World!OfNumbers |
HOME plate WON | |
||
---|---|---|---|

Facts and figures about 2002the last palindromic year... that most of us will live to see! part 1 | |||

year2002 part 2 |

The nicest number graphics I found on the www! Artwork from Numberland Is there more talent around ? | ||||||||||||||||||||||||||||||||||||||||||

3D Artwork by Niek van der Laak (sent 08/01/2002) Superb! | ||||||||||||||||||||||||||||||||||||||||||

Can you find the smallest prime with period length 2002 ? period of a prime (email from 13/01/2002) 8009 is the first prime p for which the decimal period of 1/p is 2002. Unexpectedly the answer (by Carlos Rivera) came soon & easy. It seems that the real hard question is to find the | ||||||||||||||||||||||||||||||||||||||||||

G. L. Honaker, Jr. tells a story (email 20/12/2001) about 2112... the palindromic successor of the year 2002 Actually, something happened yesterday that I'd like to share with you -- no doubt spurred on by the several e-mails sent among us and by me being in Johnson City, Tennessee of all places. You see, back in the late 70s I saw my first live rock concert there (at Freedom Hall Coliseum). Featured, was one of my all-time favorites -- a Canadian band called RUSH. One of the things I remember most about them is the title suite of the band's breakthrough album -- a futuristic rock opera called 2112. Interestingly, one of the songs of that album was entitled "Something For Nothing", and I managed to find a short clip of that song. To top things off, I've attached a scanned pic of the 2112 album itself. Curious is to why these 3 prime musicians chose this palindrome in the first place... perhaps out-of-reach to humans of our generation ? Great... maybe due to this page we can find out why they used 2112. G. L. really would enjoy knowing. 2112 has the following curious property : ^{2} + 32^{1} + 32^{1} + 32^{2}Finally using the other divisor 66 we can also make the expression ^{2})/2 - 66 | ||||||||||||||||||||||||||||||||||||||||||

The 1^{st} primenumber showing 2002 as a subset in it is...very simple to remember ... namely 20021 ! Jaime Ayala (email 26/01/2002) discovered the following 2002 property : N + 1 and (the product of every string partition into two halves of N) + 1, are primes. Well, Carlos Rivera conjectures that 2002 is the last palindromic numberwith this property... | ||||||||||||||||||||||||||||||||||||||||||

The smallest prime containing 2002 digits is
^{2001} + 5037_{(10)} = 35153_{(6)} = F9F_{(18)}^{2002} - 99 | ||||||||||||||||||||||||||||||||||||||||||

An important date is
initiative (from the 'Snark' mailing list) as well. | ||||||||||||||||||||||||||||||||||||||||||

The year 2002 will be the last palindromic year of our lives. 2003 on the other hand will be the first prime year of the next millennium. And I happen to notice that the two expressions ^{n_odd} + 1^{n_even} - 1Can you find more relations between these two successive years ? | ||||||||||||||||||||||||||||||||||||||||||

Jason Earls (email 26/12/2001) after reading some of the information on this 2002 web page, had some fun experimenting with different things and found some curios you might find interesting. Curio 1.Curio 2._{83}200609 containing 504 digits (certified with PRIMO - pdg)_{257}200229 containing 1548 digits (certified with PRIMO - pdg) | ||||||||||||||||||||||||||||||||||||||||||

Shyam Sunder Gupta (Number Recreations) and his wife, delighted to see all these new Palindromic Year 2002 curios, observed (email 27/12/2001) : The palindromic year prior to 2002 was 1991, which was only 11 years
Similarly the previous such occasion was in 999 and 1001." | ||||||||||||||||||||||||||||||||||||||||||

G. L. Honaker, Jr.'s email from 28/12/2001 EARTH DAY 2002 is on the (22) ^{nd} day of the (2 x 2)^{th} month.It will be the (2 x 2 x 2 x 2 x 2) ^{th} annual observance.
Readers are encouraged to add all the 2s in the above curio. | ||||||||||||||||||||||||||||||||||||||||||

Mark Farrar's email from 29/12/2001 ( www.markfarrar.co.uk leftclicking background pops up a handy menu! ) If you want to use a 4 x 4 Magic Square that has a magic total of 2002, then here's one you could use:
This Magic Square has 86 different combinations of four cells Alternatively, this 5 x 5 Magic Square also has a magic total of 2002:
There are, I believe, 1291 different combinations of five cells | ||||||||||||||||||||||||||||||||||||||||||

Did you know ?^{n}) with exactly 2002 digits!^{10} + 1.2^{9} + 1.2^{8} + 1.2^{7} + 1.2^{6} + 0.2^{5} + 1.2^{4} + 0.2^{3} + 0.2^{2} + 1.2^{1} + 0.2^{0}_{{10}} = 11111010010_{{2}}_{{2}} = 607_{{10}} | ||||||||||||||||||||||||||||||||||||||||||

2002 findings on the ! What's Special About This Number ? _{14}C_{5}^{603} form a 2002-bit prime. [Kulsha] | ||||||||||||||||||||||||||||||||||||||||||

2002 can be written as a sum of consecutive positive integers starting from 13 up to 64, one number for every week of 2002 !
Expressing 2002 as the sum of the consecutive positive integers
My luck has returned it seems ! (email from 05/01/2002) Thanks, Terry, for providing this smooth transition to the next feature ! | ||||||||||||||||||||||||||||||||||||||||||

The four seasons of 2002 2002 has a Sum Of Digits equal to 4 ( = 2 + 0 + 0 + 2 ) 2002 has four distinct prime factors 2 * 7 * 11 * 13 ^{2} + 501^{7} + 500^{11} + 499^{13}four times prime 29733695172086354662551885027753001 !four distinct prime factors !^{2} * 49117 * 154335341 * 164714564559993788333185532716763The first four digits form a palindrome equal to a palindromic expressionfour nontrivial integers q (q>1) sandwiched between two 2002'sso that the whole is divisible by this integer q are these four primes !2002 !!! | ||||||||||||||||||||||||||||||||||||||||||

Carlos Rivera (email 08/01/2002) found the smallest prime such that the sum of its digits is 2002. ^{222}– 1 – 10^{165}_{56}8(9)_{165}59999999999999999999999999999999999999999999999999Note its prime length of 223 digits ! | ||||||||||||||||||||||||||||||||||||||||||

Other 3D Artwork by Niek van der Laak (sent 08/01/2002) | ||||||||||||||||||||||||||||||||||||||||||

(email 08/01/2002) A friend of Terry Trotter told him she received this from her friend... Express the tautonym 20022002 as a product of three palindromes Factor 20022002 2 x 7 x 11 x 13 x 73 x 137 And 22 x 949 x 959 = 20022002 ! Who find similar solutions for longer tautonyms (2002) | ||||||||||||||||||||||||||||||||||||||||||

2002 Prime Time ! (email 05/01/2002) Enoch Haga, fond of the latest Mersenne discovery M39, counted the occurrences of substring 2002 in the decimal expansion of this 4053946 digits long (2 ^{13466917} – 1) record prime.The result is 377 times. Who can find the longest palindromic substring in M39 ? The 2002 The first time that K consecutive prime numbers concatenated in The first time that K consecutive prime numbers concatenated in The reader is invited to find the first What is the next value of K ? between successive primes happens first with pair 7 and 11 7200211 between twin primes happens first with pair 11 and 13 11200213 between consecutive integers happens first with pair 18 and 19 18200219 between an integer and its square happens first with pair 17 and 289 172002289 between two Fibonacci numbers happens first with pair 13 and 21 13200221 between three consecutive integers happens first with Can you discover other combinations ? Can you discover the | ||||||||||||||||||||||||||||||||||||||||||

Some 2002 loose ends Carlos Rivera discovered (email 15/01/2002) that two successive 2002'scan be expressed as the sum of two consecutive primes in just one way.20022002 = 10010989 + 10011013 One more property for 2002 from Carlos Rivera (email 21/01/2002) 2002 is the first Langford Number according to the definition given in his PP&P Puzzle 144. | ||||||||||||||||||||||||||||||||||||||||||

2002 puzzles with expiry date 2112. When occurs the first Prime Gap of length 2002 ?
The concatenation of the integers from 2002 down to 1 is a number - Rearrange the first 2002 integers so that you get a prime (or at least
strongpseudoprime), preferably the smallest and/or highest one !
- Rearrange the first 2002 integers so that you get a perfect square
or a perfect cube !
- Rearrange the first 2002 integers so that you get a palindrome (or
prove it is not possible) !
When produces the procedure of appending to the right of 2002 - Carlos Rivera (email 04/02/2002)
According to the Pi Search page by Dave Andersen - The string 2002 was found at position 30926 counting from the
- first digit after the decimal point. The 3. is not counted.
- But on the other hand also
- The string 20022002 did not occur in the first 100,000,000
- digits of Pi after position 0.
- This brings immediately the obvious yet interesting question
- 'Where 20022002 happens first in Pi' ?
Can you device other challenging problems around 2002 so that | ||||||||||||||||||||||||||||||||||||||||||

This is a 2002 story assembled from two lines of approach. [pdg] First lineI noted that the sum of integers from 1 to 2002 (triangular number n(n+1)/2) is 2005003. I wanted to beautify this number and inserted a zero before thelast digit three, rendering the result 20050003 more self-descriptive. Digit 2 says two zeros to its right, digit 3 now counts three zeros to its left and the middle digit 5 summed all the zeros. And what was the greatest surprise ? Well, 20050003 is prime and when split into two parts 2005 and 0003 and subtracted... the outcome is again our initial 2002.
__Highlight of the story__ In fact I was dealing twice with 8-digit numbers of the form abcd_efgh. By some stroke of inspiration I set out to look for chains of primes whereby the last part efgh would be the starting part abcd of the next member of the chain. Thus 20050003 would be a cycle_1 example and the two primes 4899__6901__and__6901__4899 constitute a cycle_2 example.- Here is an example of three primes making a cycle_3
3917 5919 5919 7921 7921 5919 - And finally when tracking for a cycle_4 example only this one popped up
so revealing quite a unique and remarkable curio! __3389__5391 5391 7393 7393 5391 5391 __3389__
Note all |abcd - efgh| equal 2002 Do you fancy exploring the 10-digit numbers of the form abcde_fghij ? | ||||||||||||||||||||||||||||||||||||||||||

2002 and Fibonacci Numbers Carlos Rivera (email 24/01/2002) discovered that F _{1558} = 1791672002678372... ...3860392002067977... ...376879contains twice 2002, one starting at position 7, the other at position 135 (positions taken from left to right). F _{2587} = 200235153260...996215944133As you can see this number starts with 2002 ! In both cases the pertinent question is 'will this ever happen again ?' of length 2002. Alas none of them contains 2002 as a substring. _{9577} = 133396879947398...639463408214257_{9578} = 215840685748080...115559275163089_{9579} = 349237565695479...755022683377346_{9580} = 565078251443559...870581958540435_{9571} = 914315817139038...625604641917781What more can be discovered in these five numbers ? | ||||||||||||||||||||||||||||||||||||||||||

2002 patterns Hugo Sánchez (email 29/01/2002) in his very own style produced some nice numerical (in)finite patterns related to 2002 and palindromes. _{n})1 = 2002(0_{n-3})2002 for n > 2_{n}) = 222(4_{n-3})222 for n > 2_{n}) = 444(8_{n-3})444 for n > 2_{n}) / (2 * 5)_{n-3})111_{n})1 = 2446(8_{n-3})6442 for n > 2^{2} = 4008004_{n})2^{2} = 4(0_{n})8(0_{n})4_{n})11 = 22022(0_{n-3})22022 for n > 2_{n})22 = 44044(0_{n-3})44044 for n > 2_{n})3 = 6006(0_{n-3})6006 for n > 2_{n})4 = 8008(0_{n-3})8008 for n > 2_{n})5 / 5)_{n-3})1001_{n})6)) / 2_{n-3})6006_{n})7)) / 2_{n-3})7007_{n})8)) / 2_{n-3})8008_{n})9)) / 6_{n-3})3003_{4} = 252_{10} = 202_{8} = 11211_{3}_{6} = 434_{10}_{8} = (100001 + 1)_{4}_{8} = (1000000001)_{2}_{9} = 2000002_{3}_{n})2 - 2) / (9_{n+2}) = 0,[2(0_{n+1})]^{rep. inf.}_{n})91 * 22 = 2(0_{2(n+1)})2_{n}) + 1) * 22 = 2(0_{2n})2_{5})91 = 909090909091^{2} + 8^{2} + 1^{2} + 1^{2}^{3} + 10^{3} + 1^{3} + 1^{3}^{4} + 5^{4} + 3^{4}_{n}) / (1000_{n-1})1 = 2002^{-2002} = 2002^{2} + 1002000^{2} = 1002002^{2}^{4} - x^{3} - x^{2} - x^{4} - 2^{3} - 2^{2} - 2 = 2^{4} - 0^{3} - 0^{2} - 0 = 0^{4} - 0^{3} - 0^{2} - 0 = 0^{4} - 2^{3} - 2^{2} - 2 = 2^{4} - 4^{3} - 4^{2} - 4 = 172^{4} - 1^{3} - 1^{2} - 1 = -2^{4} - 7^{3} - 7^{2} - 7 = 2002^{4} - 2^{3} - 2^{2} - 2 = 2^{4} - x^{3} - x^{2} - x es | ||||||||||||||||||||||||||||||||||||||||||

Carlos Rivera (email 04/02/2002) According to the N ^{th} Prime Page by Andrew BookerThe 20,022,002nd prime is 374,024,059. Well, it happens that 374,024,059 is a reversible prime (emirp) since 950,420,473 is also a prime number ! | ||||||||||||||||||||||||||||||||||||||||||

A Time Palindrome. By Wade VanLandingham (email 13/02/2002) ( maintainer of a website about the quest for a solution to 196, still moving ahead, now nearing 28 million digits ! ) A Time Palindrome... As the clock ticks over from 8:01PM on Wednesday, February 20 Similar message (email 15/02/2002) from Maggie Feeney from Australia 8:02 pm on February 20, 2002 will be an historic moment in time. Dinesh Aneja (email 20/02/2002) doesn't fully agree as there existed And another correction from Carlos Rivera (email 20/02/2002) 21:12PM, on December 21, 2112 or 21:12, 21/12, 2112. Terry Trotter (email 21/02/2002) remarks that the date part is given | ||||||||||||||||||||||||||||||||||||||||||

Fermat numbers and 2002. By Carlos Rivera (email 14/02/2002) These first few Fermat numbers F
| ||||||||||||||||||||||||||||||||||||||||||

The sixth dimension of 2002 From Number Theory List [NMBRTHRY@LISTSERV.NODAK.EDU]. Yasutoshi Kohmoto searched the solutions of 2002 ^{6} = x^{3} + y^{3} + z^{3} + u^{3} for this year.^{6}^{3} + 579012^{3} - 8925^{3} - 483^{3}^{3} + 604606^{3} + 32002^{3} + 1008^{3}^{3} + 696742^{3} + 23467^{3} + 216^{3}^{3} + 1068909^{3} + 34102^{3} + 862^{3}^{3} + 1433279^{3} + 4441^{3} + 366^{3}^{3} + 1618364^{3} + 35772^{3} + 946^{3}^{3} + 1726369^{3} + 11640^{3} + 102^{3}^{3} + 1970150^{3} - 11336^{3} - 182^{3}^{3} + 2163515^{3} - 16512^{3} - 830^{3}^{3} + 2727930^{3} + 28824^{3} + 916^{3}^{3} + 2739726^{3} - 15407^{3} - 185^{3}^{3} + 3179941^{3} - 16933^{3} - 228^{3}Yasutoshi used the algorithm described in the mail | ||||||||||||||||||||||||||||||||||||||||||

The Gabriela Puzzle This puzzle was constructed by Carlos Rivera in honor of his new (& 2nd) granddaughter Gabriela who was born 22 minutes before 20:02PM of 20/02 of the year 2002 ! Carlos, you and your family have my congratulations ! simultaneously primeor demonstrate that such number X does not exist. | ||||||||||||||||||||||||||||||||||||||||||

The Number of the Beast and 2002 The Number_Of_Digits of {666^666} = 1881 The Number_Of_Digits of {66^66} = 121 Note both results are palindromes ! And yes... 1881 + 121 = 2002 | ||||||||||||||||||||||||||||||||||||||||||

(email 27/02/2002) What do you think of this Flashy 2002 creation (one needs a Macromedia Flash Player plugged in of course) by designer Paulo Hartmann www.paulohartmann.net/2002 PS, right clicking over the animation pops up an additional menu. | ||||||||||||||||||||||||||||||||||||||||||

Keithing to 2002... an article by Terry Trotter. Prologue (01/03/2002) While reflecting on the interesting work being done by many others in their My method of search might be called " Keithing " on some numbers.
I asked myself: "Why stop at the original number ? Why not continue To see what was discovered, go to WONplate 128. | ||||||||||||||||||||||||||||||||||||||||||

A 4 x 4 Prime & Magic Square with Magic Sum 2002... (forwarded by Carlos Rivera). (email 02/03/2002) Mr. John Everett has sent the following beautiful 4 x 4 prime & magic square
For Prime-magic squares adding up to 2000 Puzzle 106 | ||||||||||||||||||||||||||||||||||||||||||

Norman Luhn's 2002 entries. (email 12/03/2002)22 Norman used an old version of DERIVE to derive this neat fact. Well, as a matter of fact, the curio is not so unique because there
are many bases b besides 22 such that b^{2002} = b mod 2002.In Europe, the EURO or € became official currency in this year 2002. even and, combine three digits of the | ||||||||||||||||||||||||||||||||||||||||||

[ TOP OF PAGE]

Patrick De Geest - Belgium - Short Bio - Some Pictures

E-mail address : pdg@worldofnumbers.com