In the above, the numbers A & B combined and also the result on the right hand side
contain the digits 1...9 (zero-less pandigital).

There are 78 such solutions ¬

A      B

A      B

A      B

A      B

1596   24837
1635   29487
1875   24693
2367   19485
2436   18975
2538   17649
2784   15369
2943   17586
3495   12768
3516   28479
3654   12897
3846   29175
3864   25179
3915   28467
3984   26715
4197   25638
4359   16278
4395   27186
4536   12987
4698   21357

4857   21396
4896   21753
4965   17382
4968   15237
5367   14928
5376   12984
5376   18429
5436   17289
5469   13728
5496   18237
5613   27948
5679   14238
5697   21384
5763   21948
5964   18273
5976   21483
6189   25374
6198   23547
6345   21987
6453   19287

6459   17328
6471   23589
6471   29538
6528   13749
6549   27831
6813   24597
6831   29745
6945   13782
6975   14823
7164   25983
7356   24189
7359   28146
7431   29658
7458   23619
7659   13248
7659   13428
7824   19356
7863   21495
7869   23514
7953   16824

8259   13476
8295   14763
8349   17562
8352   17469
8517   29643
8745   16392
8793   12546
8796   21345
8796   23541
8934   15267
8973   12546
8976   25413
9156   27483
9168   23475
9627   18543
9687   24513
9735   24861
9741   26358

Observations

There are only 5 examples where A or B appears in more than one solution: viz ¬

There are several examples where simply swapping two digits in A leads to a member of another solution ¬
E.g.:

In the solution above, A contains the digits 1...5 and B contains the digits 6...9

The above is the only example where N is duplicated.

P = 3 A³ + B³ + C³ = N

A, B & C combined and N are zero-less pandigital.

There are 101 solutions; viz ¬

 A     B     C

 A     B     C

 A     B     C

 A     B     C

123   465   897
124   389   576
124   635   789
124   639   875
126   593   784
128   634   759
129   375   468
129   365   847
132   479   865
135   297   684
136   582   749
137   286   954
138   259   746
138   264   795
139   576   824
142   396   857
145   297   863
145   296   873
145   693   728
146   397   825
147   265   983
149   258   637
153   469   728
156   492   738
159   423   768
162   354   789

165   324   789
165   234   978
167   532   894
172   348   596
175   629   843
176   549   832
178   426   593
182   379   564
184   527   936
186   395   742
189   246   537
192   548   763
192   356   847
193   468   572
194   285   736
194   376   825
194   627   835
196   273   458
196   342   785
197   364   825
198   243   675
213   467   589
213   578   649
215   378   496
216   397   485
216   387   495

219   438   567
219   436   587
231   456   789
234   619   857
237   654   819
246   539   718
249   381   675
251   486   739
253   419   768
254   391   768
258   314   697
259   641   837
261   457   938
267   384   951
269   537   841
271   398   546
271   458   639
274   591   863
278   351   496
279   413   856
312   479   658
317   489   526
318   547   692
321   498   576
321   564   897
328   519   674

342   596   871
351   468   792
362   459   718
372   519   864
374   526   918
378   512   649
381   469   572
386   519   724
394   561   728
397   458   621
398   415   672
418   693   725
429   576   831
429   671   835
462   539   718
483   597   621
487   521   693
489   613   725
493   581   627
493   512   786
514   639   782
574   639   812
591   624   738

While looking through the above list I noticed that some values occur in more than one solution
and I wondered whether I could chain several such solutions together; wherein the last number
becomes the first number in the next link: e.g. ¬

(123, 465, 897)
          (897, 321, 564) 
                    (564, 182, 379)

A slightly longer chain would be ¬

(193, 468, 572)
          (572, 381, 469)
                    (469, 153, 728)
                              (728, 693, 145)
                                        (145, 296, 873)

The longest chain that I found consists of 24 members ¬

(172, 348, 596) , (596, 871, 342) , (342, 785, 196) , (196, 273, 458) , (458, 271, 639) ,
(639, 875, 124) , (124, 389, 576) , (576, 831, 429) , (429, 671, 835) , (835, 194, 627) ,
(627, 581, 493) , (493, 786, 512) , (512, 649, 378) , (378, 215, 496) , (496, 278, 351) ,
(351, 792, 468) , (468, 193, 572) , (572, 381, 469) , (469, 153, 728) , (728, 145, 693) ,
(693, 418, 725) , (725, 613, 489) , (489, 317, 526) , (526, 918, 374)

This chain was found by 'hand' so it is possible that I may have missed a longer chain.

While searching for the above I came across the following nine-member 'pandigital sociable group'
i.e. the last member can be chained with the very first member.

	(124, 389, 576) (576, 831, 429) (429, 671, 835) (835, 627, 194) (194, 376, 825) (825, 146, 397) (397, 621, 458) (458, 271, 639) (639, 875, 124)

There is also one example of a 'pandigital amicable pair'.

(539, 246, 718)
          (718, 462, 539)

(123, 564, 879) & (124, 639, 875)
  |    |    |       |    |    |
(231, 465, 897)	& (142, 396, 857)

(231, 456, 789)

(394, 561, 728) = (394, 394+167, 394+167+167)

P = 4 A⁴ + B⁴ + C⁴ + D⁴ = N

E.g.:

A, B, C & D combined and N are zero-less pandigital.

There are only 5 solutions; viz ¬

The second and third solutions shown above are particularly noteworthy (another 'pandigital amicable pair').

P = 5 A⁵ + B⁵ + C⁵ + D⁵ + E⁵ = N

The above is the only solution.

P >= 6 A^p + B^p + C^p + D^p + E^p + F^p + ... = N

No solutions.

[ March 24, 2005 ]
The Unique Pandigital 3816547290
by John Morse

" In the header of this message is 3,816,547,290 which is the only 10-digit
pandigital number whose first N digits are a multiple of N. I first discovered
that in The Dictionary of Curious and Interesting Numbers by David Wells,
and I wrote an explanation of how one could figure out that is the number.

John Morse
Albany New York USA "

Mathematics Puzzle

Find the one and only 10-digit number such that :

This explanation is my own, but I first learned about the amazing
mystery number (381,654,729) in an entertaining book, one of my ten
favorite math-themed books of all time :

Procedure

Let the letters A thru J stand for the digits of the number. Then :

A B C D E F G H I J is the mystery number, whereas :

AB is evenly divisible by 2,
ABC is evenly divisible by 3,
ABCD is evenly divisible by 4,
ABCDE is evenly divisible by 5,
ABCDEF is evenly divisible by 6,
ABCDEFG is evenly divisible by 7,
ABCDEFGH is evenly divisible by 8,
ABCDEFGHI is evenly divisible by 9,
ABCDEFGHIJ is evenly divisible by 10.

A number is evenly divisible by 10 only if its last digit is zero.

Hence, we have found right away that J must equal zero, since the
ten-digit number ends in that digit J.

What about divisibility by 9 ? This particular number takes care of that
matter because its nine-digit 'fragment' ABCDEFGHI must use one each of
digits 1 thru 9, and the sum of the nine digits is 45, which is a multiple of 9.

A number is evenly divisible by 9 if the sum of its digits is 9 or a
multiple of 9 itself. Hence, ANY arrangement of the digits 1 thru 9
(or 0 thru 9), each digit used exactly once, will always result in the
number being a multiple of 9.

A number is evenly divisible by 5 only if its last digit is a five or
a zero.

Since all the digits of ABCDEFGHIJ are different, we cannot use zero
since J already has that value.

The first five digits of the entire number are ABCDE, hence digit E
must be 5.

So far, the mystery number looks like this :

A B C D 5 F G H I 0

Even numbers only are evenly divisible by even divisors, e.g., if you
want to divide some number N by 2 or 4 or 6, etc., that number N must
be even since dividing an odd number by an even number will always leave
(an odd) remainder.

Hence, consider the mystery number's even-numbered digits. They
correspond to letters B, D, F, and H. Since zero is being used for J, the
even digits 2, 4, 6, and 8 must correspond to those other four letters.

Therefore, the remaining letters in the number - A, C, G, and I - must
stand for odd digits.

Because the odd digit 5 is already in use via the letter E, A can only
be 1, 3, 7, or 9. The same restrictions apply to C, G, and I.

There are many ways that AB is evenly divisible by 2 : merely let B
equal 2 or 4 or 6 or 8. Hence, there are too many possible ways so
far to assign digits for A and B. How about ABC, which must be a
multiple of 3 ?

The digits A + B + C must add up to a multiple of 3, and in this
instance, B is the only even digit. Hence, the following trios are possible
for fragment ABC :

Eighteen possible permutations exist for ABC; it looks like a lot of
trial and error lies ahead if we test all these 3-digit groups in
combination with the remaining digits for D, F, G, and H. (It does not
matter yet what digit I must be, since the entire 9-digit number up to I
is already divisible by 9.)

However, note that the four-digit fragment ABCD must be a multiple of 4,
and this is possible only if the number formed by the last two digits -
here, CD - is a multiple of 4.

Since C is odd, D can only be 2 or 6. If we try 4 or 8 instead for D, then we
get numbers such as 18, 34, 78, or 94 - none of which are evenly divisible by 4.

A similar situation arises for the 8-digit fragment ABCDEFGH. Since G
is odd, H can only be 2 or 6. A number is divisible by 8 only if its
last three digits (here, FGH) is evenly divisible by 8.

So, between them, D and H use up the digits 2 and 6. We are not sure
which letter will have which digit, but we can eliminate the 3-digit
fragments above where B happens to be 2 or 6, leaving these possible
permutations for the digit group ABC :

Good. Number of permutations is now only eight instead of eighteen.

Since B and F must be even, they must use the only remaining even
digits - 4 and 8 - between them. Let's take a look at the mystery
number where we assign the digits 2 and 6 to D and H and the digits
4 and 8 to B and F :

Consider the six-digit fragment ABCDEF. Its digits must add up to a multiple
of 3. All multiples of six are even, but we already know that F must be even.

Since the digits ABC add up to a multiple of 3, so must those in fragment
DEF. Since E is 5, only two possibilities exist for digits in DEF :

So, we can eliminate two of the four 10-digit numbers above, leaving :

Let's take another look at the 8-digit fragment ABCDEFGH and see if we
can determine whether F must be 4 or 8.

If D=2, then B must be 4, F must be 8, and H must be 6. Since FGH must
be a multiple of 8, it could only end in the following three digits :

Of these four 3-digit numbers, 836 and 876 are NOT multiples of 8, so
eliminate them, leaving only 816 and 896 for FGH.

If D=6, then B must be 8, F must be 4, and H must be 2. Since FGH must
be a multiple of 8, it could only end in the following three digits :

Of these four 3-digit numbers, 412 and 492 are NOT multiples of 8, so
eliminate them, leaving only 432 and 472 for FGH.

Hence, possible arrangements of digits in the mystery number could be :

Recall that fragment ABC must be a multiple of 3. In choice [1], the
only available digits for A and C are 3, 7, and 9. No combination of
these digits will enable ABC to be a multiple of 3, so eliminate choice [1].

For choice [2], the only digits available are 1, 3, and 9. No combination
of these will make ABC a multiple of 3, so scratch choice [2].

For choice [3], the only digits left are 1, 7, and 9. The only workable
combination that will make ABC a multiple of 3 is when neither A nor C
is 7, so I must be 7.

For choice [4], the only digits available are 1, 3, and 9. Digits
1 and 3 may represent A and C, in which case I must be 9. If digits
1 and 9 stand in for A and C, then I must be 3.

So, two choices for the mystery number were removed, leaving these
possible configurations for the mystery number :

Almost there! Each of these six numbers is such that the number AB is
divisible by 2, ABC by 3, ABCD by 4, ABCDE by 5, ABCDEF by 6, ABCDEFGH
by 8, ABCDEFGHI by 9, and ABCDEFGHIJ by 10.

What about the fragment ABCDEFG ? Is it divisible by 7 ? Test each of
the six numbers above using their first seven digits :

[1] 1896543 divided by 7 leaves remainder 5. nope.
[2] 9816543 divided by 7 leaves remainder 2. nope.
[3] 1836547 divided by 7 leaves remainder 6. nope.
[4] 3816547 divided by 7 leaves remainder 0. YES !
[5] 1896547 divided by 7 leaves remainder 2. nope.
[6] 9816547 divided by 7 leaves remainder 6. nope.

Aha! Only choice [4] leaves no remainder when its fragment ABCDEFG is
divided by 7.

Finally we found the mystery number ! It is

Are all the digits different ? Yes. A number with that property is
known as "pandigital". The prefix "pan-" means "all".

How about divisibility of the first N digits by N ? Check and see :

38 divided by 2 = 19, no remainder. Yawn.
381 divided by 3 = 127, no remainder. Okay.
3816 divided by 4 = 954, no remainder. Okay!
38165 divided by 5 = 7633, no remainder. Big deal.
381654 divided by 6 = 63609, no remainder. Good.
3816547 divided by 7 = 545221, no remainder. Good!
38165472 divided by 8 = 4770684, no remainder. Great!
381654729 divided by 9 = 42406081, no remainder. Of course.
3816547290 divided by 10 = 381654729, no remainder. Wonderful!!!

There is no other 10-digit number whose first N digits are evenly
divisible by N - and where every digit occurs once !

Other internet sources discussing this topic :
http://ken.duisenberg.com/potw/archive/arch96/960919sol.html
http://www.nrich.maths.org.uk/public/viewer.php?obj_id=796&part=solution&refpage=viewer.php
http://begghilos2.ath.cx/~jyseto/Academia/Math-Problem-2.php
http://www.rodoval.com/heureka/probsnum.html
... many more when 3816547290 is entered as keyword in GOOGLE for instance !

[ February 27, 2005 ]
Nine and Ten digit squares
by Peter Kogel

" I've had a fascination for numbers and rec maths ever since I first read
Martin Gardner's column in Scientific American many years ago. My particular
fascination is for nine and ten digit number patterns and I've found your
excellent website the source of much inspiration.

This project has been to investigate patterns along the lines of 99066^2
( 9814072356 ) and have found some interesting results that you might like.

Some highlights

77772277772 = 60485271895340361729

The square contains each digit 0...9 twice and the square root is a rather
particular palindrome.

42539071862 = 1809572634_7102438596

The lefthand side is pandigital and the square on the righthand side contains
the digits 0...9 repeated in each half (there are two other such examples).

Nine and Ten digit squares

It is well known that the square of 11826 ( 139854276 ) contains all the digits 1...9
and that the square of 32043 ( 1026753849 ) contains all the digits 0...9.
Others have investigated cases where the digits in the square are repeated more than once.
I have extended this investigation up to the 6th repeat of the digits. My findings are shown below.

I haven’t yet found the largest example where all ten digits are repeated 6 times.
Perhaps someone else can find it and perhaps extend the table even further.

While searching for the above I came across the following interesting pair:

A and B differ by only two digits!

Pandigital squares

It is well known that the squares of A = 57321 and B = 60984 each contain all ten digits
and that ‘A & B’ combined also contains the ten digits. I wondered whether such results
could be extended such that ‘A & B’ contain the digits repeated twice. I very soon found
literally hundreds of solutions of which the following type is of interest:

Note that the squares of ‘A’ & ‘B’ each contain all ten digits repeated twice,
‘A’ contains no digit larger than 5 and ‘A & B’ combined contains all ten digits repeated twice.
There are many other such examples.

There are also many examples where ‘A’ and ‘B’ are pan-digital. E.g.:

Similarly the following are pan-9-digital

While searching for these examples I came across the following remarkable trio:

Notice that the digits 0...9 are contained in each ‘half’ of the square
on the right hand side. There are no pan-9-digital equivalents.
[ Tom Marlow discovered only the first two squares. He submitted it to Ed Pegg's
Mathpuzzle site. Source material added 26 January 2003.]

Complete listing of the above squared nine- and pandigitals
with the higher powers as well,
up to the fifth for ninedigitals
up to the sixth for pandigitals
available at http://blue.kakiko.com/mmrmmr/htm/eqtn11.html

Palindromes

There are many examples where N is a palindrome.

I have not found any solutions where the square is also a palindrome
and I suspect that none exist.

Higher powers

It is natural to ask whether the above results can be extended to higher powers.
The answer is of course “yes!”

E.g.

In the following, the notation (N, P, D) means that the digits 1...9 (or 0...9)
are repeated ‘D’ times in the value of N^P.

Nine digit examples:

Ten digit examples:    [ A074205 ]

The challenge above was to make ‘D’ as small as possible for any particular ‘P’.
I also have constructed a table of smallest and largest of any particular type
up to and including P = 15.

There are also examples where P = D in viz:

Nine digit examples:

I have not yet found any example where P = D = 7 and I suspect there are none
because it becomes increasingly unlikely that the power of any number will not
contain a zero when P is large.

Ten digit examples:

I have not checked any power larger than 15 as yet.

Pandigital powers

There are a few examples where ‘N’ is pandigital.

Complete listing available at http://blue.kakiko.com/mmrmmr/htm/eqtn11.html

Nine digit examples:

Ten digit examples:

There do not appear to be any examples for higher powers.

A curious pan-digital example is the following

Notice that the left hand side including the index contains the digits 0...9 and that the right hand side
contains the digits 1...9 repeated 3 times. I have found no other such example for any power.

Palindromes

To date I haven’t searched for palindrome solutions for cubes and higher powers
but I have no reason to believe that they do not exist. I’ll investigate further when
I have a bit more time.

[ May 18, 2005 ]
Nine and Ten digit powers
Here are some more results from Peter Kogel which can be added as an addendum :

333333500000125024109149626 squared, contains the digits 1..9 repeated six times (smallest)
999999944444387333120727861 squared, contains the digits 1..9 repeated six times (largest)

316227783585222352038673081356² contains digits 0..9 repeated six times (smallest)
999999944444387345603558350056² contains digits 0..9 repeated six times (largest)

3254196708² = 10589_7962143580_37264 digits 0..9 contained in the centre and each digit repeated twice overall
8591674023² = 73816_8625174930_04529 ditto
8952063174² = 80139_4350712869_54276 ditto
9076148352² = 82376_4689075123_15904 ditto
9683104752² = 93762_5176382049_81504 ditto
9705843126² = 94203_3907865214_51876 ditto

4325907³ = 80952_7368052419_17643 digits 0..9 contained in the centre (only cubic example)

8691229761¹⁶ contains the digits 0..9 repeated sixteen times (smallest)
9970902252¹⁶ contains the digits 0..9 repeated sixteen times (largest)

8800389678¹⁷ contains the digits 0..9 repeated seventeen times (smallest)
9740383767¹⁷ contains the digits 0..9 repeated seventeen times (largest)

9985819785²⁰ =
972018388122736146908354492278598571766247280628254566407260946047011341
594668735384194513571360871003364967097249794743843109249267940157739156
83917180391558815561723889985220630315208530426025390625

Ie. the 200 digit number contains each digit 0..9 repeated 20 times!
This is the largest such example.

" I used Yuji Kida's excellent UBASIC for all my calculations. The
program I wrote is not very elegant (I am only an amateur programmer)
and the algorithm I developed is probably far from optimal. Indeed,
from what I have learned from this project I would probably tackle it
from a completely different angle if I were ever to do it again. "

[ December 29, 2004 ]
Solutions for ABC + DEF = GHI

Linda Rojewski asked for all the combinations of a three digit number added
to a three digit number to equal a three digit number. You can only use digits 1-9,
and can not use any number twice. You can add to carry over a number of which
this number would not count as usage of that number.

Solution:
I (pdg) used the well known UBASIC tool and the result is listed below.
In total there are 336 solutions though not all are unique!
124 + 659 = 783 is listed as well as 659 + 124 = 783.

Question:
From this list it should be easy to extract only the distinct solutions.
How many do you count ?
Send them in and I will highlight them.

 1 	124659783	 124 + 659 = 783 
 2 	125739864	 125 + 739 = 864 
 3 	127359486	 127 + 359 = 486 
 4 	127368495	 127 + 368 = 495 
 5 	128367495	 128 + 367 = 495 
 6 	128439567	 128 + 439 = 567 
 7 	129357486	 129 + 357 = 486 
 8 	129438567	 129 + 438 = 567 
 9 	129654783	 129 + 654 = 783 
 10 	129735864	 129 + 735 = 864 
 11 	134658792	 134 + 658 = 792 
 12 	135729864	 135 + 729 = 864 
 13 	138429567	 138 + 429 = 567 
 14 	138654792	 138 + 654 = 792 
 15 	139428567	 139 + 428 = 567 
 16 	139725864	 139 + 725 = 864 
 17 	142596738	 142 + 596 = 738 
 18 	142695837	 142 + 695 = 837 
 19 	143586729	 143 + 586 = 729 
 20 	145692837	 145 + 692 = 837 
 21 	146583729	 146 + 583 = 729 
 22 	146592738	 146 + 592 = 738 
 23 	152487639	 152 + 487 = 639 
 24 	152784936	 152 + 784 = 936 
 25 	154629783	 154 + 629 = 783 
 26 	154638792	 154 + 638 = 792 
 27 	154782936	 154 + 782 = 936 
 28 	157329486	 157 + 329 = 486 
 29 	157482639	 157 + 482 = 639 
 30 	158634792	 158 + 634 = 792 
 31 	159327486	 159 + 327 = 486 
 32 	159624783	 159 + 624 = 783 
 33 	162387549	 162 + 387 = 549 
 34 	162783945	 162 + 783 = 945 
 35 	163782945	 163 + 782 = 945 
 36 	167328495	 167 + 328 = 495 
 37 	167382549	 167 + 382 = 549 
 38 	168327495	 168 + 327 = 495 
 39 	173286459	 173 + 286 = 459 
 40 	173295468	 173 + 295 = 468 
 41 	175293468	 175 + 293 = 468 
 42 	176283459	 176 + 283 = 459 
 43 	182367549	 182 + 367 = 549 
 44 	182394576	 182 + 394 = 576 
 45 	182457639	 182 + 457 = 639 
 46 	182493675	 182 + 493 = 675 
 47 	182754936	 182 + 754 = 936 
 48 	182763945	 182 + 763 = 945 
 49 	183276459	 183 + 276 = 459 
 50 	183492675	 183 + 492 = 675 
 51 	183546729	 183 + 546 = 729 
 52 	183762945	 183 + 762 = 945 
 53 	184392576	 184 + 392 = 576 
 54 	184752936	 184 + 752 = 936 
 55 	186273459	 186 + 273 = 459 
 56 	186543729	 186 + 543 = 729 
 57 	187362549	 187 + 362 = 549 
 58 	187452639	 187 + 452 = 639 
 59 	192384576	 192 + 384 = 576 
 60 	192483675	 192 + 483 = 675 
 61 	192546738	 192 + 546 = 738 
 62 	192645837	 192 + 645 = 837 
 63 	193275468	 193 + 275 = 468 
 64 	193482675	 193 + 482 = 675 
 65 	194382576	 194 + 382 = 576 
 66 	195273468	 195 + 273 = 468 
 67 	195642837	 195 + 642 = 837 
 68 	196542738	 196 + 542 = 738 
 69 	214569783	 214 + 569 = 783 
 70 	214659873	 214 + 659 = 873 
 71 	215478693	 215 + 478 = 693 
 72 	215748963	 215 + 748 = 963 
 73 	216378594	 216 + 378 = 594 
 74 	216738954	 216 + 738 = 954 
 75 	218349567	 218 + 349 = 567 
 76 	218376594	 218 + 376 = 594 
 77 	218439657	 218 + 439 = 657 
 78 	218475693	 218 + 475 = 693 
 79 	218736954	 218 + 736 = 954 
 80 	218745963	 218 + 745 = 963 
 81 	219348567	 219 + 348 = 567 
 82 	219438657	 219 + 438 = 657 
 83 	219564783	 219 + 564 = 783 
 84 	219654873	 219 + 654 = 873 
 85 	234657891	 234 + 657 = 891 
 86 	235746981	 235 + 746 = 981 
 87 	236718954	 236 + 718 = 954 
 88 	236745981	 236 + 745 = 981 
 89 	237654891	 237 + 654 = 891 
 90 	238419657	 238 + 419 = 657 
 91 	238716954	 238 + 716 = 954 
 92 	239418657	 239 + 418 = 657 
 93 	241596837	 241 + 596 = 837 
 94 	243576819	 243 + 576 = 819 
 95 	243675918	 243 + 675 = 918 
 96 	245673918	 245 + 673 = 918 
 97 	245718963	 245 + 718 = 963 
 98 	245736981	 245 + 736 = 981 
 99 	246573819	 246 + 573 = 819 
 100 	246591837	 246 + 591 = 837 
 101 	246735981	 246 + 735 = 981 
 102 	248319567	 248 + 319 = 567 
 103 	248715963	 248 + 715 = 963 
 104 	249318567	 249 + 318 = 567 
 105 	251397648	 251 + 397 = 648 
 106 	254619873	 254 + 619 = 873 
 107 	254637891	 254 + 637 = 891 
 108 	257391648	 257 + 391 = 648 
 109 	257634891	 257 + 634 = 891 
 110 	259614873	 259 + 614 = 873 
 111 	264519783	 264 + 519 = 783 
 112 	269514783	 269 + 514 = 783  

 113 	271593864	 271 + 593 = 864
 114 	271683954	 271 + 683 = 954 
 115 	273186459	 273 + 186 = 459 
 116 	273195468	 273 + 195 = 468 
 117 	273546819	 273 + 546 = 819 
 118 	273591864	 273 + 591 = 864 
 119 	273645918	 273 + 645 = 918 
 120 	273681954	 273 + 681 = 954 
 121 	275193468	 275 + 193 = 468 
 122 	275418693	 275 + 418 = 693 
 123 	275643918	 275 + 643 = 918 
 124 	276183459	 276 + 183 = 459 
 125 	276318594	 276 + 318 = 594 
 126 	276543819	 276 + 543 = 819 
 127 	278316594	 278 + 316 = 594 
 128 	278415693	 278 + 415 = 693 
 129 	281394675	 281 + 394 = 675 
 130 	281673954	 281 + 673 = 954 
 131 	283176459	 283 + 176 = 459 
 132 	283671954	 283 + 671 = 954 
 133 	284391675	 284 + 391 = 675 
 134 	286173459	 286 + 173 = 459 
 135 	291357648	 291 + 357 = 648 
 136 	291384675	 291 + 384 = 675 
 137 	291546837	 291 + 546 = 837 
 138 	291573864	 291 + 573 = 864 
 139 	293175468	 293 + 175 = 468 
 140 	293571864	 293 + 571 = 864 
 141 	294381675	 294 + 381 = 675 
 142 	295173468	 295 + 173 = 468 
 143 	296541837	 296 + 541 = 837 
 144 	297351648	 297 + 351 = 648 
 145 	314658972	 314 + 658 = 972 
 146 	316278594	 316 + 278 = 594 
 147 	317529846	 317 + 529 = 846 
 148 	317628945	 317 + 628 = 945 
 149 	318249567	 318 + 249 = 567 
 150 	318276594	 318 + 276 = 594 
 151 	318627945	 318 + 627 = 945 
 152 	318654972	 318 + 654 = 972 
 153 	319248567	 319 + 248 = 567 
 154 	319527846	 319 + 527 = 846 
 155 	324567891	 324 + 567 = 891 
 156 	324657981	 324 + 657 = 981 
 157 	327159486	 327 + 159 = 486 
 158 	327168495	 327 + 168 = 495 
 159 	327519846	 327 + 519 = 846 
 160 	327564891	 327 + 564 = 891 
 161 	327618945	 327 + 618 = 945 
 162 	327654981	 327 + 654 = 981 
 163 	328167495	 328 + 167 = 495 
 164 	328617945	 328 + 617 = 945 
 165 	329157486	 329 + 157 = 486 
 166 	329517846	 329 + 517 = 846 
 167 	341586927	 341 + 586 = 927 
 168 	342576918	 342 + 576 = 918 
 169 	346572918	 346 + 572 = 918 
 170 	346581927	 346 + 581 = 927 
 171 	348219567	 348 + 219 = 567 
 172 	349218567	 349 + 218 = 567 
 173 	351297648	 351 + 297 = 648 
 174 	352467819	 352 + 467 = 819 
 175 	354618972	 354 + 618 = 972 
 176 	354627981	 354 + 627 = 981 
 177 	357129486	 357 + 129 = 486 
 178 	357291648	 357 + 291 = 648 
 179 	357462819	 357 + 462 = 819 
 180 	357624981	 357 + 624 = 981 
 181 	358614972	 358 + 614 = 972 
 182 	359127486	 359 + 127 = 486 
 183 	362187549	 362 + 187 = 549 
 184 	362457819	 362 + 457 = 819 
 185 	364527891	 364 + 527 = 891 
 186 	367128495	 367 + 128 = 495 
 187 	367182549	 367 + 182 = 549 
 188 	367452819	 367 + 452 = 819 
 189 	367524891	 367 + 524 = 891 
 190 	368127495	 368 + 127 = 495 
 191 	372546918	 372 + 546 = 918 
 192 	376218594	 376 + 218 = 594 
 193 	376542918	 376 + 542 = 918 
 194 	378216594	 378 + 216 = 594 
 195 	381294675	 381 + 294 = 675 
 196 	381546927	 381 + 546 = 927 
 197 	382167549	 382 + 167 = 549 
 198 	382194576	 382 + 194 = 576 
 199 	384192576	 384 + 192 = 576 
 200 	384291675	 384 + 291 = 675 
 201 	386541927	 386 + 541 = 927 
 202 	387162549	 387 + 162 = 549 
 203 	391257648	 391 + 257 = 648 
 204 	391284675	 391 + 284 = 675 
 205 	392184576	 392 + 184 = 576 
 206 	394182576	 394 + 182 = 576 
 207 	394281675	 394 + 281 = 675 
 208 	397251648	 397 + 251 = 648 
 209 	415278693	 415 + 278 = 693 
 210 	418239657	 418 + 239 = 657 
 211 	418275693	 418 + 275 = 693 
 212 	419238657	 419 + 238 = 657 
 213 	428139567	 428 + 139 = 567 
 214 	429138567	 429 + 138 = 567 
 215 	438129567	 438 + 129 = 567 
 216 	438219657	 438 + 219 = 657 
 217 	439128567	 439 + 128 = 567 
 218 	439218657	 439 + 218 = 657 
 219 	452187639	 452 + 187 = 639 
 220 	452367819	 452 + 367 = 819 
 221 	457182639	 457 + 182 = 639 
 222 	457362819	 457 + 362 = 819 
 223 	462357819	 462 + 357 = 819 
 224 	467352819	 467 + 352 = 819 

 225 	475218693	 475 + 218 = 693 
 226 	478215693	 478 + 215 = 693 
 227 	482157639	 482 + 157 = 639 
 228 	482193675	 482 + 193 = 675 
 229 	483192675	 483 + 192 = 675 
 230 	487152639	 487 + 152 = 639 
 231 	492183675	 492 + 183 = 675 
 232 	493182675	 493 + 182 = 675 
 233 	514269783	 514 + 269 = 783 
 234 	517329846	 517 + 329 = 846 
 235 	519264783	 519 + 264 = 783 
 236 	519327846	 519 + 327 = 846 
 237 	524367891	 524 + 367 = 891 
 238 	527319846	 527 + 319 = 846 
 239 	527364891	 527 + 364 = 891 
 240 	529317846	 529 + 317 = 846 
 241 	541296837	 541 + 296 = 837 
 242 	541386927	 541 + 386 = 927 
 243 	542196738	 542 + 196 = 738 
 244 	542376918	 542 + 376 = 918 
 245 	543186729	 543 + 186 = 729 
 246 	543276819	 543 + 276 = 819 
 247 	546183729	 546 + 183 = 729 
 248 	546192738	 546 + 192 = 738 
 249 	546273819	 546 + 273 = 819 
 250 	546291837	 546 + 291 = 837 
 251 	546372918	 546 + 372 = 918 
 252 	546381927	 546 + 381 = 927 
 253 	564219783	 564 + 219 = 783 
 254 	564327891	 564 + 327 = 891 
 255 	567324891	 567 + 324 = 891 
 256 	569214783	 569 + 214 = 783 
 257 	571293864	 571 + 293 = 864 
 258 	572346918	 572 + 346 = 918 
 259 	573246819	 573 + 246 = 819 
 260 	573291864	 573 + 291 = 864 
 261 	576243819	 576 + 243 = 819 
 262 	576342918	 576 + 342 = 918 
 263 	581346927	 581 + 346 = 927 
 264 	583146729	 583 + 146 = 729 
 265 	586143729	 586 + 143 = 729 
 266 	586341927	 586 + 341 = 927 
 267 	591246837	 591 + 246 = 837 
 268 	591273864	 591 + 273 = 864 
 269 	592146738	 592 + 146 = 738 
 270 	593271864	 593 + 271 = 864 
 271 	596142738	 596 + 142 = 738 
 272 	596241837	 596 + 241 = 837 
 273 	614259873	 614 + 259 = 873 
 274 	614358972	 614 + 358 = 972 
 275 	617328945	 617 + 328 = 945 
 276 	618327945	 618 + 327 = 945 
 277 	618354972	 618 + 354 = 972 
 278 	619254873	 619 + 254 = 873 
 279 	624159783	 624 + 159 = 783 
 280 	624357981	 624 + 357 = 981 
 281 	627318945	 627 + 318 = 945 
 282 	627354981	 627 + 354 = 981 
 283 	628317945	 628 + 317 = 945 
 284 	629154783	 629 + 154 = 783 
 285 	634158792	 634 + 158 = 792 
 286 	634257891	 634 + 257 = 891 
 287 	637254891	 637 + 254 = 891 
 288 	638154792	 638 + 154 = 792 
 289 	642195837	 642 + 195 = 837 
 290 	643275918	 643 + 275 = 918 
 291 	645192837	 645 + 192 = 837 
 292 	645273918	 645 + 273 = 918 
 293 	654129783	 654 + 129 = 783 
 294 	654138792	 654 + 138 = 792 
 295 	654219873	 654 + 219 = 873 
 296 	654237891	 654 + 237 = 891 
 297 	654318972	 654 + 318 = 972 
 298 	654327981	 654 + 327 = 981 
 299 	657234891	 657 + 234 = 891 
 300 	657324981	 657 + 324 = 981 
 301 	658134792	 658 + 134 = 792 
 302 	658314972	 658 + 314 = 972 
 303 	659124783	 659 + 124 = 783 
 304 	659214873	 659 + 214 = 873 
 305 	671283954	 671 + 283 = 954 
 306 	673245918	 673 + 245 = 918 
 307 	673281954	 673 + 281 = 954 
 308 	675243918	 675 + 243 = 918 
 309 	681273954	 681 + 273 = 954 
 310 	683271954	 683 + 271 = 954 
 311 	692145837	 692 + 145 = 837 
 312 	695142837	 695 + 142 = 837 
 313 	715248963	 715 + 248 = 963 
 314 	716238954	 716 + 238 = 954 
 315 	718236954	 718 + 236 = 954 
 316 	718245963	 718 + 245 = 963 
 317 	725139864	 725 + 139 = 864 
 318 	729135864	 729 + 135 = 864 
 319 	735129864	 735 + 129 = 864 
 320 	735246981	 735 + 246 = 981 
 321 	736218954	 736 + 218 = 954 
 322 	736245981	 736 + 245 = 981 
 323 	738216954	 738 + 216 = 954 
 324 	739125864	 739 + 125 = 864 
 325 	745218963	 745 + 218 = 963 
 326 	745236981	 745 + 236 = 981 
 327 	746235981	 746 + 235 = 981 
 328 	748215963	 748 + 215 = 963 
 329 	752184936	 752 + 184 = 936 
 330 	754182936	 754 + 182 = 936 
 331 	762183945	 762 + 183 = 945 
 332 	763182945	 763 + 182 = 945 
 333 	782154936	 782 + 154 = 936 
 334 	782163945	 782 + 163 = 945 
 335 	783162945	 783 + 162 = 945 
 336 	784152936	 784 + 152 = 936

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(All rights reserved) - Last modified : November 20, 2011.

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Patrick De Geest - Belgium - Short Bio - Some Pictures
E-mail address : pdg@worldofnumbers.com

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