# Which is bigger 13 999999999 or 14

**What is the checksum?**

The checksum of a natural number is the sum of its digit values. If the number is one-digit, the number itself is defined as a cross sum. |

The checksum depends on the number system.

This website only deals with checksums in the decimal system.

**Follow the checksums **Top

A checksum is clearly assigned to each natural number. Therefore the assignment is a function. Since the domain is D = | N, the function is also a sequence.

The range of values is W = | N.

That is the sequence of the first 100 numbers with their checksums.

11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, numbers | 02, 03, 04, 05, 06, 07, 08, 09, 10, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17, 09, 10, 11, 12, 13, 14, 15, 16, 17, 18, 01, Checksums |

Graph to it

The special thing is that the terms of the sequence always fall back to the number 1, because the powers of ten 1.10, 100, 1000, ... have the cross sum 1. Always after that the numbers repeat themselves in a typical way. There is self-similarity. This becomes clear when more elements of the sequence are graphically represented, e.g. on the website

*Checksum*at de.wikipedia.

**Inverse of the checksum function **Top

This means that you specify a cross sum and then ask yourself which numbers have the same cross sum.

Here is an example

The number 234 is given. How many three-digit numbers have the same cross sum?

solution

The checksum is 2 + 3 + 4 = 9. One looks for all the decompositions of the number 9 into three summands.

3 decompositions have 3 different digits: 126, 135 and 234. For each number there are 5 additional numbers by permutating the digits, e.g. for 234 still 243, 324, 342, 423 and 432. |

3 decompositions have 2 different digits: 144, 117 and 225. There are 2 additional numbers for each number, e.g. 414 and 441 for 144. |

Then there are the numbers that contain a zero. There are 4 decompositions of the number 9 into two summands: 18, 27, 36, 45. There are 4 additional numbers for each decomposition, E.g. for 18 the numbers 180, 108, 801 and 810. |

Finally there are the numbers 333 and 900 ............................... |

Result: There are 18 + 9 + 16 + 2 = 45 numbers with the cross sum 9.

My computer confirms the result.

Only with the addition of "three digits" does the task have a finite number of solutions. Otherwise, for example, you could add any number of zeros to the numbers without changing the checksum.

**Divisibility rules **Top

Most people are familiar with the checksum due to some divisibility rules.

(1) Divisibility of a number by 9

A number is divisible by 9 if and only if its checksum is divisible by 9.

The rule is explained using the example "9 divides 347.265".

The following applies: 347265/9 = (3 * 100000 + 4 * 10000 + 7 * 1000 + 2 * 100 + 6 * 10 + 5) / 9

=[3*(99999+1)+4*(9999+1)+7*(999+1)+2*(99+1)+6*(9+1)+5]/9

=(3*99999+3+4*9999+4+7*999+7+2*99+2+6*9+6+5)/9

=(3*99999+4*9999+7*999+2*99+6*9+3+4+7+2+6+5)/9

=(3*1111+4*1111+7*111+2*11+6)+(3+4+7+2+6+5)/9.

It follows that 347265/9 is an integer if and only if (3 + 4 + 7 + 2 + 6 + 5) / 9 is an integer.

The following statement is behind the calculation.

If you divide the powers of 10 by 9, the remainder remains 1.

(2) Divisibility of a number by 3

A number is divisible by 3 if and only if its sum is divisible by 3.

The considerations for rule (2) correspond to those for rule (1).

(3) Divisibility of a number by 11

A number is divisible by 11 if and only if its alternating checksum is divisible by 11.

The alternating checksum is obtained by alternately subtracting and adding the numerical values starting from the right.

The rule (3) is explained using the example "11 divides 124542".

The following applies: 124542/11 = (1 * 100000 + 2 * 10000 + 4 * 1000 + 5 * 100 + 4 * 10 + 2) / 11

=([1*(100001-1)+2*(9999+1)+4*(1001-1)+5*(99+1)+4*(11-1)+2]/11

=(1*100001-1+2*9999+2+4*1001-4+5*99+5+4*11-4+2)/11

=(1*100001+2*909+4*91+5*9+4)+(-1+2-4+5-4+2)/11.

It follows that 124542/11 is an integer if (-1 + 2-4 + 5-4 + 2) / 11 is an integer.

The following statements are behind the calculation.

> If you divide the even powers of 10 by 11, the remainder remains 1.

> If you divide the odd powers of 10 by 11, the remainder remains -1.

(4) Divisibility of a number by 7

A number is divisible by 7 if and only if the checksum weighted with the sevens remainder is divisible by 7.

The rule is explained using the example "7 divides 857423".

857423/7=(8*100000+5*10000+7*1000+4*100+2*10+3)/7

=[8*(99995+5)+5*(9996+4)+7*(994+6)+4*(98+2)+2*(7+3)+3]/7

=(8*99995+8*5+5*9996+5*4+7*994+7*6+4*98+4*2+2*7+2*3+3]/7

=(8*14285+5*1428+7*142+4*14+2*1)+(8*5+5*4+7*6+4*1+2*3+1*3)/7

It follows that 124542/11 is an integer if (5 * 8 + 4 * 5 + 6 * 7 + 2 * 4 + 3 * 2 + 3) / 7 is an integer.

In the term 5 * 8 + 4 * 5 + 6 * 7 + 2 * 4 + 3 * 2 + 1 * 3, the summands of the checksum are linked with sevens remainders. The term is a checksum weighted with the sevens residues.

The following facts are behind the invoice.

10/7 is remainder 3, 100/7 is remainder 2, 1000/7 is remainder 6, 10000/7 is remainder 4, 100000/7 is remainder 5.

If you divide a power of ten by 7, you get the remainders 1, 3, 2, 6, 4, 5, 1, 3, ...

Perhaps it is cheaper to allow negative residues as well. Then it is easier to remember the sequence of the sevens residues: 1, 3, 2, -1, -3, -2, 1, 3, ...

But even then, the rule remains unwieldy.

Test of nine

In the days before computers, there was a great risk of miscalculation when multiplying in writing, for example. The statements "Every number leaves the same remainder when divided by 9" and "Calculating with remainders follows the rules of calculating with numbers" were used to check the calculation.

Example:

Does 345 * 739 = 254956 apply?

345 has the checksum 12 and the remainder of the nine is 3. 739 has the checksum 19 and the remainder of the nine is 1.

254956 has the checksum 31 and the remainder of nine has 4.

The test of nine 3 * 1 = 4 is wrong. (The product is 254955, then 3 * 1 = 3.)

**Single digit checksums **Top

(English: digit roots)

Above in the investigation on "9 divides 347.265" it was found that the cross sum 3 + 4 + 7 + 2 + 6 + 5 = 27 stands for the divisibility of the number 347.265. M an can go one step further and determine the checksum of 27 and thus ensure divisibility by 2 + 7 = 9.

With larger numbers you could calculate the checksum of the checksum of the checksum ...

You can always be sure that you will end up with a single-digit number.

Persistence

So you need two steps for the number 347.265 to get to the one-digit number 9. The number of steps is called the persistence of the number. The persistence of 347,265 is 2.

In general, the question arises as to the smallest number for which n steps are required.

There are the following numbers.

number persistence | 1 | 2 | 3 | 4 | ... |

More at Mathworld under

*Additive persistence*(Url below)

lucky number

Everyone has a one-digit lucky number ;-).

Here is a suggestion on how to uniquely calculate a number from the date of birth.

In January 23, 1977 there is the number 23,011,977. The checksum is 2 + 3 + 0 + 1 + 1 + 9 + 7 + 7 = 30 and 30 has the checksum 3 + 0 = 3.

Result: The lucky number is 3.

Check digit calculation

Sequences of digits play an increasingly important role in many areas of life.

I call

> Serial numbers on euro banknotes,

> ISBN - International Standard Book Number,

> ID card number.

On the website

*http://www.pruefzifferberechnung.de/*you can read how the checksums of these and many other key figures are included in the calculations of a check number (URL below).

**Follow the cross products **Top

The cross product of a natural number is the product of the numerical values. If the number is one-digit, the number itself is defined as a cross product. |

A cross product is clearly assigned to each natural number. Therefore the assignment is a function. Since the domain is D = | N, the function is also a sequence.

The range of values is W = | N.

This is the result of the first 100 numbers with their cross products.

11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 90, 100, numbers | 01, 02, 03, 04, 05, 06, 07, 08, 09, 00, 02, 04, 06, 08, 10, 12, 14, 16, 18, 00, 03, 06, 09, 12, 15, 18, 21, 24, 27, 00, 04, 08, 12, 16, 20, 24, 28, 32, 36, 00, 05, 10, 15, 20, 25, 30, 35, 40, 45, 00, 06, 12, 18, 24, 30, 36, 42, 48, 54, 00, 07, 14, 21, 28, 35, 42, 49, 56, 63, 00, 08, 16, 24, 32, 40, 48, 56, 64, 72, 00, 09, 18, 27, 36, 45, 54, 63, 72, 81, 00, Cross products |

Graph to it

The special thing is that the terms of the sequence always fall back to the number 0, because even a zero in the number leads to the product 0. In between, the numbers repeat themselves in a typical way. There is self-similarity. This becomes clear when more elements of the sequence are graphically represented, e.g. on the website

*Cross product*at de.wikipedia.

**Reversal of the cross product function **Top

This means that you specify a cross product and then ask yourself which numbers have the same cross product.

Here is an example.

The number 234 is given. Which three-digit numbers have the same cross product?

solution

The cross product is 2 * 3 * 4 = 24.

There are 3 more decompositions of the number 24 into three factors, namely 24 = 1 * 3 * 8 = 1 * 4 * 6 = 2 * 2 * 6.

Each breakdown requires further adjustments.

Result

There are 21 three-digit numbers, viz

My computer confirms the result.

Only with the addition of "three digits" does the task have a finite number of solutions. Otherwise you could add any number of ones to the number without changing the cross product.

**Single-digit cross products **Top

(English: Multiplicative digit roots)

The concept of the single-digit checksum was introduced above. This can be transferred to the cross product. In general, the cross product has several digits. If you continue to form the cross product of the cross product ..., you finally arrive at a single-digit number.

example

The number 896 has the cross product 432, the number 432 has the cross product 24 and the number 24 has the cross product 8.

Persistence

To find the one-digit cross product of the number 896, you need 3 steps.

The number of steps is called the multiplicative persistence of a number. 896 has persistence 3.

In general, the question arises as to the smallest number for which n steps are required.

There are the following numbers.

number persistence | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ... |

More at Mathworld under

*Multiplicative persistence*(Url below)

**Numbers games **Top

Numerous, more or less cute questions have been thought up about the digits of a number and especially the checksums.

**01**

Which three-digit numbers can be represented as the sum of the third powers of their numerical values?

These and many of the following tasks can be investigated using a simple computer program.

for x = 0 to 9 for y = 0 to 9 for z = 0 to 9 if x * x * x + y * y * y + z * z * z = 100 * x + 10 * y + z then print x; y; z next z next y next x | ... | Result 153=1 ^{3}+5^{3}+3^{3}370=3 ^{3}+7^{3}+0^{3}371=3 ^{3}+7^{3}+1^{3}407=4 ^{3}+0^{3}+7^{3} |

The problem can be extended to n-digit numbers.

^{4}+6^{4}+3^{4}+4^{4}8208=8 ^{4}+2^{4}+0^{4}+8^{4}9474=9 ^{4}+4^{4}+7^{4}+4^{4} | ^{5}+4^{5}+7^{5}+4^{5}+8^{5}92727=9 ^{5}+2^{5}+7^{5}+2^{5}+7^{5}93084=9 ^{5}+3^{5}+0^{5}+8^{5}+4^{5} |

More at Mathworld under

*Narcissistic Number*(Url below)

Variations

^{5}+1^{5}+5^{5}+0^{5} | 4151=4^{5}+1^{5}+5^{5}+1^{5} |

3435=3^{3}+4^{4}+3^{3}+5^{5} | More at Mathworld under Muenchhausen Number (Url below) |

**02**

> Form the cross sum of a number.

> Break down the number into its prime factors and form the cross sum of the digit values of the prime factors.

For which number do the checksums match?

1st example

> 852 has the checksum 8 + 5 + 2 = 15.

> On the other hand, 852 = 2 * 2 * 3 * 71. The checksum is 2 + 2 + 3 + 7 + 1 = 15.

2nd example

> The famous number 666 has the checksum 6 + 6 + 6 = 18.

The sum of the digits of all prime divisors is 2 + 3 + 3 + (3 + 7) = 18.

More at Mathworld under

*Smith Number*(Url below)

**03**

"A happy number is a natural number which, as the starting value for a certain iteration rule, leads to the numerical value 1 after a finite number of iteration steps, similar to the (3n + 1) problem."

(Quote from de.wikipedia)

example

19 is a happy number because it takes 3 steps to get to one.

More at Mathworld under

*Happy number*(Url below)

**04**

Which number can be represented as the power of the checksum?

512=(5+1+2)³ 4913=(4+9+1+3)³ 5832=(5+8+3+2)³ 17576=(1+7+5+7+6)³ 19683=(1+9+6+8+3)³ | 81=(8+1)^{2}2401=(2+4+0+1) ^{4}^{...}20047612231936=(2+0+0+4+7+6+1+2+2+3+1+9+3+6) ^{8} |

variation

(8+1)²=81 | (20+25)²=2025 | (30+25)²=3025 | (98+01)²=9801 |

More at Mathworld under

*Kaprekar Number*(Url below)

**05**

Which number is divisible by its checksum?

Three examples

More at Mathworld under

*Harshad Number*(Url below)

**06**

miscellaneous

135=(1*3*5)*(1+3+5) 144=(1*4*4)*(1+4+4) | 2+2=2*2 1+2+3=1*2*3 (9+9+9)²=9*9*9 (3+3+3)³=(3*3*3)² | 1³+2³=(1+2)² 1³+2³+3³=(1+2+3)² 2³+2³=(2+2)² 1³+2³+3³=(1+2+3)² 3³+3³+3³=(3+3+3)² | 37*(3+7)=3³+7³ 48*(4+8)=4³+8³ |

**07**

For which numbers are the cross sum and cross product the same?

Example: 321 has the cross sum 3 + 2 + 1 and the cross product 3 * 2 * 1

The first numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 99, 123, 132, 213, 231, 312, 321, 1124, 1137, 1142, 1173, 1214, 1241, 1317, 1371, 1412, 1421, 1713, 1731, 2114, 2141, 2411, 3117, 3171, 3344, 3434, 3443, 3711, 4112, 4121, 4211, 4334, 4343, 4433, 7113, 7131, 7311, 11125,

Source OESIS A249334

**08**

222

Another curiosity on my part: The Kaprekar number and other number games

A three-digit number consisting of non-identical digits is given, e.g. 369. You create 5 more numbers by rearranging the digits in every possible way [396, 639, 693, 936, 963]. Add the six numbers [369 + 396 + 639 + 693 + 936 + 963 = 3996]. You get 222 times the checksum [222 * (3 + 6 + 9) = 3996].

**Check sum on the internet **Top

German

Holger Krug

A workable divisibility rule for the 7 (.pdf file)*(Put up for discussion)*

NN

Check digit calculation in practice

Wikipedia

Cross sum, cross product, test of nine, narcissistic number, happy number

English

Dr. Math (The Math Forum)

Products of Digits

Eric W. Weisstein (MathWorld)

Digit Sum, Digital Root, Additive Persistence, Multiplicative Persistence, Multiplicative Digital Root, Narcissistic Number, Muenchhausen Number, Smith Number, Kaprekar Number

N. J. A. Sloane (On-Line Encyclopedia of Integer Sequences)

Integer sequences

Digital sum (i.e. sum of digits) of n.A007953

Smallest number of additive persistence according to A006050

Product of decimal digits of n. A007954

Smallest number of multiplicative persistence. A003001

a (n) is a power of the sum of its digits. A023106

Numbers n for which the digital sum contains the same distinct digits as the digital product. A249334

Numbers n such that n = (product of digits of n) * (sum of digits of n). A038369

Wikipedia

Digit sum, Digital root, Persistence of a number, Vedic square, Happy number, Sum-product number

**credentials**Top

(1) Walter Lietzmann: Oddities in the Realm of Numbers, Bonn 1948

(2) Walter Lietzmann: Funny and strange things about numbers and shapes, Göttingen 1969

(3) Martin Gardner: Geometry with Taxis, the heads of the Hydra and other mathematical gimmicks, Basel 1997 [ISBN 3-7643-5702-9]

**Feedback:**Email address on my main page

URL of my homepage:

http://www.mathematische-basteleien.de/

© 2012 Jürgen Köller

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