210 is VERY Goldbachy - Numberphile

Carl Pomerance discusses 210 and its special relationship with Goldbach's Conjecture.
More links & stuff in full description below ↓↓↓
Goldbach's Conjecture: • Goldbach Conjecture - ...
The paper discussed in this video: math.dartmouth.edu/~carlp/PDF...
Professor Pomerance is based at Dartmouth College.
More Prime number videos: bit.ly/primevids
Discuss on Brady's subreddit: redd.it/6duckx
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
We are also supported by Science Sandbox, a Simons Foundation initiative dedicated to engaging everyone with the process of science.
NUMBERPHILE
Website: www.numberphile.com/
Numberphile on Facebook: / numberphile
Numberphile tweets: / numberphile
Subscribe: bit.ly/Numberphile_Sub
Videos by Brady Haran
Patreon: / numberphile
Brady's videos subreddit: / bradyharan
Brady's latest videos across all channels: www.bradyharanblog.com/
Sign up for (occasional) emails: eepurl.com/YdjL9

[@Simoneister]

"…and myself" *drops mic*

[@SportsStacker16]

Simoneister SIMON

[@lamrhariilias173]

Simoneister first time said at a numberphile video :o

[@BubbaYoga]

Yeah, I was thinking 'Why would he memorize those names?' Oh, that makes sense.

[@Djorgal]

I never remembered the names of my coworkers. I simply call them "meatbags".

[@SHUBHAMGI]

Djorgal , LOLZ... :)

[@clover7359]

This is why we should count in base 210.

[@imfrommanndame]

Which symbols do you propose we use for this system? We'd need 210 of them.

[@fudgesauce]

emojis

[@lucianodebenedictis6014]

what is pidgeon times hospital minus knobs equal to?

[@lucianodebenedictis6014]

Well this is actually a pigeon 🕊️

[@themanhimself1229]

that's a dove

[@iteragami5078]

210 is the sum of consecutive whole numbers from 1 to 20.

[@stulora3172]

210 in base 3 is 21 in base 10.

[@Taintain101]

Great video, I want to see more of this guy

[@dongphandinh3722]

Tain plm xet my

[@sillysausage4549]

Hello random people watching old Numberphile videos!

[@Triantalex]

??.

[@peppybocan]

Carl Pomerance! You are digging out some real gold nuggets, Brady!

[@nataliekanakova9496]

Finally someone who approves the name Brady gave his discovery. Not like #ParkerSquare

[@joshmyer9]

210: that's ultra numberwang!

[@Jayohennn]

Josh Myer YES THANKYOU

[@SiddharthJhakaas]

Brady seems like the most likeable guy ever :D Thanks man!

[@ryukaiser3]

all of the old men in this videos are just so damn cute that sounds weird but there's something about old men showing childlike excitement or joy that's just so wonderful, yknow? and it's because of MATH of all things!

[@Xezlec]

♫ 210 is the Goldbachiest number that you'll ever dooooo ♫

[@kobilica999]

210*210 = 44100, which is common sampling rate for audio signals

[@danteregianifreitas6461]

this guy is really cool bring more videos with him pls

[@renzoaragon4523]

Astounding claim at 4:42, GREAT video.

[@Halosty45]

103+107 is the best one of those because it's the closest pair equidistant from half of 210.

[@pabloagsutinnavavieyra2308]

I like your sense of symmetry :)

[@ruinenlust_]

I lke 105+105 more.

[@MauveTendingToBeige]

105 is not prime.

[@Deciheximal]

Fun fact: 105 is a centered pentagonal number. These guys, when multiplied by 2, tend to have above average prime symmetry.

[@khoitrinh1]

Any two numbers that sum up to 210 will necessarily be equidistant from half of 210.

[@stevefrandsen]

Very interesting. Thank you.

[@Robi2009]

6:02 - kudos for how well he pronounced obviously Polish name like Władysław Narkiewicz

[@alandouglas2789]

Robi_CK Umm, he worked with the guy...

[@Robi2009]

IKR, but you can work with a guy called Zhu Chung Tan or whatever and just call him "John"... He actually knew Polish "ł", "cz" etc.

[@paulbarbat1926]

But not the slightest kudos for how he pronounced the frenchman's name

[@briandeschene8424]

@@paulbarbat1926 C’est rien. :-)

[@guitaoist]

210 is also one of the only triangle numbers you can half and that number (105) is also triangular. plus its my bday 2/10 and thats the most important thing;>

[@GravelLeft]

"It works for 36, but that's boring, 36 is small". But.. 36 is my favorite number T~T

[@StreuB1]

This was very very elegant. :-)

[@TheCarsonAdams]

Video Suggestion: Catalan's Conjecture. Very easy to understand, but a surprising result. Theorem: There are no consecutive powers of natural numbers besides 8 and 9 (i.e. 3^2 - 2^3 = 9 - 8 = 1). That is, x^a - y^b = 1 has no natural solutions x, y for integers a, b > 0.

[@00bean00]

Why is that surprising? Your definition is a little bit unclear, as both the base numbers, the powers, AND the resulting numbers are consecutive (in either way, however), and this makes it impractically narrow. If you mean there are no natural numbers that give consecutive numbers by *any* powers in the (positive) integers, then that is a very unusual conjecture.

[@NoNameAtAll2]

Carson Adams Released

[@Arycke]

00bean00 just search in text, the OIES, or general internet to find more information on Catalan's Conjecture. Yes the OP should be more clear and concise, as everyone should be in defining things, so as to not mislead others or spread misinformation in the mathematical community so I agree there.

[@GellyGelbertson]

It's a,b > 1. Otherwise, you have examples like 2^5-31^1=1

[@maxvilla5005]

typo @ 1:54 shows "3 + 21" summing to the even number 34. Should be "3 + 31"

[@johnfmartin2576]

The "Very Goldbachy" quality of the number 210 can be used to create a fast mental primality test for any number in the interval X=[111,199]. Recollection of the first 25 primes is all that is needed Begin Method-- To test the primality of any number x in X=[111,199], follow these 3 steps: [A] Exclude multiples of 11, (x = 121, 143 or 187). If the right most digit of x is 1 less than its middle digit, then 11|x and x is not prime [B] Exclude multiples of 13. If x = 169, then 13|x and x is not prime [C] If x is not exluded above, then calculate y=210-x. This mental subtraction is easy, since each digit of y is simply the ten's complement of its corresponding digit of x The numbers x and y necessarily have the same primality, assuming x+y=210 and x is not a multiple of 11 or 13. Moreover, all 17 primes in the interval X=[111,199] are Goldbach addends of 210 --End of method

[@johnfmartin2576]

Here are a few examples... Let x=191, not a multiple of 11 or 13. y=19 is prime. So, 191 is also prime Let x=187. Since 8-7=1, 11|x. So, 187 is not prime Let x=159, not a multiple of 11 or 13. y=51 is not prime. So, 159 is also not prime Let x=139, not a multiple of 11 or 13. y=71 is prime. So, 139 is also prime Let x=119, not a multiple of 11 or 13. y=91 is not prime. So, 119 is also not prime Let x = 109. x not in [111,199]. But, 109 is prime (see "full house," below) As an aside, the interval [100,109] is a so called "full house" because all of 101, 103, 107 and 109 are prime. By contrast, the interval [200,209] is an "empty house" because none of 201, 203, 207 or 209 are prime.

[@ben1996123]

oh cool its pomerance. you should do a video with him about integer factorization algorithms

[@ruinenlust_]

A brony talking about integer factorization algorithms... cool cool

[@willferrous8677]

EGGHEAD!

[@malcomdiene7745]

Fascinating...

[@DimaVaulin]

Cool! Unexpected ending!)

[@aitor2623]

min 1:55 34 is not equal to 3+21 xD it should be 3+31

[@furkell]

tienes buen ojo jeje

[@Dadik2580]

21 Also isn't prime (3x7)

[@marceloa.8881]

they messed up a digit, yeah

[@SHUBHAMGI]

deadEye * _*

[@blanmorrison1794]

Aaaaand here we have an ocd person that checks everything to make sure it is perfect.

[@dominikstepien2000]

Shouldn't 2*3*5*7*11 be more Goldbachy than 210? Btw as a Polish guy I love how professor said Wladyslaw Narkiewicz.

[@Kaepsele337]

13*13 < 2*3*5*7*11; 4:00

[@dominikstepien2000]

peterwhy oh okay

[@stulora3172]

@@Kaepsele337 shouldn't the argument be: 13*13 < 3*5*7*11 ?

[@rylaczero3740]

Yup, the higher the Primorial, the more goldbachy it will be.

[@kapilanbavananthan5457]

Make a video about the golden ratio!

[@MrRabix007]

it is interesting to see that the other primes less than 105 which miss in the list like 67 must match with a composite number containing 11 13 17 or 19. Example 210 = 67 + 11*13 or 210 = 89 + 11*11 . the combinaition of these new generated primes is the reason why we cannot give the exact number of prime less than an number

[@MrRabix007]

in other terms human are not capable to predict if a*b is bigger or not than c*d even if all prperty of a b c and d are given. 2*3*5*7 bigger or smaller than 13*19 or 17*17 etc...

[@shivanand.r.p8960]

beautiful

[@jrsleao]

In my formula I plug 210 (or any other even number) and i get two primes that represent it. Not only that. I change a parameter and aplly the formula recursevely and I get ALL the pairs of primes that add up to that even number.

[@nO_d3N1AL]

Dunno about anyone also, but I found this video to be far better at explaining the Goldbach conjecture than the previous one

[@MrPictor]

The wording "every even number starting with 6" is a bit ambiguous to me. For a minute I thought they the conjecture only applied to numbers starting with a 6!

[@johnchancey3941]

MrPictor Same here - I was about to comment the same thing!

[@johnchancey3941]

maybe "Beginning at 6" would have been more clear

[@TaohRihze]

Just wondering, why are 4 not counted, does 2+2 not matter?

[@jrjihu]

2 is not odd.

[@izimsi]

I thought that 6 has to be the first digit of the number...

[@DRD363]

is there a counter example to the conjecture that states that every even number can be written as the product of 2s, 3s, and 5s and some other prime?

[@alexbarrena3013]

math challenge: what's the formula for the numbers with the next two propeties: -the summe of te digits is 13 -if you change the number in the third place starting from the right with the first one starting from the right. the number must be 180 units smaller will you do a video anout this?

[@jrsleao]

I have a formula to find all the primes representations (as a sum of two primes) for any even number. Would that be interesting? Does this formula proves the Goldbach Conjecture ? I mean..If I can find all combinations of numbers for any even number.

[@Deciheximal]

23, 41, 67, and 89 must fell really left out. So we've got every q matching with a p. Are there any numbers where every p above 7 matches up with a q?

[@cecasiahaan6801]

6:03 nailed it

[@psyk2642]

Nice

[@skidsune363]

Cool!

[@rubenscabral2657]

Hello I'm from Brazil the Goldbach conjctura works because of the wildcards 2 that goes through all pairs also the 5 that doubles in itself 5.10.15.20.25...to infinity.

[@EricsShitposting]

ok, amazing

[@williamwebb9724]

Brady, I want to know if all the clean & simple mathematical proofs have already been discovered. Are they all going to be 100+ page calculus ridden nightmare proofs going forward?

[@lucromel]

What's the highest (known) even number with only 1 pair?

[@timen.space.]

12

[@timen.space.]

and if you write it as 012, it's 210 backwards, which makes sense

[@lucromel]

There *could* be a higher number with exactly 1 pair though, correct? As it's unproven that there's at least one pair for all numbers, I'm assuming it's not proven that their couldn't be a higher number with only 1 pair? Or is there a proof showing that if a number greater than (let's call it x) has at least 1 pair, then it must have multiple pairs?

[@TheBalfrog]

I'm personally unsure but as numbers get larger you simply have more possible primes of the form X/2 +/- a, because every prime below half the composite numbers could be part of a pair, so the chances just get larger as you progress through larger composite numbers. Another cool thing with Terence Tao and some others work on the Twin Prime conjecture you can prove there are an infinite amount of primes that differ by I believe a number greater than 244, 246, 248 ect... So if we did solve the twin prime conjecture and some further conjectures could you use it to say X+/- 122 (For 244), X +/- 123 (For 246) ect, could that be used to prove Goldbachs? All fun stuff in number theory!

[@jsmith754]

12 (5,7) is the right answer.

[@Trockenshampooleopard]

I want a t-shirt that says "VERY GOLDBACHY". Where can I buy it?

[@PM-gp5pq]

A bit off the topic here - Since your last video with Terry Tao was really great, I suspect you could do some interviews with say someone like Serre, Deligne, Milnor,..with them people getting really old and very few of their interviews available on youtube

[@Rotaermel]

Where does this spooky music 4:40 come from?

[@sambam97]

Can one of the professors answer the question, is a circle a straight line? Cause I believe it is but my friends don't

[@sethapex9670]

is it possible that the goldbach conjecture is related to the fact that between any two primes greater than 2 the gap between them will be even?

[@jsmith754]

Seth Apex, not quite. Although that is also an open conjecture, that any even number can be the difference of 2 primes. I suspect Goldbach would imply this one as truth as well.

[@Doktor_Vem]

Question: How do you *prove* that a number is the largest number with a certain property or possibility? Don't numbers go on forever?

[@samueldeandrade8535]

With arguments.

[@supernovaaust]

This leads to another conjecture : If you -+ 210 from any non prime which has a multiplication of 2,3, 5 or 7 you will always get a non prime. However this will not be the case for primes which will always have an associated twin primes as long as the multiplication doesn't go for 11 or above starting at 11squared. Therefore this should in any case be true for all numbers 210+210 420. In the infinite realm, any factor of 2,3,5,7 minus or plus 210 The result will always be it's associated factor.

[@yuyiya]

Define the Goldbach function on natural numbers, g(n), as the number of representations of n as the sum of two primes p + q, with p >= q. The paper mentioned in this KZbin video ("10 is VERY Goldbachy - Numberphile") and hosted at Dartmouth, "An Upper bound in Goldbach's Problem", Deshouillers et al (1993), shows - using some advanced methods - that 210 is the largest value of even n > 2 for which g(n) = pi(n-2) - pi(n/2 - 1/2), where pi(x) represents the number of primes up to x. For any larger n, g(n) is strictly less than pi(n-2) - pi(n/2 - 1/2). My question is this: What else is known about how fast g(n) grows for larger n?

[@larrycornell240]

All the primorials are 'goldbachy', it has to be so since they have locally the largest concentration of primes in their r.r.s. And the residues add in pairs across p#/2. More primes, more matches.

[@yudhistirs]

Does this has any significance on any real world application or usage ?

[@almondtart]

Every prime video will be my favourite video!

[@SkytoM11]

Why can't it (210-q) be divisible by 3? Is it explained properly so I should "get it" just from the Video or is there something missing?

[@msolec2000]

Because 210 is a multiple of 3, but q is not, which means that the difference cannot be. In other words, if 210-q was a multiple of 3, since 210 also is, then q would have to be a multiple of 3 too, which cannot be because q is a prime bigger than 105.

[@MeuleDomino]

210 times 12 is 2520. Someone also familiar with this number?

[@ferretyluv]

How can 210 not divisible by 3? Shouldn't 210/3=70?

[@willferrous8677]

it's (210-p), not 210

[@AlexanderNaumenko-bf7hn]

If Goldbach's Conjecture is wrong then for all primes p below some n: 2n - p is complex or is divisible by q < square root of 2n - p. This is true for p congruent to 2n mod q. Let's call primes q below the square root of 2n - 3 basis primes. If 2n is divisible by any q then there are no primes congruent to 2n mod q. It increases the number of Goldbach pairs for 2n. 210 is divisible by 3, 5, 7 - lowest q, no wonder it has a big number of Goldbach pairs. Multiply 210 by 11 and you will get another number with an unusually big number of Goldbach pairs relative to neighbor numbers.

[@awtactrl]

So this goldbach theory only works with even numbers up to (and including) 210?

[@JimmyLundberg]

Nono, nonono. No. The Goldbach conjecture (any even number is the sum of two primes) possibly works for integers of any size. 210 is just the biggest one where _all_ the primes add up like this. I would suggest rewatching the video.

[@awtactrl]

De va som fan

[@HansPeter-wo1ub]

could someone explain the 210-q isn't dividible by 2,3,4,5 part again? the very much

[@MrRyanroberson1]

what has ever been so special about even numbers? its a multiple of 2. so why not have names for multiples of 3, 5, so on? also -1^2=1

[@CorrectHorseBatteryStaple472]

The blinking text with the funky sound is a mistake.

[@sphakamisozondi]

06:04, Mic Drop!

[@echo5delta286]

But what's the last even number with only ONE unique pair of prime addends???

[@saidatulhusna1533]

nice video 210 is a good number

[@concorde__]

I didn’t know Stan Sitwell was into math!

[@GrandTheftAutowkthru]

Very unrelated but why does 4065 / 5 = the digits before 5 ((406 * 2) + 1) = 813 why when you double the digits before the digit 5 and add 1 u get the answer

[@nassima1141]

With that music i felt like i was having a math lesson in church 😝😝😝😝😝

[@astroash]

so like, goldbach is a better version of nikelback!

[@oz_jones]

Au > Ni

[@crazydrummer4827]

I have a problem for you, if anyone would like to work it out. This is actual problem in my school. A class of 30 students need to choose a subject for next year to listen to. They have 5 subjects to choose from, but one group must have at least 11 students, so the maximum number of groups is 2. How to make this work with biggest number of satisfied students?

[@ryanlatterell7850]

i guess as long as we don't come out with a sequel for "2", this conjecture will hold

[@rubenscabral2657]

The primes are not doubled by the numbers 3 and 7 also the perfect squares odd minus ending 5 example 3+3+3+3... to infinity and its opposite 7+7+7... to infinity after the perfect squares odd

[@pizzamannetje79]

1:10 Goldbach's conjecture about numbers starting with six. Then the rest of the video goes on about 210, which doesn't start with six. I'm sorry, but i don't see the link.

[@jgregorygraves5792]

any number =>6

[@taswelll]

Someone alternaned Capital and lowercase In the Subtitles and I don't Know Why.

[@mvivian100]

I would have used the phrase "Very Goldbachian"

[@trdi]

Very interesting.

[@6infinity8]

Typo at 3:40

[@christopherlee3143]

Is it possible to program an A.I. to solve the toughest math conjectures?

[@aitor2623]

Christopher Lee Programming an A.I. to solve the toughest math conjetures would be like a very tough math conjeture soooo I gess I should not comment on this stuff and start studying

[@lubomirsalgo7638]

It's not possible. If you're thinking about neural networks, they operate with a certainty level, meaning they can't be 100% correct. You can't prove anything, If you can't be sure that you're 100% correct. If you're thinking about conventional programming, it's impossible for programmer to develop A.I. for solving this problem without knowing the solution beforehand.

[@martinepstein9826]

Carbon-based intelligences have been solving tough math problems for millennia so silicon-based intelligences ought to be able to do the same.

[@wingracer1614]

Ľubomír Šalgo, not necessarily true. There are plenty of math conjectures that we find the answers to quite easily, we just can't prove it. In theory, a computer program could come up with a proof for such a conjecture.

[@lubomirsalgo7638]

Can you provide some examples?

[@volbla]

At first i figured the same reasoning should work for 2310 = 2 * 3 * 5 * 7 * 11, but that is so big that p doesn't have to be prime. 2310 - q will not be divisible by any of the factors of 2310, but it could be divisible by 221 = 13 * 17, and that aint goldbachy.

[@-fitzy-3335]

I love this professor :p

[@RoGeorgeRoGeorge]

If 210 is so mind-blowing, just imagine 420!

[@ItumelengS]

2*210... mind blown

[@zeke7209]

There are four Very Goldbachy numbers: 10, 16, 36 and 210.

[@disgruntledtoons]

So now we need the sequence of even values which have a greater number of Goldbach representations than any smaller even number, and the sequence of even values which have fewer Goldbach representations than any greater even number.

[@TommyBeckerAndStuff]

2,300 years ago? @ 0:32

[@anon6514]

Fairly intuitive that there will be a last number with this property. After all, the cut-off point is where the square of the next prime is less than half of the product of the first primes. p(n+1) ^ 2 < 0.5 * p1 x p2 x p3 x ... x p(n+1) As n increases, p(n+1)^2 / pn ^ 2 tends to 1 and so the left hand side grows by a smaller and smaller amount, where as the right-hand side is growing large very quickly. It's therefore plain to see why there has to be a largest number with this property.

[@joefagan9335]

Very sweet! And it doesn't work for 2310 = 2.3.5.7.11 because the last step requires that 13^2

[@killerdevil980]

This number is my area code

[@Pr1est0fDoom]

Haha, I agree with Mr. Pomerance there, Brady: that's an excellent name for 210!

[@RonWolfHowl]

"fully dense Goldbach number"

[@wuorson5111]

i say 5 is very goldbachy and cuz not only it fits that every p has a pair q, and that it includes an even prime!!

[@leo17921]

he looks like 55 but he's actually 73 either way great vid (also i think it could work for bigger numbers if the 1 doesnt matter)

[@jsmith754]

On the other hand, the numbers with the fewest Goldbach pairs are of the type n=2p (2 x prime). You can find my proof to Goldbach's conjecture on vixra.org/abs/1701.0618. Comments are welcome.

[@andreweberlein1509]

jsmith754 There is an error at 2.6 where you convert from the typical twin prime constant that had a fraction in a single limit (since an infinite product is implicitly a limit of a product as the number of terms goes to infinity) that you convert to a fraction of limits. You had stated in 2.5 that the denominator converges to 0 (as x-> infinity), so the fraction you claim holds true has division by zero.

[@jsmith754]

Andrew Eberlein not an issue when the numerator also converges to 0. See lhospital's rule.

[@andreweberlein1509]

jsmith754 Counterexample: lim(x)/lim(x/x^2)=0/0 is indeterminate, lim(x/x^2)=lim(1/x) does not exist. lim(x^2)/lim(x)=0/0 is indeterminate, lim(x^2/x)=lim(x)=0. lim(x)/lim(x)=0/0 is indeterminate, lim(x/x)=lim(1)=1. Three examples with the numerator and denominator equaling zero with drastically different answers. If you were to use l'Hôpital's rule as you suggested, you would have needed to differentiate the numerator and denominator separately. I do not remember seeing any derivatives.

[@jsmith754]

Andrew Eberlein the cases you mention diverge to infinity, or 0. The case I use converges to a constant which is the twin prime constant. So no need for lhospital's. Also, although the products tend to 0 at infinity, I only need to evaluate then at sqrt of n, not up to infinity.

[@andreweberlein1509]

First, not all my cases diverse to infinity or 0. One of my cases went to 1. Second, 1 and 0 are constants just like the twin prime constant, only more "regular" in a colloquial sense. Third, you brought up l'Hopital's rule, not me. Fourth, when you mention Merten's third theorem, you reference [10] and [11], where [10] has no mention of the theorem except for a footnote on page 4 and both [10] and [11] contain an error term. So if you aren't taking the infinite product, you must include the error term. Fifth, I found a problem even earlier. Algorithm 1 and Theorem 3 have serious errors. First, you initially include a pair that includes 1 and arbitrarily remove it, as the program would not have removed that pair in the first step as you show in Figure 2.1 (nothing in the column is shaded, so why remove it?). Further, your width value does not hold for n=8 or n=10. If we include (1,7) or (1,9), then for n=8 the size of the array halves (1-1/2), but for n=10 the size of the array does not half since the array has an odd length. If we do not include (1,7) or (1,9), then we have the same problem for n=8 because the array size is 3 which we cannot halve. Further, let's say we fix that problem by not including any even numbers. Then it still fails for n=18. Passing through the prime 3, we remove 2 of the four columns, and 1/2 is not (1-1/3) as claimed. Even more, in the example given in Figure 2.1, the widths go 53->26->10->7->6 and, using Theorem 3, the widths should go 53->26.5->8.8333->5.3->3.786. Please let me know if I am wrong about any and all of this.

[@ali99_82]

OMG 210 X 2 OR [210+210] = 420 HALF LIFE 3 CONFIRMED

[@wingracer1614]

That's the parker square version of 210

[@TheGamer2554_]

divided by 10, that’s 42 THE ANSWER OF LIFE CONFIRMED

[@Triantalex]

false.

[@alexandertownsend3291]

For any computer programmers out there. I challenge you to find the least Goldbachy numbers. More specifically, I challenge you to write a computer program that will determine which even numbers can be written as the sum of two primes in only one way. I do not know how to write a computer program so I am leaving this to the programmers.

[@masynlithgo717]

whats 9 + 10, 210

[@marsyasian]

Go Spurs Go

[@rylaczero3740]

Time is not right or I would tell you more about this story.

[@leonardomaranon]

It also happens with 60 (apart for 1 and 59, but who cares).

[@HistoricaHungarica]

4:40 poor 41 don't have a pair :(

[@ffggddss]

As there are 19 primes between 105 and 210, and 26 odd primes less than 105, it follows that there are 26-19=7 of the latter that don't pair up to do the Goldbach thing for 210. And since the last prime below 210 is 199, and 210-199 = 11, the primes 3, 5, 7 don't participate in this. Altogether, the non-participants are: 3, 5, 7, 23, 41, 67, 89 whose "co-addends" are: 207 = 3²·23 205 = 5·41 203 = 7·29 187 = 11·17 169 = 13² 143 = 11·13 121 = 11²