World!Of Numbers 
HOME plate WON  


The Nine Digits Page 1 with some Ten Digits (pandigital) exceptions  
Page 2 Page 3 Page 4 Page 5 Page 6 Page 7 
When I use the term ninedigital in these articles I always refer to a strictly zeroless pandigital (digits from 1 to 9 each appearing just once).
Topic 1.8
Albert H. Beiler's book provides an excellent start for this page
( Source : “Recreations in the Theory of Numbers” )
"The following curious table shows how to arrange the 9 digits so that the product
of 2 groups is equal to a number represented by the remaining digits."
"The product of 51249876 and 3 (all the digits used once only) is 153749628,
which again contains all 9 digits used once."
[ Chapter XIV : The eternal triangle ]
"There are many Pythagorean triangles whose area utilizes 9 different digits.
Using generators 149 and 58, the area is 162789354 ;
using 224 and 153, the area is 917358624."
[ Chapter XV : On the square ]
"The following table shows the difference of two squares equal to a number containing
all the nine digits used only once."
Readers should recall that x^{2} – y^{2} = (x + y) x (x – y), hence the second expression !
"Express the numbers whose digits are in regularly ascending or descending order
as the difference of two squares."


Topic 1.7
Martin Gardner introduced the following puzzle
( Source : “Further Mathematical Diversions” )
[ Chapter 15 : Nine Problems : nr.8 : NINE TO ONE EQUALS 100 ]
"An old numerical problem that keeps reappearing in puzzle books as though it had never been analyzed before is the problem
of inserting mathematical signs wherever one likes between the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 to make the expression equal 100.
The digits must remain in the same sequence. There are many hundreds of solutions, the easiest to find perhaps being "
"The problem becomes more of a challenge if the mathematical signs are limited to plus and minus. Here again there are many solutions, for example"
"In view of the popularity of this problem it is surprising that so little effort seems to have been spent on the problem in reverse form.
That is, take the digits in descending order, 9 through 1, and form an expression equal to 100 by inserting the smallest possible number of plus and minus signs.
Answer : To form an expression equal to 100, four plus and minus signs can be inserted between the digits, taken in reverse order, as follows ¬"
"There is no other solution with as few as four signs !"
[ May 10, 2002 ] Not able to answer Barry satisfactorily he threw the topic into

Topic 1.6
Variations on a theme
by the same Martin Gardner ( Source : “Puzzles from other Worlds” )
playing with The Number of the Beast
"Three plus signs can be inserted within the sequence 123456789 to make a sum of 666"
"If minus signs are also allowed (except that a minus sign is not permitted in front of the sequence),
what is the only other way to obtain 666 with just three signs ? The sum cannot be achieved with fewer signs."
"Find the only way to insert four signs (each may be plus or minus) inside 987654321 to make a sum of 666.
Again, there is no solution with fewer signs."
"A common interpretation among today's fundamentalists is that 666 represents a falling short of 777,
which is taken to be a symbol of perfection. I believe there is no way to insert any number of plus and minus
signs in 123456789 to get a sum of 777, and just one way to do it with 987654321. How ? "
666 has eight solutions for the ascending sequence ¬ 

777 has two solutions for the ascending sequence ¬ 

[ January 10, 2001 ] Replace the three 6 digits with the 6^{th} prime or 13. The juxtaposed digits from 666 become thus (13)(13)(13) and we can make the following fraction ¬
Too bad for those who are a bit triskaidekaphobic and/or suffering from aibohphobia. 
Topic 1.5
Here is a beautiful numerical expansion. Guess where it leads us to ?
Websource : Number 8
Booksource : “Numbers: Facts, Figures and Fiction” by Richard Phillips
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321
Topic 1.4
Ever wondered how the ancient Chinese represented 123456789
on the Abacus ?
Click on the image to get an introduction to the abacus.
Topic 1.3
Nine Digits forming Unity 1
This comes straight out of the archive of the newsgroup rec.puzzles
==> arithmetic/digits/equations/123456789.p <==
In how many ways can "•" be replaced with "+", "–", or "" (concatenate) in
•1•2•3•4•5•6•7•8•9=1 to form a correct equation ?
==> arithmetic/digits/equations/123456789.s <==
12 3+4 5+6 78 9 = 1 +12 3+4 5+6 78 9 = 1 1+2 3+45+6 78 9 = 1 +1+2 3+45+6 78 9 = 1 1+2 34+5+6 78 9 = 1 1+2 34 56 7+8 9 = 1 +1+2 34 56 7+8 9 = 1 12 34+56 7+8 9 = 1 +12 34+56 7+8 9 = 1 1234 5+6 789 = 1 +1234 5+6 789 = 1 1+23 4+5 6789 = 1 +1+23 4+5 6789 = 1 1+2 3+4+56789 = 1 1 2+3 456+789 = 1 1+2+3+45+6+789 = 1 +1+2+3+45+6+789 = 1 1+2+34+5+6+789 = 1 123+4+5+6+789 = 1 +123+4+5+6+789 = 1 1+2 3+4 56 7+89 = 1 +1+2 3+4 56 7+89 = 1 1+2 34567+89 = 1  +1+2 34567+89 = 1 1+2+3+4+567+89 = 1 +1+2+3+4+567+89 = 1 1+2+3+45+67+89 = 1 12+34+5+67+89 = 1 +12+34+5+67+89 = 1 123+4+5+67+89 = 1 12+3+456+7+89 = 1 +12+3+456+7+89 = 1 1+234+56+7+89 = 1 +1+234+56+7+89 = 1 12+34+56+7+89 = 1 1+2345+6+7+89 = 1 1+2 3+4 56 78+9 = 1 12 34 5+6 78+9 = 1 +12 34 5+6 78+9 = 1 1+2 345678+9 = 1 1+2+3+4+5678+9 = 1 12+3+45+678+9 = 1 +12+3+45+678+9 = 1 1+234+5+678+9 = 1 +1+234+5+678+9 = 1 12+34+5+678+9 = 1  1+23+456+78+9 = 1 +1+23+456+78+9 = 1 12+3+456+78+9 = 1 1+234+56+78+9 = 1 12345+6+78+9 = 1 +12345+6+78+9 = 1 12 3+4+5+6+78+9 = 1 +12 3+4+5+6+78+9 = 1 1+2+3+4 56 7+8+9 = 1 +1+2+3+4 56 7+8+9 = 1 1 2+3 4+56 7+8+9 = 1 +1 2+3 4+56 7+8+9 = 1 1+2+34567+8+9 = 1 +1+2+34567+8+9 = 1 1+23+4567+8+9 = 1 1234+567+8+9 = 1 +1234+567+8+9 = 1 12345+67+8+9 = 1 12 3+4+5+67+8+9 = 1 12+3 45 6+7+8+9 = 1 +12+3 45 6+7+8+9 = 1 1 23 4+56+7+8+9 = 1 +1 23 4+56+7+8+9 = 1 
Total solutions = 69
(26 of which have a leading "+", which is redundant)
69/19683 = 0.35 %
Topic 1.2
Commissioned for BBC Education as part of the Count Me In Numeracy Site
The sum of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
Multiply 123456789 by 2 and you get 246913578, the sum of which is 45
Divide 123456789 by 2 and you get 61728394.5, the sum of which is 45
Topic 1.1
Puzzle : Olympic Ring Addition
This is another puzzle found on the web dealing with the digits one trought nine.
For instance https://donsteward.blogspot.com/2012/04/olympicrings.html
Given the five olympic rings how can the digits one through nine be placed within the nine regions
(five nonoverlapping ring regions and four overlapping regions shared between two rings)
so that each ring contains the same total ?
Or alternatively, using the labels A through I for the regions, how can the numbers one through nine
be assigned to the variables such that :
Can you solve it ?
SPOILER : it is hidden somewhere in the source code of this page !
For a logical and detailed approach other than brute force consult
Ken Duisenberg's PuzzleOfTheWeek Archive  Olympic Ring Addition.
For reference goals and easy searching I list here all the nine & pandigitals implicitly displayed in these topics.
Topic 1.8 → 124835796, 421385796, 182975346, 271985346, 391867254, 481597632, 281574396, 417386952, 419637852, 512498763, 165837429, 325478916
Topic 1.7 → 123456789, 987654321
Topic 1.6 → 123456789, 987654321
Topic 1.5 → 123456789, 987654321
Prime Curios!  site maintained by G. L. Honaker Jr. and Chris Caldwell
3
5
89
999
3187
5077
5986
28651
32423
14368485
15618090
36543120
40578660
45269999
394521678
1123465789
1227182861
1531415939
2543568463
5897230146
10123457689
1023456987896543201
5021837752995317770489
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