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Palindromic M3


Jakkk0

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Welcome to the forum !!!  And don't worry for your odd question… inthis section we are a lot of passionates on everything related to old Leitz gear… and "numerology" is one of the many details that gives some interest to collectors ;) : for instance, there are people who are very proud to have 2 lenses (or 2 bodies) with consecutive numbers… (and such "pairs" are time to time offered for sale at auctions- at a premium price). Generally speaking, " nice numbers" are appreciated… and usually they can bring some extra price in the market…. I remember several examples like 6 digits numbers with the first 3/4 (or the last 3/4) digits identical… or with many zeros in the middle like x0000y, and of course the first or last number of a certain batch of cameras/lenses are specifically noticed, and valued… ;).  Frankly, I don't remember a case of an item with a palindromic number…. but is surely an oddity which is worth to evidence if you put your item for sale : probably, the right way is to try a tentative sale on ebay and see which kind of offer come in… keep in mind that an old M3 (expecially in the 7xx.xxx) is Always a valuable item, and surely the conditions are commanding the value more than the particularity of the s/n...  Another way could be to write some significant and specialized Auction House (WestLicht, Rahn…) and ask to them an opinion : they are Always well disposed towards items like old M3, which has a stable demand by collectors.

Btw… how did you compute that 216 is the number of 6-digits M3 with palindromic s/n ?  Or do you include also the (possible) 7 digits M3 (*)  ?  Just a curios…  looking at the published batches of M3 (taken for good… it's NOT the Bible… :rolleyes:) one can do the computation… which isn't a quick task B)  (in theory, there ought to be 1999 palindrome with 6 digits… but of course M3 started at 700.000 , and not all the 299.000 s/n to 999.999 (**)   were used for M3s... there ought to be something around 158.000 - 160.000.

(*) By feeling - not verified - with the 7 digits they are many … 1000001 was an M3, and also 1001001 1002001 1003001 … 1014101... 1015101..1022201... B)

(**) THIS would be a VERY valued item !!! :D

Edited by luigi bertolotti
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Hello Luigi and thank you for your reply. The 216 was a rough calculation for 6 digit serial numbers beginning with 7. I should have stated this in my post. The first 3 digits would all need to be different ie; 712. If you were to include all numbers up to and including 798 there would in theory be a calculation of 3 x 8 x 9 =  216. However it would in fact be less because some 7 numbers are not used. I think the number of palindromic for numbers beginning with 7 would be 159. If you are to include 8 and 9 three digit serial numbers a rough number is another 358 (i think)? As you say the M3 numbers seem to be not set in stone. Ps, how did you arrive at 158000 to 160000 if the first 3 digits are different.

Best regards and thank you.

Jack

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9 hours ago, Jakkk0 said:

. Ps, how did you arrive at 158000 to 160000 if the first 3 digits are different.

Best regards and thank you.

Jack

Sorry, maybe my phrase wasn't clear… 158-160.000 is (roughly) the number of M3 cameras built, considering only the 6 digits s/n... is a not difficult computation, based on the numbers af any batch from 700.000 to 999.999... of course, as always, with several doubts about the exact number built (our Wiki area does report a list - other lists do exist, with slight differences)

Edited by luigi bertolotti
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14 hours ago, luigi bertolotti said:

Sorry, maybe my phrase wasn't clear… 158-160.000 is (roughly) the number of M3 cameras built, considering only the 6 digits s/n... is a not difficult computation, based on the numbers af any batch from 700.000 to 999.999... of course, as always, with several doubts about the exact number built (our Wiki area does report a list - other lists do exist, with slight differences)

Hello Luigi and thank you for your reply. The 216 was a rough calculation for 6 digit serial numbers beginning with 7. I should have stated this in my post. The first 3 digits would all need to be different ie; 712. If you were to include all numbers up to and including 798 there would in theory be a calculation of 3 x 8 x 9 =  216. However it would in fact be less because some 7 numbers are not used. I think the number of palindromic for numbers beginning with 7 would be 159. If you are to include 8 and 9 three digit serial numbers a rough number is another 358 (i think)? As you say the M3 numbers seem to be not set in stone. Ps, how did you arrive at 158000 to 160000 if the first 3 digits are different.

Best regards and thank you.

Jack

 

sorry Luigi, I completely mis-read your meaning.

thanks again, Jack

 

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