Palindromic Year 2002

World!Of Numbers		HOME plate WON \|
Facts and figures about 2002 the 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 second prime
with this property !

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 :

2112 = 32 * 66

and 2112 can be reconstructed as sums of powers of this divisor 32

32² + 32¹ + 32¹ + 32²

Note the exponents form again our palindrome !
Finally using the other divisor 66 we can also make the expression

2112 = (66²)/2 - 66

A small curio is also 2112

21 * 12 = 252 a palindrome !

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 :

2002 + 1 = prime

2 * 002 + 1 = prime

20 * 02 + 1 = prime

200 * 2 + 1 = prime

This property can be phrased into the following way :
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 number
with this property...

The smallest prime containing 2002 digits is

10²⁰⁰¹ + 5037

or 10000000000... ...00000000000005037

Certified prime using PRIMO after 62h 36mn 30s (20/03/2002) [pdg].

And yes... 5037 has relation with palindromes since

5037₍₁₀₎ = 35153₍₆₎ = F9F₍₁₈₎

5037 = 3 * 23 * 73 and 3 + 23 + 73 = 99.

And yes... we need this 99 for the following statement !

The largest prime containing 2002 digits is

10²⁰⁰² - 99

or 99999999999... ...99999999999999901

Also certified prime using PRIMO after 103h 6mn 16s (01/04/2002) [pdg].

An important date is

and stay alert at 20:02 of that day

That special day has been declared

The Universal Day of the Symmetry

Read all about it - this site is in 'castellano' in case it's more convenient - at

Día Universal de la Simetría

No doubt you like to contribute or participate with Ivan Skvarca's
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

2002^n_odd + 1

and

2002^n_even - 1

are divisible by 2003

Yep, there is a connection between palindrome 2002 and prime 2003 !
Can 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.

The sum of 'distinct' primes (Sloane sequence A008472) dividing

palindrome 8277728 is sopf(8277728) = 2 + 139 + 1861

which is

2002 !

Curio 2.

Some large primes with an overall 2002 pattern

(200200)₈₃200609 containing 504 digits (certified with PRIMO - pdg)

(200200)₂₅₇200229 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

before. So we (born before 1991) are lucky to live in two palindromic

years, which is rare, as next such occasion will be in 2992 and 3003.

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.

www.earthday.net

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:

501	498	495	508
496	507	502	497
506	493	500	503
499	504	505	494

This Magic Square has 86 different combinations of four cells
that sum to 2002.

Alternatively, this 5 x 5 Magic Square also has a magic total of 2002:

404	413	388	395	402
412	392	394	401	403
391	393	400	407	411
397	399	406	410	390
398	405	414	389	396

There are, I believe, 1291 different combinations of five cells
that sum to 2002.

Did you know ?

703^703 is the only power of the form n^n (or nⁿ) with exactly 2002 digits!

2002 can be expressed as a sum of two primes in 44 (is a palindrome) ways.

3 + 1999, 5 + 1997, 23 + 1979, etc.

Note 1979 - 23 = 1956 which is my birthyear ! (pdg)

By the way, I'm entering the new year aged 45 which is the sum of the

exponents in the binary expression of 2002

1.2¹⁰ + 1.2⁹ + 1.2⁸ + 1.2⁷ + 1.2⁶ + 0.2⁵ + 1.2⁴ + 0.2³ + 0.2² + 1.2¹ + 0.2⁰

45 = 10 + 9 + 8 + 7 + 6 + 4 + 1

Prime 607 is the reverse of the binary expression of

2002_{10} = 11111010010_{2}

and (0)1001011111_{2} = 607_{10}

2002 findings on the

What's Special About This Number ?

2002 = ₁₄C₅

Prime Curios! 603

The first 603 digits of 603⁶⁰³ form a 2002-bit prime. [Kulsha]

966 bits are "1" and 1036 bits are "0" in the binary expansion.

Prime Curios! 2002
Prime Curios! 99899...99899 (223-digits)

2002 can be written as a sum of consecutive positive integers starting from
13 up to 64, one number for every week of 2002 !

Unlucky seems this number 13 to me now (I am getting supersticious...) as just
today (30/12/2001) my computer started producing an ominous compressor-like
loud noise. To all who read this, if you don't hear a while from me it's because
the whole system went berserk or crashed!

Expressing 2002 as the sum of the consecutive positive integers
from 148 up to 160 doesn't help either as we have to sum 13 items ! Aahrgh...

Good news, my pc is repaired! (04/01/2002).
The electrical feed(ing) has been replaced and the silence has returned.

My luck has returned it seems !
2002 can be written as the sum of 4 consecutive integers
starting from 499 up to 502.
2002 can be written as the sum of 7 consecutive integers
starting from 283 up to 289.
2002 can be written as the sum of 11 consecutive integers
starting from 177 up to 187.
2002 can be written as the sum of 28 consecutive integers
starting from 58 up to 85. (Note that 5885 is a palindrome.)
2002 can be written as the sum of 44 consecutive integers
starting from 24 up to 67.

(emails from 05-06/01/2002)
Carlos Rivera and Terry Trotter provided me with much more
interesting material (formulas and methods) for finding these
sequences. I will dedicate WONplate 122 to this topic.

(email from 19/01/2002)
Terry Trotter noticed that 2002 can also be expressed in the following manner
13 + 14 + ... + 63 + 64 = 2(65 + 66 + ... + 77 + 78)
Note the left expression ends with 64, then the right expression between
parentheses begins with the next consecutive integer 65.

(email from 05/01/2002)
Terry Trotter noticed the first two sequences have resp. 4 and 7 terms
and that 4 + 7 = 11 and 4 * 7 = 28
Note that 11 and 28 are the number of terms (#) in the next two sequences.
13 shows up twice, once as #, once as 1^st term and that 1 + 3 = 4
The digital roots of 499, 283, 148 and 13 (starting numbers of 4 different
sequences) is 4 or as the saying goes: "what goes around, comes around."

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

All combined yield this prime 2002271113 !

And 2 + 7 + 11 + 13 = 33, another palindrome !

Thanks Jaime Ayala (dd. 12/01/2002) for mailing this ignored property.

2002 is the sum of four consecutive primes 491 + 499 + 503 + 509

2002 is the sum of four consecutive integers 499 + 500 + 501 + 502

Note the reversed concatenation 502501500499 is prime!

502² + 501⁷ + 500¹¹ + 499¹³

is four times prime 29733695172086354662551885027753001 !

The HOMEPRIME of 2002 is found in 4 steps !

2 * 7 * 11 * 13

3 * 90371

17 * 22963

3 * 11 * 109 * 479 and 311109479 is already a prime !

2002 has this many PARTITIONS

4994467742183366148074839035447416380393781644

a number with exactly four distinct prime factors !

2² * 49117 * 154335341 * 164714564559993788333185532716763
The first four digits form a palindrome equal to a palindromic expression

4994 = 227.22

Note that the partition number and the palindrome starts and ends with...

2002q2002

The first four nontrivial integers q (q>1) sandwiched between two 2002's
so that the whole is divisible by this integer q are these four primes !

200222002 = 2 * 100111001

200272002 = 7 * 28610286

2002112002 = 11 * 182010182

2002132002 = 13 * 154010154

And we already knew that 2 * 7 * 11 * 13 = 2002 !

An amazing cycle coincidence, wouldn't you agree !

Can you give the next four primes q ?

2002525792002 = 52579 * 38086038

2002999900012002 = 99990001 * 20032002 !!!

2002?2002 = ? * ?

What is the smallest palprime q (q>11) with this property ?

Carlos Rivera (email 08/01/2002) found the smallest prime
such that the sum of its digits is 2002.

6*10²²²– 1 – 10¹⁶⁵

5(9)₅₆8(9)₁₆₅

59999999999999999999999999999999999999999999999999
99999998999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999
99999999999999999999999

Note 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
each of two or more digits in length. Here is Terry's solution:

Factor 20022002 2 x 7 x 11 x 13 x 73 x 137

959 = 7 x 137

949 = 13 x 73

22 = 2 x 11

And 22 x 949 x 959 = 20022002 !

Who find similar solutions for longer tautonyms (2002)_n (n>2) ?
(email 09/01/2002)
Terry himself explored longer 2002 concatenations and discovered
that 2002₄ or 2002200220022002 is the nicest because its factorization
2 x 7 x 11 x 13 x 17 x 73 x 137 x 5882353
yield four SEMIPRIMES, all palindromic, in the following manner

100000001 = 17 x 5882353

959 = 7 x 137

949 = 13 x 73

22 = 2 x 11

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 ?
Who can find the longest prime substring in M39 ?
Who can find a prime substring of length 2002 in M39 ?

(email 10/01/2002 & 14/04/2002)
Carlos Rivera submitted some prime related 2002 divertimenti !

The 2002^th odd number is 4003, a prime number.

The first time that K consecutive prime numbers concatenated in
'ascending' order provides a prime number of 2002 digits is
with K = 286, the first prime is 1058263, the last prime is 1062367.
(Certified prime using PRIMO after 70h 47mn 43s (14/04/2002) [pdg].)

The first time that K consecutive prime numbers concatenated in
'descending' order provides a prime number of 2002 digits is
with K = 286, the first prime is 1036957 downto the last prime 1033069.
(Currently only strongpseudoprime.)

The reader is invited to find the first reversible prime of this type.
A hard task... Carlos warns me!

Patrick De Geest
started appending to the left and right side of 2002 the consecutive primes
from 3 up to K in a wing-like manner :
... 11 7 5 3 2002 3 5 7 11 ...
and got a prime of length 62 when K reached 59.
59534743413731292319171311753200235711131719232931374143475359
is another prime with 2002 in the center!
(ps. this concatenated number is NOT palindromic!)

What is the next value of K ?

2002 acting as primeglue

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
         11200212200213
between three consecutive odd numbers happens first with
         35200237200239
        (Note 35 * 37 * 39 = 50505, another palindrome)

Can you discover other combinations ?

Let 2002 be followed by its next K primes and a new prime
emerges for K = 3, 6 and 22.

2002200320112017

2002200320112017202720292039

20022003201120172027202920392053...214121432153

Can you discover the fourth K value ?

Some 2002 loose ends

Carlos Rivera discovered (email 15/01/2002) that two successive 2002's
can 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 ?

Carlos Rivera wrote me that this is an extremely hard question and suggests to
encourage people to find smaller and smaller occurrences of Prime Gaps of 2002.
He added that I may be rewarded getting some interesting answers before the
puzzle expires...

Carlos Rivera (email 25/01/2002) informed me that Thomas R. Nicely maintains
a page on the First known occurrence prime gaps (exceeding 1132)
and we read there that no other person less than Harvey Dubner himself
already found the first prime gap of length 2002.
The lower prime contains 57 digits
(154395297513931047577420767367029385148800231451509075009)
and was found in this very same year 2002 !!

The concatenation of the integers from 2002 down to 1 is a number
of 6901 digits alas... divisible by 11959.

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
its following consecutive integers a prime number ?
200220032004200520062007...

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
setting the expiry date at 2112 for the solution is not a luxury ?

This is a 2002 story assembled from
two lines of approach. [pdg]

First line
I 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 the
last 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.

Second line
It was already known that the string formed by the concatenation of the
consecutive integers from 1 to 2002 yielded a number of 6901 digits.
After playing around a while with this number I subtracted 2002 from it
and came out of with number 4899 but what's more these two composites
transformed miraculously into primes when combined in either way.
48996901
69014899

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 48996901 and 69014899 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
and as a bonus the last (prime) part 3389 equals the very first one!

Do you fancy exploring the 10-digit numbers of the form abcde_fghij ?
Who knows, there might be lurking a cycle_5 somewhere...

2002 and Fibonacci Numbers

Carlos Rivera (email 24/01/2002) discovered that

F₁₅₅₈ = 1791672002678372... ...3860392002067977... ...376879
contains twice 2002, one starting at position 7, the other at position 135
(positions taken from left to right).

F₂₅₈₇ = 200235153260...996215944133
As you can see this number starts with 2002 !

In both cases the pertinent question is 'will this ever happen again ?'

Not very important but maybe worth mentioning :

2587 – 2002 = 585

2002 – 1558 = 444

two palindromes !

I (pdg - 28/02/2002) found that there exactly five Fibonacci numbers
of length 2002. Alas none of them contains 2002 as a substring.

F₉₅₇₇ = 133396879947398...639463408214257

F₉₅₇₈ = 215840685748080...115559275163089

F₉₅₇₉ = 349237565695479...755022683377346

F₉₅₈₀ = 565078251443559...870581958540435

F₉₅₇₁ = 914315817139038...625604641917781

What more can be discovered in these five numbers ?
Surely, one of these five must be the odd one out...

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.

2002 * 1 = 2002

2002 * 11 = 22022

2002 * 101 = 202202

2002 * 1001 = 2004002

2002 * 10001 = 20022002

2002 * 100001 = 200202002

2002 * 1(0_n)1 = 2002(0_n-3)2002 for n > 2

2002 * 11 = 22022

2002 * 111 = 222222

2002 * 1111 = 2224222

2002 * (1_n) = 222(4_n-3)222 for n > 2

2002 * 22 = 44044

2002 * 222 = 444444

2002 * 2222 = 4448444

2002 * (2_n) = 444(8_n-3)444 for n > 2

2002 * 3 = 6006

2002 * 33 = 66066

2002 * 333 = 666666

2002 * 4 = 8008

2002 * 44 = 88088

2002 * 444 = 888888

2002 * (5_n) / (2 * 5)

if n = 1 then 1001

if n = 2 then 11011

if n = 3 then 111111

if n = 4 then 1112111

for n > 3 then 111(2_n-3)111

2002 * 111 = 222222

2002 * 222 = 444444

2002 * 333 = 666666

2002 * 444 = 888888

2002 * 555 = 1111111 - 1

2002 * 666 = 1333331 - 1

2002 * 777 = 1555551 - 1

2002 * 888 = 1777771 - 1

2002 * 999 = 1999991 - 1

2002 * 121 = 242242

2002 * 1221 = 2444442

2002 * 12221 = 24466442

2002 * 122221 = 244686442

2002 * 1(2_n)1 = 2446(8_n-3)6442 for n > 2

2002 * 12321 = 24666642

2002 * 123321 = 246888642

2002 * 12421 = 24866842

2002 * 242 = 484484

2002 * 2442 = 4888884

2002 * 343 = 686686

2002² = 4008004

2(0_n)2² = 4(0_n)8(0_n)4

2002 * 11011 = 22044022

2002 * 110011 = 220242022

2002 * 1100011 = 2202222022

2002 * 11(0_n)11 = 22022(0_n-3)22022 for n > 2

2002 * 22022 = 44088044

2002 * 220022 = 440484044

2002 * 2200022 = 4404444044

2002 * 22(0_n)22 = 44044(0_n-3)44044 for n > 2

2002 * 3(0_n)3 = 6006(0_n-3)6006 for n > 2

2002 * 4(0_n)4 = 8008(0_n-3)8008 for n > 2

(2002 / 2) * (5(0_n)5 / 5)

if n = 0 then 11011

if n = 1 then 101101

if n = 2 then 1002001

if n = 3 then 10011001

for n > 2 then 1001(0_n-3)1001

(2002 * (6(0_n)6)) / 2

if n = 0 then 66066

if n = 1 then 606606

for n > 3 then 6006(0_n-3)6006

(2002 * (7(0_n)7)) / 2

if n = 0 then 77077

if n = 1 then 707707

for n > 2 then 7007(0_n-3)7007

(2002 * (8(0_n)8)) / 2

if n = 0 then 88088

if n = 1 then 808808

for n > 2 then 8008(0_n-3)8008

(2002 * (9(0_n)9)) / 6

if n = 0 then 33033

if n = 1 then 303303

if n = 2 then 3006003

for n > 2 then 3003(0_n-3)3003

Bases Numéricas (2,3,...,10) con el Símbolo 2002 y los cambios de Bases,

que generan palindrómicos

2002₄ = 252₁₀ = 202₈ = 11211₃

2002₆ = 434₁₀

2002₈ = (100001 + 1)₄

(2002 / 2)₈ = (1000000001)₂

2002₉ = 2000002₃

Fracción y Decimales periodicos Puros

(2(0_n)2 - 2) / (9_n+2) = 0,[2(0_n+1)]^{rep. inf.}

if n = 0 then 0,20202020...

if n = 1 then 0,200200200200...

if n = 2 then 0,2000200020002000...

(90_n)91 * 22 = 2(0_2(n+1))2

((90_n) + 1) * 22 = 2(0_2n)2

Nota los n, 2(n+1) y el 2n son subindices e indican la repetición

del número entre paréntesis (90₅)91 = 909090909091

Observar el patrón : 2002, 20002,... ,2000000000000002

obtenido con los números y operaciones anteriores.

2002 =

44² + 8² + 1² + 1²

10³ + 10³ + 1³ + 1³

6⁴ + 5⁴ + 3⁴

(2002_n) / (1000_n-1)1 = 2002

Potencia de un Complejo puro

-2002i^-2002 = 2002

i = unidad imaginaria

Tripletas Pitagóricas y el 2002

2002² + 1002000² = 1002002²

Su Perímetro = 1002 * 2002

Su Area = 501000 * 2002

Recurrencia

2002 | x⁴ - x³ - x² - x

tomar el valor de x de cada dígito del 2002 y sumar los resultados,

el valor obtenido aplicar el mismo proceso y... regresa al 2002

x = 2

2⁴ - 2³ - 2² - 2 = 2

x = 0

0⁴ - 0³ - 0² - 0 = 0

x = 0

0⁴ - 0³ - 0² - 0 = 0

x = 2

2⁴ - 2³ - 2² - 2 = 2

sumando = 4

x = 4

4⁴ - 4³ - 4² - 4 = 172

...recurrencia

x = 1

1⁴ - 1³ - 1² - 1 = -2

x = 7

7⁴ - 7³ - 7² - 7 = 2002

x = 2

2⁴ - 2³ - 2² - 2 = 2

sumando = 2002 ! el valor inicial !

En síntesis la Iteración del 2002 con x⁴ - x³ - x² - x es

2002

172

2002

Carlos Rivera (email 04/02/2002)
According to the N^th Prime Page by Andrew Booker
The 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^th,
2002, time will (for sixty seconds only) read in perfect symmetry.
To be more precise: 20:02, 20/02, 2002.
It is an event which has only ever happened once before,
and is something which will never be repeated. The last occasion
that time read in such a symmetrical pattern was long before the days
of the digital watch (or the 24-hour clock) :
10:01AM, on January 10, 1001 or 10:01, 01/10, 1001.
And because the clock only goes up to 23.59, it is something that will
never happen again.

Similar message (email 15/02/2002) from Maggie Feeney from Australia
forwarded to me by Enoch Haga.

8:02 pm on February 20, 2002 will be an historic moment in time.
It will not be marked by the chiming of any clocks or the ringing of bells,
but at that precise time, on that specific date, something will happen
which has not occurred for 1001 years and will never happen again.
As the clock ticks over from 8:01 pm on Wednesday, February 20, time
will, for sixty seconds only, read in perfect symmetry 2002, 2002, 2002,
or to be more precise - 20:02, 20/02, 2002.

Dinesh Aneja (email 20/02/2002) doesn't fully agree as there existed
another more recent occasion of a symmetrical time.
11:11AM, on November 11, 1111 or 11:11, 11/11, 1111.
Very well spotted, Dinesh !

And another correction from Carlos Rivera (email 20/02/2002)
There exist a counterexample to the statement
"Because the clock only goes up to 23.59, it is something that will
never happen again."
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
as mm/dd, i.e. the so-called American style, which is not being done for
the 2002 trivia.
Note that even using the European/Latin American style of dd/mm,
10/01 also is palindromic. Hence, this moment in time is "ambidextrous" as
         100110011001 and 100101101001
Another curious thing here is, why say 'AM' ? Wouldn't 'PM' provide
a more recent moment ? Albeit just a few hours later...
Then the Nov 11 idea really is "smashingly clever!"
Terry especially liked Carlos' correction about the "never again" idea...

Visit also
        Harvey Heinz's Palindromes webpage.
        Site OPUNDO time (email 12/01/2003 by Terry Trotter (†)).

Fermat numbers and 2002.

By Carlos Rivera (email 14/02/2002)

These first few Fermat numbers F_k
of the form 2^(2^k)+1
with k resp. equal to 3, 4, 5, 6, 7, 8, 11 & 12
relate to our year number 2002.
Do you know how ?

Hint: calculate F₃.F₄.F₅.F₆.F₇.F₈.F₁₁.F₁₂ and count the digits !

The sixth dimension of 2002
From Number Theory List [NMBRTHRY@LISTSERV.NODAK.EDU].

Yasutoshi Kohmoto searched the solutions of
2002⁶ = x³ + y³ + z³ + u³ for this year.

2002⁶

= 4003972³ + 579012³ - 8925³ - 483³

= 4003412³ + 604606³ + 32002³ + 1008³

= 4000973³ + 696742³ + 23467³ + 216³

= 3982499³ + 1068909³ + 34102³ + 862³

= 3945952³ + 1433279³ + 4441³ + 366³

= 3918046³ + 1618364³ + 35772³ + 946³

= 3898263³ + 1726369³ + 11640³ + 102³

= 3842592³ + 1970150³ - 11336³ - 182³

= 3785773³ + 2163515³ - 16512³ - 830³

= 3532614³ + 2727930³ + 28824³ + 916³

= 3525536³ + 2739726³ - 15407³ - 185³

= 3182368³ + 3179941³ - 16933³ - 228³

Yasutoshi used the algorithm described in the mail
"a^3+b^3+c^3+d^3=0" posted on Jun 2001.

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 !

Find the number X such that

2002 + X

20022002 + X

200220022002 + X

are simultaneously prime
or demonstrate that such number X does not exist.

On October 2, 2002 Nathan Russell (email) came to the rescue :

"This puzzle is impossible because one of those 3 numbers, for

any positive X, will be divisible by 3 and thus cannot be prime.

Note that 2002 is 1 mod 3, 20022002 is 2 mod 3, and

200220022002 is 0 mod 3."

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
search for patterns and structures that involve the current palindromic year
2002, I decided to do the same. This report will show some of the things that
were found by Patrick and myself.

My method of search might be called " Keithing " on some numbers.
Recall that a Keith Number is defined as follows:

an n-digit integer N with the following property: If a Fibonacci-
like sequence (in which each term in the sequence is the sum of
the n previous terms) is formed, with the first n terms being
the decimal digits of the number N, then N itself occurs as a
term in the sequence. For example, 197 is a Keith number since
it generates the sequence
1, 9, 7, 17, 33, 57, 107, 197, ...

I asked myself: "Why stop at the original number ? Why not continue
the sequence of sums ? Maybe 2002 would occur later." Unfortunately,
it does not appear for 197. Nor, as Patrick's ubasic program showed,
does any 3-digit number produce 2002. (I had earlier verified that no
2-digit number would either.) But things really begin to pick up once
we turned our attention to seed numbers of 4, 5, 6, digits, and so on.

To see what was discovered, go to WONplate 128.