From Jan-Willem Tel, a puzzle of efficient breaking and entering:

You have a gate that requires a passcode to open. There are 10 buttons: the digits 0 to 9. You have forgotten the passcode, but you remember it has four digits. You have no choice but to try them all.

Since there are 10^4 = 10,000 four-digit passcodes, you might think this would take you 40,000 button presses to guarantee an opened gate. However, this gate’s keypad never resets: The gate opens as soon as the last four buttons you’ve pressed are the correct code, so you can be more efficient. For example, you can try two different codes by pressing just five buttons: The combination “12345” tries both “1234” and “2345.” Of course, pressed for time, you want to press as few buttons as possible while still trying different codes and eventually opening the gate.

So the question is: What’s the smallest number of buttons you need to press to make sure you open the gate — i.e., that you’ve tried every possible four-digit combination?

Extra credit:How do things change if you didn’t remember the passcode’s length?

I thought about this at the time, but didn’t come up with anything worth writing about. Having now seen the solutions I went back to it.

The site’s a little terse about the proof that the answer’s 10003. Let’s see if I can clarify it. Suppose you’re building up a string of digits to enter, you have say 3141 digits so far and you’re about to add the 3142nd. Call the last 4 digits of the string you already have *a, b, c, *and* d:*

*zyxwv*… (3133 digits) …*abcd*

and you’re going to add a new digit, call it *e*. What you want, if possible, is for the last four digits of the result, *bcde*, to be unique, not having occurred already. That is, you’d like each four digit combination to occur once and only once. That means you want each three digit combination to occur ten times, followed by a unique one of the ten digits. (Mostly. I’ll get to that.)

So when you go to add the 3142nd digit, will there be one available that will do that? Well, if *bcd* has already occurred ten times before this, with each of the ten digits following it, then no. But in that case the *bcd* occurring at the end is the 11th instance of *bcd*. Meaning the *abcd* at the end must be the second (or later) occurrence of that string. Meaning that if adding a digit is impossible without creating a duplicate, then a duplicate already exists. If there are no duplicates already, then you can add a digit without creating a duplicate. The only exception to this argument is if the string both starts and ends with *bcd*; then the terminating *bcd* can be the eleventh one without *abcd* being a duplicate, and then you can’t add a digit.

That means you can always add a digit without creating a duplicate four digit code, until you have a string that contains every code, and that string has to end with the same three digits you start with. Each digit from the fourth on ends a new unique code, of which there are 10,000, so the string you end up with will be 10,003 digits long.

Or 10^*n*+(*n*-1), for *n*-digit codes.

I wrote a Python script to generate these strings. It starts with a string of length *n* (all zeros for instance), then repeatedly tries to add a digit to that which will make a new code: first 0, then 1, then 2, and so on, until it succeeds, after which it goes back and tries to add another digit. For *n* = 2:

00

001 # 01 is new

0010 #10 is new

00102 #02 is new

001020 #20 is new

…

0010203040506070809 #09 is new

00102030405060708090 #90 is new

But here we’re stuck, because we’ve used 0 11 times already and we can’t add a new digit without creating a duplicate. The script’s not that smart, it just tries 0 through 9 and finds none of them work. So now it gets rid of the last digit in the string (the 0) and tries the next one:

00102030405060708091 #91 is new

001020304050607080911 #11 is new

0010203040506070809112 #12 is new

and so on. Obviously if it gets stuck with a string ending in 9 there isn’t a next digit to try instead, so it has to backtrack further. But eventually it finds the 101 digit solution

001020304050607080911213141516171819223242526272829334353637 38394454647484955657585966768697787988990

and it works the same way for larger values of *n*. If you’re skeptical the solution has to end with the same *n*-1 digits it starts with, here’s the 10,003 digit solution it finds for *n*=4, if it starts with the somewhat arbitrary four digit string “3141”:

314100001001100200021003000310040004101011101200121013001310 140014110201021103010311040104111120112111301131114011412020 212030203120402041212130213121402141303031304030413131403142 002200320042012201320142102210321042112211321142202220322042 212221322142300230123022303230423102311231223132314300330043 013301431033104311331143202320332043212321332143302330333043 312331333144002400340044012401340144102410341044112411341144 202420342044212421342144302430343044312431343145000500150025 003500450105011501250135014510051015102510351045110511151125 113511452005201520252035204521052115212521352145300530153025 303530453105311531253135314600060016002600360046010601160126 013601461006101610261036104611061116112611361146200620162026 203620462106211621262136214630063016302630363046310631163126 313631470007001700270037004701070117012701370147100710171027 103710471107111711271137114720072017202720372047210721172127 213721473007301730273037304731073117312731373148000800180028 003800480108011801280138014810081018102810381048110811181128 113811482008201820282038204821082118212821382148300830183028 303830483108311831283138314900090019002900390049010901190129 013901491009101910291039104911091119112911391149200920192029 203920492109211921292139214930093019302930393049310931193129 313931502050215030503151205121513051315220522152305231532053 215330533154005401541054115420542154305431550055015510551155 205521553055315600560156105611562056215630563157005701571057 115720572157305731580058015810581158205821583058315900590159 105911592059215930593160206021603060316120612161306131622062 216230623163206321633063316400640164106411642064216430643165 006501651065116520652165306531660066016610661166206621663066 316700670167106711672067216730673168006801681068116820682168 306831690069016910691169206921693069317020702170307031712071 217130713172207221723072317320732173307331740074017410741174 207421743074317500750175107511752075217530753176007601761076 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399609961996299639970997199729973998099819982998399909991999 299944445444644474448444944554456445744584459446544664467446 844694475447644774478447944854486448744884489449544964497449 844994545464547454845494555455645574558455945654566456745684 569457545764577457845794585458645874588458945954596459745984 599464647464846494655465646574658465946654666466746684669467 546764677467846794685468646874688468946954696469746984699474 748474947554756475747584759476547664767476847694775477647774 778477947854786478747884789479547964797479847994848494855485 648574858485948654866486748684869487548764877487848794885488 648874888488948954896489748984899494955495649574958495949654 966496749684969497549764977497849794985498649874988498949954 996499749984999555565557555855595566556755685569557655775578 557955865587558855895596559755985599565657565856595666566756 685669567656775678567956865687568856895696569756985699575758 575957665767576857695776577757785779578657875788578957965797 579857995858595866586758685869587658775878587958865887588858 895896589758985899595966596759685969597659775978597959865987 598859895996599759985999666676668666966776678667966876688668 966976698669967676867696777677867796787678867896797679867996 868696877687868796887688868896897689868996969776978697969876 988698969976998699977778777977887789779877997878797888788978 9878997979887989799879998888988998989999314

Yep, it ends with 314. Go ahead, look for the last four digits of your Social Security number, they’re there.

]]>

for and positive real numbers

can be tightened up on the left side. I used

but in fact this can be improved to

which leads to

.

For , the left side is , while for where , the left side is .

]]>Here’s an inequality which, if you’re like me, looks pretty mysterious:

for and positive real numbers.

I mean, the difference between and doesn’t “know” anything about or , right? And neither does their ratio. But the difference over the log of the ratio is always between and ? Weird!

(Okay, maybe you’re not so easily impressed, since if you know and you can get and . Fine. *I* think it’s mysterious anyway.)

Well, let’s consider a function with derivative monotonically decreasing (we could do increasing too, similarly). Let and be in the range where and exist and . Then

But that integral is the area under the curve from to and so

.

Then and or

.

But that makes sense given the definition of the derivative and the monotonic property of ; with a little thought you could’ve written this down directly. As obvious as it is, it takes a decidedly less obvious form if you specify , , which gives

or the formerly mysterious

QED.

I called the original post “Power trip” and this one “Power trip (part 2)”, and there haven’t been any powers here. Sorry. Here:

so

.

That form is… neither obvious nor interesting, though. Unless in which case the left part is

or

which is where we came in.

]]>Ha ha. No, seriously, I like it. A visual proof that .

But as Math with Bad Drawings pointed out someone else pointed out, the proof doesn’t depend on but can be generalized to any number , and as Math with Bad Drawings pointed out, a similar proof works for any positive number . That is, for all and , .

Proof here if you don’t want to click through to MwBD:

If ,

so

or

or

and similarly, if ,

so

or

or

QED

]]>

From Dave Moran, this is the final boarding call for flight RDLR 100:

Michelle decided to get some extra exercise at the airport by walking toward her departure gate at her normal pace, but on a lengthy moving walkway … going the wrong way. Gotta get those steps in.

After going 100 meters (that is, getting 100 meters closer to her departure gate), Michelle dropped her boarding pass onto the walkway. But she didn’t notice at first and continued walking toward her gate. After walking another 90 seconds, she finally realized that she had dropped her boarding pass. She then immediately turned around and jogged in the direction of the walkway’s movement. Michelle’s jogging pace is exactly twice as fast as her walking pace. She caught up with her boarding pass 10 meters from the start of the moving walkway.

How fast does the walkway move?

You could set this up as a complicated algebra problem, or you can be lazy enough to be clever.

Think about this from the boarding pass’s point of view. As far as it’s concerned it’s sitting still while Michelle walks at her normal walking speed away from it for 90 seconds, and then jogs back to it at twice that speed. She’s covering the same distance both times, so it takes her 45 seconds to return to where she dropped the pass after she turns around; that is, the pass is lying on the walkway for 135 seconds.

Now, from the airport’s point of view, the pass is moving at the speed of the walkway and it covers 90 meters in those 135 seconds. So that speed is 90 / 135 = 2/3 meters per second.

]]>

From Ben Gundry via Eric Emmet, find and replace with a twist:

Riddler Nation has been enlisted by the Pentagon to perform crucial (and arithmetical) intelligence gathering. Our mission: decode two equations. In each of them, every different letter stands for a different digit. But there is a minor problem in both equations.

In the first equation, letters accidentally were smudged on their clandestine journey to a safe room within Riddler Headquarters and are now unreadable. (These are represented with dashes below.) But we know that all 10 digits, 0 through 9, appear in the equation.

What digits belong to what letters, and what are the dashes?

In the second equation, our mathematical spies have said that one of the letters in the equation is wrong. But they can’t remember which one. Which is it?

First puzzle:

The smudged letter business seems to be an obvious red herring; all 10 digits are used, but 6 letters appear and 4 dashes, so each dash is a different letter and might as well be replaced by four different letters, say: . Right away we know .

and , so . But is or . So with and . .

Now , so or .

. The only possibility still available is and with = 8. The two digits left are and and those are indeed what and work out to respectively. The sum is .

Second puzzle:

We’ll assume everything’s okay until we hit a contradiction. By exactly the same logic as before , , . In column 7, so . Since in column 5 cannot be 1, there must be a carry into that column, meaning .

But in column 9, so which is not available.

If one of the letters column 5 is wrong then there could be no carry into that column, in which case we could have , and then . But from column 4 we’d still need neither of which is possible. So the problem lies elsewhere.

Maybe a letter in column 9 is wrong. Then , and . . If there’s a carry into column 8 then , which doesn’t work, so there’s no carry and means , . We’re left with which gives the incorrect sum ; if the first two numbers are right the sum should be 1369109876. Or if the first number is right and the sum is right the second number should be 673596653. Or if the second number is right and the sum is right the first number should be 695513263. We can’t determine which letter is incorrect.

This doesn’t rule out other possibilities, such as that one of the letters in column 7 or column 10 is wrong.

]]>5 = 3 + 2

7 = 5 + 2

12 = 7 + 5

17 = 12 + 5

and so on. Specifically each number is the sum of the immediately preceding one and one of the two numbers summed to make the immediately preceding one.

If you vary the generator by a small amount (and it’s still irrational) you get the same sequence to a point, and then it differs: for a musically significant example, a quarter comma tempered fifth, cents, the sequence is 2, 3, 5, 7, 12, 19… . Notice that the two numbers summed to make 12 are 7 and 5, and the two numbers summed to make 17 are 12, the immediately preceding one, and 5 which is one of the numbers in the sum for 12, while the two numbers summed to make 19 are 12 and 7, the *other* number in the sum for 12. Of course the sequence terminates if the generator is rational: for a generator of 700 cents, for example, it goes 2, 3, 5, 7; then the 8- through 11-note scales are all two-gap, and the 12-note scale is one-gap, i.e., an equal division of the octave. It stops there because further applications of the generator just give back notes you already have.

A generator for a 17-note equal division is cents and a generator for a 19-note equal division is cents. Generators between these two values (other than 700 cents) will give 2, 3, 5, 7, and 12 note two-gap scales, with ones below 700 going on to 19 and ones above 700 going on to 17. Generators a little outside that range will not give 12-note scales; they’ll go 2, 3, 5, 7, 9… if below the range or 2, 3, 5, 8, 13… if above. A diagram from which one can read off the 2-gap scales for generators from 685 to 720 cents is here, and for all generators from 600 to 1200 cents is here (the diagram for 0 to 600 cents is just a mirror image).

Stare at the first of those diagrams and you see 47 appears in two places. That is, the 47-note equal division can be generated by two different generators in that range: cents and cents.

In fact, looking at the whole range from 0 to 1200 cents, there must be 46 generators for a 47-note equal division: for . That’s because 47 is prime. For a 46-note equal division, though, will generate the scale only for odd ; even values will generate only a 23-note scale. And in general, the generators for an *m*-note equal division are for values of where and *n* and *m* are relatively prime. The familiar 12-note equal division has only four generators: , , , and cents.

Examine that diagram some more and you can see how it relates to a sequence of numbers developed as follows: Start with

5, 7

and interpolate the sum of the adjacent numbers

5, 12, 7

(in the diagram, 12 is linked to the horizontal lines associated with 5 and 7); and again

5, 17, 12, 19, 7

(17 is linked to 5 and 12, 19 to 12 and 7); and again

5, 22, 17, 29, 12, 31, 19, 26, 7

(22 is linked to 5 and 17, and so on); and again

5, 27, 22, 39, 17, 46, 29, 41, 12, 43, 31, 50, 19, 45, 26, 33, 7

ad infinitum. Those two values of 47 arise in later iterations of the this sequence from 27+5+5+5 and 33+7+7.

Likewise the big diagram and its mirror image relate to:

1, 1

1, 2, 1

1, 3, 2, 3, 1

1, 4, 3, 5, 2, 5, 3, 4, 1

1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1

et cetera. We can concatenate the rows of this, dropping the 1s from one end, into a single sequence:

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, …

which we can define with the recurrence relation:

This is known as Stern’s diatomic series, A002487 in OEIS.

From that definition it’s not at all obvious that, for instance, 47 will arise as a sum 46 times while 12 will arise as a sum only 4 times. It’s true though. In fact, consider the following sequence of rational numbers:

1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1…

That’s just each entry in Stern’s series divided by the subsequent entry. But there’s a theorem: *The nth rational number, in reduced form, can be taken to be b(n)/b(n + 1), for n = 0, 1, 2 … That is, b(n) and b(n + 1) are relatively prime, and each positive reduced rational number occurs once and only once in the list b(0)/b(1), b(1)/b(2), … *Or to put it another way, Stern’s series provides a way to enumerate the rational numbers.

The theorem’s proved in Calkin, N., & Wilf, H. (2000). Recounting the Rationals. *The American Mathematical Monthly, 107*(4), 360-363. doi:10.2307/2589182, if you want to look at it. Another fact about the series, also discussed there, is *b(n)* is the number of ways of writing the integer *n* as a sum of powers of 2, each power being used at most twice (i.e., once more than the legal limit for binary expansions).

Every positive integer *n *appears in the list surrounded by two smaller numbers, which are relatively prime and sum to *n*, exactly once for each such possible distinct sum (counting *m, n, n-m* and *n-m, n, m* as distinct). So one finds 1, 12, 11; 5, 12, 7; 7, 12, 5; and 11, 12, 1; and no other instances of 12 adjacent to smaller numbers. On the other hand, one finds all 46 of 1, 47, 46; 2, 47, 45; 3, 47, 44; … 45, 47, 2; and 46, 47, 1.

Mind you, you have to look through a lot of numbers to find them. After all, 1 appears adjacent to and at position in the series: 1 followed by 2 and 1 is at position 2, 1 then 3 and 2 at position 4, 1 then 4 and 3 at position 8, and so on. 1 followed by 47 and 46 appears at position (!) So if you want to figure out how many numbers are smaller than and relatively prime to 47, examining Stern’s series may not be the best approach.

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From Charlie Drinnan, find the letters’ numbers:

If A, B, C, D and E are all unique digits, what values would work with the following equation?

ABC,CDE × 4 = EDC,CBA

No problem. must be even, and , so . Then so , and so and , or and . But so .

Now so . so .

Finally or , so .

.

This is fairly mechanical although I tripped up several times on the way.

Let be the number of coconuts found by the th pirate, and be the number of coconuts found in the morning. is divisible by 7 so write .

But is what was left when the seventh pirate discarded one coconut out of and hid one seventh of the rest, so . For that to be an integer must be a multiple of 6, so write . Then .

Now . For integer value, , so write . Then .

You see how this goes.

. For integer value, , so write . Then .

. For integer value, , so write . Then .

. For integer value, , so write . Then .

. For integer value, , so write . Then .

. For integer value, , so write . Then . The smallest value corresponds to so .

These pirates were either very sleepy, or incredibly quick at counting coconuts.

]]>Let’s suppose we have three different gap sizes. For instance, the 8 note scale:Here the first note, the note we started with, is F, and the last note added is F♯. That means the Type I rigid gaps are FF♯ and F♯G (the yellow gap on the left, and the purple one immediately to its right, surrounding the last note). AB (the blue gap closest to the center) is a Type II rigid gap, because shifting it by *X* makes it coincide with EF♯, which isn’t a gap, it’s the two gaps surrounding the first note (EF and FF♯). Clearly that means the size of gap AB is the size of EF (which is the same as the size of rigid gap F♯G) plus the size of FF♯. So in this case the size of the largest gap is the sum of the sizes of the two smaller gaps.

But is that always true? Can you get a Type II gap that isn’t the same size as the sum of the gaps surrounding the first point? Or can you get a scale in which the two gaps surrounding the first point are both the size of one or the other of the Type I gaps?

Let’s try to look at all the possibilities. For the Type I (rigid) gaps:

- There are no Type I gaps
- There is only one Type I gap
- There are two Type I gaps, and they’re the same size
- There are two Type I gaps, and they’re different sizes

And for the Type II (rigid) gap:

- There is no Type II gap
- There is a Type II gap, and it is also a Type I gap
- There is a Type II gap, and it is not a Type I gap

Type I gaps surround the last note. If the gaps adjacent to the last note aren’t rigid, that means there’s a note *X* above the last note, which would have to be the first note, and the only way that can happen is if . That is, *X* has to be rational, of the form (in lowest terms). Then the *n* note scale divides the octave into *n* equal intervals, and we have a 1-gap scale. In that case there aren’t any Type II rigid gaps, either. Every gap coincides with a gap when shifted.

If there is no note *X* above the last note, then the gap to the left of the last note is rigid and so is the gap to the right, and there are two Type I gaps *unless those two gaps are one and the same*. In other words, it’s a one-note scale with one (one octave) gap. Which is a degenerate sort of equal division of the octave, so in fact there aren’t any rigid gaps. In other words, if it’s not an equal division of the octave (or a one-note scale), then there must be two distinct Type I gaps. They can be the same size, or not.

Now suppose there is no Type II gap. No gap contains the first note in its shifted interior. The only way that can happen is if there is a note *X* to the left of the first note, and that would have to be the last note. Again, this means it’s an equal division of the octave, and there are no Type I gaps either.

So suppose there is a Type II gap, but it is also a Type I gap. That is, one of its end notes is the last note. Then when shifted, that end will not coincide with a note. If the two Type I gaps are different sizes, then all we can say is we have a 2-gap scale, and there is no particular relationship between the two sizes.

Here gap AB (the blue gap closest to the center) is a Type II rigid gap: if you shift it, it goes from E to where F♯ would be if there were an F♯, and it contains F. It’s also one of the Type I rigid gaps, since B is the last note; the other is BC. So it’s a 2-gap scale.

If the Type II gap is also Type I, and the two Type I gaps are the same size, well… then obviously we have a 1-gap scale, which means an equal division of the octave, but that has no rigid gaps at all, so by contradiction that case is impossible.

Finally, suppose there is a Type II gap, and it is not a Type I gap. That means there’s a note *X* to the right of each of its end notes, so when you shift the gap it’ll coincide with two gaps, the two surrounding the first note. The Type II gap is the sum of the gaps surrounding the first note. One of the two gaps surrounding the first note has its “older” end note (namely the first note) on the right end, the other has the older note on the left end. If you shift the first of these as many times as you can, the “newer” note on the left end reaches the last note first, so this gap matches up with the Type I rigid gap in which the last note is on the left end. But if you shift the second one, the “newer” note on the right end reaches the last note first, so that gap matches up with the Type I rigid gap in which the last note is on the right end. So if the two Type I gaps are different sizes, then so are the two gaps surrounding the first note, and we have a 3-gap scale in which the Type II rigid gap is the same size as the sum of the sizes of the two Type I gaps.

The only way the two gaps surrounding the first note can be the same size is if both Type I gaps are the same size, in which case the Type II gap is exactly twice that size and it’s a 2-gap scale. That would happen, for instance, if instead of ~702 cents, *X* were 700 cents, and you generated an 8-note scale. It would look like the 8-note scale above, except the gaps surrounding the first note both would be 100 cents in size, and so would the gaps surrounding the last note.

Summing up:

- If there are no Type I gaps, there are no Type II gaps, and vice versa; we have a 1-gap scale
- If there is a Type II gap, and it is also a Type I gap, then the two Type I gaps are different sizes, and we have a 2-gap scale, with no particular relationship between the two sizes
- If there is a Type II gap, and it is not a Type I gap, and the Type I gaps are the same size, we have a 2-gap scale with the large gap twice the small gap
- If there is a Type II gap, and it is not a Type I gap, and the Type I gaps are not the same size, we have a 3-gap scale with the large gap the sum of the small gaps

So there you go. I’ve framed this discussion in terms of musical scales, putting notes within an octave, at positions ranging from 0 to 1200 cents, but of course this all could be restated in terms of putting points on a unit line segment, at positions from 0 to 1, or on a circle, at angular positions from 0 to 2π. The Wikipedia article says in the latter form it has applications to phyllotaxis, although I’m not sure it has anything significant to say in that field. The theorem also has applications in the theory of Sturmian words, it says, and if I ever come to grips with what Sturmian words are and why one would care, maybe I’ll write about them here… don’t hold your breath, though…

]]>Here’s the 6-note scale again:

We’re going to look for gaps that are “rigid”, by which we mean this: A rigid gap is one that, if shifted (right) by *X*, does not coincide with another gap.

Think about the gap from F to G (the leftmost blue gap). If you shift it (right) by *X*, you get the gap from C to D. Shift gap CD by *X* and you get the gap GA; shift that by *X* and you get the gap DE. But if you shift DE by *X* you don’t get a gap: you get a segment from A to where B would be if there were a B. The gap DE is rigid. EF (the purple gap) is rigid too, because shifting it by *X* doesn’t give you a gap (it goes from where B would be to C). And AC (orange) is rigid, because shifting it by *X* gives you the segment EG, which is two adjacent gaps (EF and FG), not one.

Generally, there are only two ways a gap can be rigid. One is if when you do the shift one or the other of its end notes doesn’t map to a note. That happens only when one or the other of the end notes is the last note added (A in this case), in which case there isn’t a note *X* to the right of it. We’ll call this a Type I rigid gap.

The other way is if when you do the shift there is a note, or more than one note, in the middle of the shifted segment. But that happens only if the note(s) in the middle have no notes *X* to the left of them. Otherwise there’d be one or more notes in the middle of the unshifted gap, and by the definition of a gap, there isn’t. And the only note with no note *X* to its left is the first note (F in this case), so there can be only one note in the middle of the shifted gap and there can be only one rigid gap of this sort. This is a Type II rigid gap.

And that means there are no more than three rigid gaps: two Type I gaps that start or end on the last point, and one Type II whose shifted version contains the first point. All the other gaps are nonrigid. By shifting any one of them one or more times, you can make it coincide with one of the rigid gaps, and so it must be the same size as one of the rigid gaps. Therefore there can be at most three gap sizes.

That’s the first part of the theorem. Second part, coming up soon.

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