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

The Nine Digits Page 1with some Ten Digits (pandigital) exceptions | |||

Most Recent Page 5 Page 4 Page 3 Page 2 |

- 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).

Albert H. Beiler's book provides an excellent start for this page.

( Source : *"Recreations in the Theory of Numbers"* )

[ Chapter VIII : Digits - and the magic of 9 ] [ I found the same exposé in Shakuntala Devi's book "Figuring : The Joy 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."12 x 483 = 5796 42 x 138 = 5796 18 x 297 = 5346 27 x 198 = 5346 39 x 186 = 7254 48 x 159 = 7632 28 x 157 = 4396 4 x 1738 = 6952 4 x 1963 = 7852 "The product of 51249876 and 3 (all the digits used once only) is 153749628,

which again contains all 9 digits used once. Again ¬ "16583742 x 9 = 149253678 32547891 x 6 = 195287346 [ 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 !

11113 ^{2}– 200^{2}= 11313 x 10913 = 12345876931111 ^{2}– 200^{2}= 31311 x 30911 = 96785432111117 ^{2}– 200^{2}= 11317 x 10917 = 12354768911356 ^{2}– 2000^{2}= 13356 x 9356 = 12495873612695 ^{2}– 6017^{2}= 18712 x 6.678 = 12495873616260 ^{2}– 11808^{2}= 28068 x 4452 = 12495873612372 ^{2}– 300^{2}= 12672 x 12072 = 152976384"Express the numbers whose digits are in regularly ascending or descending order

as the difference of two squares."

123456789 Factorisation = 3 ^{2}x 3607 x 3803= 61728395 ^{2}– 61728394^{2}= 20576133 ^{2}– 20576130^{2}= 6858715 ^{2}– 6858706^{2}= 18917 ^{2}– 15310^{2}= 18133 ^{2}– 14330^{2}= 11115 ^{2}– 294^{2}987654321 Factorisation = 3 ^{2}x 17^{2}x 379721= 493827161 ^{2}– 493827160^{2}= 164609055 ^{2}– 164609052^{2}= 54869689 ^{2}– 54869680^{2}= 29048665 ^{2}– 29048648^{2}= 9682911 ^{2}– 9682860^{2}= 3227705 ^{2}– 3227552^{2}= 1708889 ^{2}– 1708600^{2}= 570015 ^{2}– 569148^{2}= 191161 ^{2}– 188560^{2}

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 "1 + 2 + 3 + 4 + 5 + 6 + 7 + ( 8 x 9 ) = 100 "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"

1 + 2 + 34 – 5 + 67 – 8 + 9 = 100 12 + 3 – 4 + 5 + 67 + 8 + 9 = 100 123 – 4 – 5 – 6 – 7 + 8 – 9 = 100 123 + 4 – 5 + 67 – 89 = 100 123 + 45 – 67 + 8 – 9 = 100 123 – 45 – 67 + 89 = 100 "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 ¬"98 – 76 + 54 + 3 + 21 = 100 "There is no other solution with as few as four signs !"

[ *May 10, 2002* ]

Barry Sanders (email) asked himself if there is a solution using each of the digits 1 to 9 once, but

they don't have to be in consecutive order, in combination with only the PLUS sign ( + ) sums to 100 ?

Is there an answer ? Is it unique ?

Is there a program that would show all possibilities ?

Not able to answer Barry satisfactorily he threw the topic into

a discussion newsgroup alt.math where very soon afterwards

Terry Moriarty (email) from Northern Ireland gave the following interesting reply :

Consider the following:

The sum of all numbers 1-9 is 45.

Since this is too low (to reach 100) you must use some 2 figure numbers (23, 34, etc.)

but *all* of these give you an addition of a multiple of 9

for instance use '12' instead of '1+2'

( creates an additional 9 since you have to remove the 1 and 2 )

your total is now 45 - 1 - 2 + 12 = 54.

No matter what doubles you choose, they will always add some multiple of 9.

Assume that the high part of your double is 'a' and the low part is 'b' (10a + b) - a - b = 9a For you to succeed you need to add 55 (to reach 100) to your original total.

You can't do this with multiples of 9.

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 ¬"123 + 456 + 78 + 9 = 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."

Answer: 1234 – 567 + 8 – 9 = 666"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."

Answer: 9 + 87 + 6 + 543 + 21 = 666"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 ? "

Answer: 98 + 7 + 654 – 3 + 21 = 777

666has eight solutions for theascendingsequence ¬– 1 + 2 – 3 + 4 – 5 + 678 – 9 + 1 – 2 – 3 – 4 + 5 + 678 – 9 + 1 + 2 + 3 – 4 – 5 + 678 – 9 + 1 – 23 – 4 + 5 + 678 + 9 + 1 + 2 + 3 + 4 + 567 + 89 + 1 + 23 – 45 + 678 + 9 + 123 + 456 + 78 + 9 + 1234 – 567 + 8 – 9 And five solutions for the descending¬– 9 + 8 + 7 + 654 + 3 + 2 + 1 + 9 – 8 + 7 + 654 + 3 + 2 – 1 + 9 + 8 – 7 + 654 + 3 – 2 + 1 + 9 – 8 – 7 + 654 – 3 + 21 + 9 + 87 + 6 + 543 + 21

777has two solutions for theascendingsequence ¬– 12 – 3 + 4 + 5 – 6 + 789 – 12 + 3 – 4 – 5 + 6 + 789 And two for the descending¬– 9 – 8 + 765 – 4 + 32 + 1 + 98 + 7 + 654 – 3 + 21

[ *January 10, 2001* ]

A nice relation between two repdigits

by P. De Geest

Replace the three

The juxtaposed digits from

131313 13 * 13 | = 777 |
---|

Too bad for those who are a bit triskaidekaphobic and/or suffering from aibohphobia .

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

Ever wondered how the ancient Chinese represented 123456789 on the Abacus ?

Click on the image to get an introduction to the abacus.

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 <==

1-2 3+4 5+6 7-8 9 = 1 +1-2 3+4 5+6 7-8 9 = 1 1+2 3+4-5+6 7-8 9 = 1 +1+2 3+4-5+6 7-8 9 = 1 -1+2 3-4+5+6 7-8 9 = 1 1+2 3-4 5-6 7+8 9 = 1 +1+2 3-4 5-6 7+8 9 = 1 1-2 3-4+5-6 7+8 9 = 1 +1-2 3-4+5-6 7+8 9 = 1 1-2-3-4 5+6 7-8-9 = 1 +1-2-3-4 5+6 7-8-9 = 1 1+2-3 4+5 6-7-8-9 = 1 +1+2-3 4+5 6-7-8-9 = 1 -1+2 3+4+5-6-7-8-9 = 1 -1 2+3 4-5-6+7-8-9 = 1 1+2+3+4-5+6+7-8-9 = 1 +1+2+3+4-5+6+7-8-9 = 1 -1+2+3-4+5+6+7-8-9 = 1 1-2-3+4+5+6+7-8-9 = 1 +1-2-3+4+5+6+7-8-9 = 1 1+2 3+4 5-6 7+8-9 = 1 +1+2 3+4 5-6 7+8-9 = 1 1+2 3-4-5-6-7+8-9 = 1 | +1+2 3-4-5-6-7+8-9 = 1 1+2+3+4+5-6-7+8-9 = 1 +1+2+3+4+5-6-7+8-9 = 1 -1+2+3+4-5+6-7+8-9 = 1 1-2+3-4+5+6-7+8-9 = 1 +1-2+3-4+5+6-7+8-9 = 1 -1-2-3+4+5+6-7+8-9 = 1 1-2+3+4-5-6+7+8-9 = 1 +1-2+3+4-5-6+7+8-9 = 1 1+2-3-4+5-6+7+8-9 = 1 +1+2-3-4+5-6+7+8-9 = 1 -1-2+3-4+5-6+7+8-9 = 1 -1+2-3-4-5+6+7+8-9 = 1 -1+2 3+4 5-6 7-8+9 = 1 1-2 3-4 5+6 7-8+9 = 1 +1-2 3-4 5+6 7-8+9 = 1 -1+2 3-4-5-6-7-8+9 = 1 -1+2+3+4+5-6-7-8+9 = 1 1-2+3+4-5+6-7-8+9 = 1 +1-2+3+4-5+6-7-8+9 = 1 1+2-3-4+5+6-7-8+9 = 1 +1+2-3-4+5+6-7-8+9 = 1 -1-2+3-4+5+6-7-8+9 = 1 | 1+2-3+4-5-6+7-8+9 = 1 +1+2-3+4-5-6+7-8+9 = 1 -1-2+3+4-5-6+7-8+9 = 1 -1+2-3-4+5-6+7-8+9 = 1 1-2-3-4-5+6+7-8+9 = 1 +1-2-3-4-5+6+7-8+9 = 1 1-2 3+4+5+6+7-8+9 = 1 +1-2 3+4+5+6+7-8+9 = 1 1+2+3+4 5-6 7+8+9 = 1 +1+2+3+4 5-6 7+8+9 = 1 1 2+3 4+5-6 7+8+9 = 1 +1 2+3 4+5-6 7+8+9 = 1 1+2+3-4-5-6-7+8+9 = 1 +1+2+3-4-5-6-7+8+9 = 1 -1+2-3+4-5-6-7+8+9 = 1 1-2-3-4+5-6-7+8+9 = 1 +1-2-3-4+5-6-7+8+9 = 1 -1-2-3-4-5+6-7+8+9 = 1 -1-2 3+4+5+6-7+8+9 = 1 1-2+3 4-5 6+7+8+9 = 1 +1-2+3 4-5 6+7+8+9 = 1 1 2-3 4+5-6+7+8+9 = 1 +1 2-3 4+5-6+7+8+9 = 1 |

Total solutions = **69**

(26 of which have a leading "+", which is redundant)

69/19683 = 0.35 %

Commissioned for BBC Education as part of the __Count Me In__ Numeracy Site

https://nrich.maths.org/2007

The sum of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Multiply 123456789 by2and you get 246913578, the sum of which is 45

Divide 123456789 by2and you get 61728394.5, the sum of which is 45

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/olympic-rings.html

Given the five olympic rings how can the digits one through nine be placed within the nine regions

(five non-overlapping 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 Puzzle-Of-The-Week Archive - Olympic Ring Addition.

- A034277 Decimal part of a(n)^(1/2) starts with a 'nine digits' anagram. - Patrick De Geest
- A034278 Decimal part of a(n)^(1/3) starts with a 'nine digits' anagram. - Patrick De Geest
- A034279 Decimal part of a(n)^(1/4) starts with a 'nine digits' anagram. - Patrick De Geest
- A034280 Decimal part of a(n)^(1/5) starts with a 'nine digits' anagram. - Patrick De Geest
- A034281 Decimal part of a(n)^(1/6) starts with a 'nine digits' anagram. - Patrick De Geest
- A034282 Decimal part of a(n)^(1/7) starts with a 'nine digits' anagram. - Patrick De Geest
- A034283 Decimal part of a(n)^(1/8) starts with a 'nine digits' anagram. - Patrick De Geest
- A034284 Decimal part of a(n)^(1/9) starts with a 'nine digits' anagram. - Patrick De Geest
- A034285 Decimal part of a(n)^(1/10) starts with a 'nine digits' anagram. - Patrick De Geest
- A034286 Decimal part of a(n)^(1/11) starts with a 'nine digits' anagram. - Patrick De Geest
- A034306 Fibonacci iteration F(1,palindrome a(n)) leads to a 'nine digits' anagram. - Patrick De Geest
- A034587 Fibonacci iteration F(1,a(n)) leads to a 'nine digits' anagram. - Patrick De Geest
- A034588 Fibonacci iteration F(1,prime p(n)) leads to a 'nine digits' anagram. - Patrick De Geest
- A034589 Fibonacci iteration F(1,lucky number a(n)) leads to a 'nine digits' anagram. - Patrick De Geest
- A034590 Decimal part of square root of 'nine digits anagram' starts with a 'nine digits anagram'. - Patrick De Geest
- A035136 Decimal part of a(n)^(1/n) starts with a 'nine digits' anagram. - Patrick De Geest
- A035304 Decimal part of a(n)^(1/n) starts with a pandigital anagram (digits 0 through 9 in some order). - Erich Friedman
- A036744 Penholodigital squares: includes each digit once. - dww (wilson)
- A036745 Holodigital squares: includes each digit once. - dww (wilson)
- A039667 Pandigital sums: n such that n+x=y with x=2n and y=3n. - Felice Russo
- A046043 Autobiographical (or curious) numbers: n = x0 x1 x2...x9 such that xi is the number of digits equal to i in n. - Robert Leduc
- A049442 Sum of first n consecutive prime numbers is pandigital (includes all 10 digits exactly once). - G. L. Honaker, Jr.
- A049443 Sum of consecutive primes to p is pandigital (includes all 10 digits exactly once). - G. L. Honaker, Jr.
- A049446 Pandigital sums associated with A049442. - G. L. Honaker, Jr.
- A050278 Pandigital numbers: numbers containing the digits 0-9. - Eric W. Weisstein
- A050288 Pandigital primes. - Eric W. Weisstein
- A050289 Zeroless pandigital numbers: numbers containing the digits 1-9 and no 0's. - Eric W. Weisstein
- A050290 Zeroless pandigital primes. - Eric W. Weisstein
- A053654 Multiples of 123456789. - Klaus Strassburger
- A054037 n^2 contains exactly 9 different digits. - Asher Auel
- A054038 n^2 contains every digit at least once. - Asher Auel
- A054383 Number of pandigital fractions for 1/n. - Eric W. Weisstein
- A058760 Integers whose set of prime factors (taken with multiplicity) uses each digit exactly once (i.e. is pandigital) in some base b>1. Numbers are expressed in base 10. - G. L. Honaker, Jr. and Mike Keith
- A058908 Six prime numbers in arithmetical progression with a common difference of 9876543210. - G. L. Honaker, Jr.
- A058909 Integers whose set of prime factors (taken with multiplicity) uses each digit exactly once (i.e., is pandigital) in base 10. - Mike Keith and G. L. Honaker, Jr.
- A071519 Numbers whose squares are zeroless pandigitals (i.e. use the digits 1 through 9 once). - Lekraj Beedassy
- A074205 Smallest positive integer whose n-th power contains an equal number of each digit (0-9) when represented in base 10. - Jack Brennen

[ TOP OF PAGE]

Patrick De Geest - Belgium - Short Bio - Some Pictures

E-mail address : pdg@worldofnumbers.com