||

But this is obvious?

(741 + 963)/2 = (700+900)/2 + (40+60)/2 + (1+3)/2, it's just average in each decimal place.

jvanderbot 10/30/2025||

Obvious now and really cool in hindsight.

tempodox 10/30/2025|||

The decimal digits clearly have a conspiracy going on.

deepsun 10/30/2025||

That would work in any base, I even think we would find way more interesting coincidences in base 12 (as Sumerians preferred), because it's divisible by 2,3,4,6.

It's unfortunate that we have 5 fingers.

kapitar 10/31/2025||

If you count the sections of your four fingers with your thumb, you can count up to 12 on one hand!

ghoul2 10/31/2025|||

I have always counted to 20 on one hand. even as a kid. base, lower joint, upper joint, top. times 5 - including the thumb: my motor memory is trained so that i switch seamlessly from keeping the curse on top of the finger using my thumb, and then, once i cross 16, switch to using the index finger to "cursor" the thumb.

susam 10/31/2025|||

Same here. I have always counted 20 on one hand, so 40 with both. That's how my parents taught me to count when I was little. I used this method so often as a kid that, even though I don't count like this anymore, every number up to 40 still has its own place on my fingers.

It was only as an adult that I realised nobody around me counted this way. You are the first person I have found who talked about this method, so I am glad to find this comment of yours.

kevindamm 10/31/2025||||

If you count with each finger as a binary digit you can count up to 15 on one hand!

255 if you use both hands!

More like 1023 if you also use thumbs but I prefer to use them as carry, overflow bits.

soulofmischief 10/31/2025||

I trained myself to do this by default a very long time ago and I can't imagine counting any other way.

It's so natural, useful and lends well to certain numerical tricks. We should explicitly be teaching binary to children earlier.

BrandoElFollito 11/1/2025||||

I am French and we cont extending our fingers from a closed fist. Typically to 2x5=10.

When I was a kid I relized that I can count the fives on the right hand (1 finger for each 5 on the left), which brought me to 25.

It is only when I was traveling in Asia and watched people on markets that I realized that I can use my thumb to count my 12 other finger phalanges, which brought the total to 144. You just need to know your multiplication table of 12 :)

euroderf 11/1/2025|||

A system I read about uses the thumb for 5, so that each hand can count (thumb down) 0..4 (thumb up) 5..9.

This gives you the range 0..99. Sweeet.

nasvay_factory 10/30/2025|||

i remember the 1110 thing on a calc as well.

741 + 369 & 963 + 147 | 123 + 987 & 321 + 789 (left right | up down)

159 + 951 & 753 + 357 | 258 + 852 & 456 + 654 (diagonally | center lines)

the design of a keypad... it unintentionally contains these elegant mathematical relationships.

i call this phenomena: outcomes of human creations can be "funny and odd", and everybody understand that eventually there will be always something unpredictable.

sim7c00 10/30/2025||

ita intuited knowledge, which is only later understood. (itzhak bentov, stalking the wild pendulum)

nashashmi 10/30/2025|||

14789 + 36987 / 2 would do the same thing. Why trace back?

dfee 10/30/2025|||

So would 147 and 369. As it’s just an average, per digit, I’m not sure this is very interesting.

gowld 10/30/2025||

Being curious is delightful.

huflungdung 10/30/2025||

[dead]

tetris11 10/30/2025|||

Just to show that you could - 14861 and 36843 gives 25852

bitwize 10/30/2025|||

Great, now I'm getting Carrot Top flashbacks. "Dial right down the center of the phone!"

For non-Americans and/or those too young to remember when landline service was still dominant, in the 90s and early 2000s AT&T ran a collect-call service accessible through the number 1-800-CALL-ATT (1-800-225-5288) and promoted it with ads featuring comedian Carrot Top. And if you don't know who Carrot Top is, maybe that's for the best.

firefax 10/31/2025|||

now there's some solid ascii, great work sir

a13n 10/31/2025|||

how did you submit this table in HN??

_Microft 10/31/2025||

Use two or more spaces at the beginning of a line and it will be formatted as code ("<code>") and then use symbols as you like.

  ┌──────╖
  │  OK  ║
  ╘══════╝

https://news.ycombinator.com/formatdoc

BoorishBears 10/31/2025||

  ┌──────────╖
  │  CANCEL  ║
  ╘══════════╝

(for posterity)

alyxya 10/30/2025||

I like to think of 0.987654... and 0.123456... as infinite series which simplify to 80/81 and 10/81, hence the ~8 ratio.

oersted 10/30/2025||

I didn't get where this comes from until I saw the second answer from the StackOverflow question another commenter shared.

https://math.stackexchange.com/a/2268896

Apparently 1/9^2 is well known to be 0.12345679(012345679)...

EDIT: Yes it's missing the 8 (I wrote it wrong intially): https://math.stackexchange.com/questions/994203/why-do-we-mi...

Interesting how it works out but I don't think it is anywhere close to as intuitive as the parent comment implies. The way its phrased made me feel a bit dumb because I didn't get it right away, but in retrospect I don't think anyone would reasonably get it without context.

alyxya 10/30/2025|||

It actually skips the 8 in its repeating decimal. It’s better to think of 1/9^2 as the infinite sum of k * 10^-k for all positive integers k. The 8 gets skipped because you have something like ...789(10)(11)... where the 1 from the “10” and “11” digits carry over, increment the 9 digit causing another carry, so the 8 becomes a 9.

dpacmittal 10/30/2025||||

Also 12345679*x*9 = xxxxxxxxx

Eg 12345679*6*9 = 666666666

madcaptenor 10/30/2025||

I think your formatting is off.

iso1631 10/30/2025|||

9^2 is 81

1/81 is 0.012345679012345679....

no 8 in sight

madcaptenor 10/30/2025||

The 8 is there but then it's followed by a 9 and a 10, and the carry from the 10 ends up bumping it up.

zkmon 10/30/2025|||

Shouldn't wee see two zeros then?

madcaptenor 10/30/2025|||

The reason you don't see two zeroes is as follows: you have

  .123456789

then add 10 on the end, as the tenth digit after the decimal point, to get

  .123456789(10)

where the parentheses denote a "digit" that's 10 or larger, which we'll have to deal with by carrying to get a well-formed decimal. Then carry twice to get

  .12345678(10)0

  .1234567900

So for a moment we have two zeroes, but now we need to add 11 to the 11th digit after the decimal point to get

  .1234567900(11)

or after carrying

  .12345679011

and now there is only one zero.

zkmon 10/30/2025||

Ah, that's cool. Thanks!

oersted 10/30/2025|||

This illustrates it nicely: https://math.stackexchange.com/a/994214

Stolpe 10/30/2025||

Care to elaborate? Why does 0.987654 simplify to 80/81 and 0.123456 to 10/81?

madcaptenor 10/30/2025|||

.123456... = x + 2 x^2 + 3 x^3 + ... with x = 1/10.

Then you have (x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...) + (x^3 + x^4 + x^5 + x^6 + ...) (count the number of occurrences of each power of x^n on the right-hand side)

and from the sum of a geometric series the RHS is x/(1-x) + x^2/(1-x) + x^3/(1-x) + ..., which itself is a geometric series and works out to x/(1-x)^2. Then put in x = 1/10 to get 10/81.

Now 0.987654... = 1 - 0.012345... = 1 - (1/10) (10/81) = 1 - 1/81 = 80/81.

gowld 10/30/2025|||

Don't need the clutter of infinite series and polynomials:

    1/9 = 0.1111...

    1/81 = 1/9 * 1/9 = 0.111... * 0.111... =

    Sum of:
       0.0111...
       0.00111...
       0.000111...
       ...
    
    =  0.012345...

GuB-42 10/30/2025|||

Isn't it essentially the same thing, but less formal

0.1111... is just a notation for (x + x^2 + x^3 + x^4 + ...) with x = 1/10

1/9 = 0.1111... is a direct application of the x/(1-x) formula

The sum of 0.0111... + 0.00111... ... = 0.012345... part is the same as the "(x + 2 x^2 + 3 x^3 + ...) = (x + x^2 + x^3 + x^4 + ...) + (x^2 + x^3 + x^4 + x^5 + ...)" part (but divided by 10)

And 1/81 = 1/9 * 1/9 ... part is the x/(1-x)^2 result

madcaptenor 10/30/2025|||

This is better than my answer, at least if you can get your brain to interpret it in base b. In that case the first two lines would become

  1/(b-1) = 0.1111...
  1/((b-1)^2) = 1/b * 1/b = 0.111... * 0.111... =

gus_massa 10/30/2025|||

I don't know who downvoted this, but it's correct.

The use of series is a little "sloppy", but x + 2 x^2 + 3 x^3 + ... has absolute uniform convergence when |x|<r<1, even more importantly that it's true even for complex numbers |z|<r<1.

The super nice property of complex analysis is that you can be almost ridiculously "sloppy" inside that open circle and the Conway book will tell you everything is ok.

[I'll post a similar proof, but mine use -1/10 and rounding, so mine is probably worse.]

alyxya 10/30/2025|||

If you set x = 0.123456..., then multiplying it by (10 - 1) gives 9x = 1.111111..., and multiplying it by (10 - 1) again gives 81x = 10, or x = 10/81. I’m not writing things formally here but that’s the rough idea, and you can do the same procedure with 0.987654... to get 80/81.

jamesmaniscalco 10/30/2025||

This is fun! but not so surprising to me:

987,654,321 + 123,456,789 = 1,111,111,110

1,111,111,110 + 123,456,789 = 1,234,567,899 \approx 1,234,567,890

So 987,654,321 + 2 x 123,456,789 \approx 10 x 123,456,789

Thus 987,654,321 / 123,456,789 \approx 8.

If you squint you can see how it would work similarly in other bases. Add the 123... equivalent once to get the base-independent series of 1's, add a second time to get the base-independent 123...0.

kazinator 10/30/2025||

Here is a correction which m akes it exactly 8.0:

  > 987654320 / 123456790
  8.0

I've decremented the numerator and incremented the denominator:

   ( 987654321 - 1 )
   -----------------  = 8
   ( 123456789 + 1 )

Works in other bases. TXR Lisp, base 4:

  1> (/ (poly 4 '(3 2 1)) (poly 4 '(1 2 3)))
  2.11111111111111
  2> (/ (poly 4 '(3 2 0)) (poly 4 '(1 2 4)))
  2.0

It also works for base 2, which is below the lowest base used in the article: the Python code goes from 3.

For base 2, the ratio is 1/1. When we apply the correction, we get (1 - 1) / (1 + 1) = 0, which is 2 - 2.

throw0101c 10/30/2025||

See perhaps various "What every programmer / CSist should know about floating-point arithmetic" papers and articles:

* David Goldberg, 1991: https://dl.acm.org/doi/10.1145/103162.103163

* 2014, "Floating Point Demystified, Part 1": https://blog.reverberate.org/2014/09/what-every-computer-pro... ; https://news.ycombinator.com/item?id=8321940

* 2015: https://www.phys.uconn.edu/~rozman/Courses/P2200_15F/downloa...

MrOrelliOReilly 10/30/2025|

As someone who has recently been fighting bugs from representing very simple math with floats... thank you!

msuvakov 10/30/2025|

Why the b > 2 condition? In the b=2 case, all three formulas also work perfectly, providing a ratio of 1. And this is interesting case where the error term is integer and the only case where that error term (1) is dominant (b-2=0), while the b-2 part dominates for larger bases.

listeria 10/30/2025|

in the b=2 case, you get:

  1 / 1 = 1 = b - 1
  1 % 1 = 0 = b - 2

they are the other way around, see for example the b=3 case:

  21 (base 3) = 7
  12 (base 3) = 5
  7 / 5 = 1 = b - 2
  7 % 5 = 2 = b - 1

gowld 10/31/2025||

In the b=2 case, 1/1 = 1 = (b-2) + (b-1)/denom(b) = (b-2) + (b-1)/1 = 2b - 3 = (b-1)*b^1 -1 (b-1)

In base 2 (and only base 2), denom(b) >= b-1, so the "fractional part" (b-1)/denom(b) carries into the 1's (units) place, which then carries into the 2's (b's) place, flipping both bits.

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