HOME plateWON | World!OfNumbers 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 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
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 1st 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
Read all about it - this site is in 'castellano' in case it's more convenient - at
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
2002n_odd + 1
and
2002n_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 !

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 1st 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 = ? * ?
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*10222110165
or
5(9)568(9)165
or
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 factorisation
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 (213466917 – 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 2002th 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 :
25872002 = 585
20021558 = 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(0n)1 = 2002(0n-3)2002 for n > 2

2002 * 11 = 22022
2002 * 111 = 222222
2002 * 1111 = 2224222
2002 * (1n) = 222(4n-3)222 for n > 2

2002 * 22 = 44044
2002 * 222 = 444444
2002 * 2222 = 4448444
2002 * (2n) = 444(8n-3)444 for n > 2

2002 * 3 = 6006
2002 * 33 = 66066
2002 * 333 = 666666

2002 * 4 = 8008
2002 * 44 = 88088
2002 * 444 = 888888

2002 * (5n) / (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(2n-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(2n)1 = 2446(8n-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(0n)22 = 4(0n)8(0n)4

2002 * 11011 = 22044022
2002 * 110011 = 220242022
2002 * 1100011 = 2202222022
2002 * 11(0n)11 = 22022(0n-3)22022 for n > 2

2002 * 22022 = 44088044
2002 * 220022 = 440484044
2002 * 2200022 = 4404444044
2002 * 22(0n)22 = 44044(0n-3)44044 for n > 2

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

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

(2002 / 2) * (5(0n)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(0n-3)1001

(2002 * (6(0n)6)) / 2
if n = 0 then 66066
if n = 1 then 606606
for n > 3 then 6006(0n-3)6006

(2002 * (7(0n)7)) / 2
if n = 0 then 77077
if n = 1 then 707707
for n > 2 then 7007(0n-3)7007

(2002 * (8(0n)8)) / 2
if n = 0 then 88088
if n = 1 then 808808
for n > 2 then 8008(0n-3)8008

(2002 * (9(0n)9)) / 6
if n = 0 then 33033
if n = 1 then 303303
if n = 2 then 3006003
for n > 2 then 3003(0n-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(0n)2 - 2) / (9n+2) = 0,[2(0n+1)]rep. inf.
if n = 0 then 0,20202020...
if n = 1 then 0,200200200200...
if n = 2 then 0,2000200020002000...

(90n)91 * 22 = 2(02(n+1))2
((90n) + 1) * 22 = 2(02n)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

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

Potencia de un Complejo puro
-2002i-2002 = 2002

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 4 172 2002

Carlos Rivera (email 04/02/2002)
According to the Nth 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.

( 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 20th,
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 Fk
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

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.

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

 367 523 673 439 421 547 577 457 727 499 379 397 487 433 373 709

For Prime-magic squares adding up to 2000 Puzzle 106

Norman Luhn's 2002 entries. (email 12/03/2002)

222002 = 22 mod 2002

Norman used an old version of DERIVE to derive this neat fact.

Carlos Rivera (email 21/03/2002) reacted
Well, as a matter of fact, the curio is not so unique because there
are many bases b besides 22 such that b2002 = b mod 2002.
Who can get all of them ?

Furthermore Norman Luhn wrote
I was born on 02/13/1976.
In the year 2002 I am 26 years old.
Note that 26 = 02 * 13 and 1976 = 76 * 02 * 13.
These dates are very very rare.

In Europe, the EURO or became official currency in this year 2002.

Here is a test for the question if a number n is divisible by 2002 ?
The number must be even and, combine three digits of the
number in order (from right to left) alternating them with + then -.
If the result is 0 or a multiple of 1001, then the number 2002 divides it.
E.g. is 1976756782 divisible by 2002 ? 782 - 756 + 976 - 001 = 1001.
Yes, it is !

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