﻿ Decimal Numbers of Concatenated Blocks Which are Reciprocals of Integers
San José State University

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Thayer Watkins
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 Decimal Numbers of Concatenated Blocks Which are Reciprocals of Integers

Consider the decimal representations of the reciprocals of seven and thirteen:

#### 1/7 = 0.142857142857… 1/13 = 0.076923076923…

These are concatenated blocks of 142857 and 076923. Some questions immediately arise, such as

• What is special about the concatenation blocks of 142857 and 076923?
• Are there other integers such that their reciprocals have this concatenated block form?

Well, for starters the digit sums of 142857 and 076923 are both 9 which indicates that they are multiples of 9. Their ratios when divided by 9 are 14873 and 8437, respectively.

As to the second question consider the following:

But also

#### 1/3 = 0.3333… 1/9 = 0.1111…

And less obviously

#### 1/1 = 0.9999…

Note the products of integers of the denominators on the LHS with its concatenation block;

#### 7(142857) = 999999 13(76923) = 999999 142857(7) = 999999 76923(13) = 999999 3(3) = 9 9(1) = 9 1(9) = 9

Proposition: Let n be the integer and K be the concatenation block. Then nK is equal to 99…99, where the number of 9's in this product is equal to m, the number digits in K.

Proof:

The equation

#### 1/n = 0.KKK… means 1/n = (K/10m) + (K/102m) + (K/103m) +… or, equivalently 1/n = (K/10m)[1 + (1/10m) + (1/10m)² + (1/10m)³ + … ;

On the right there is the geometric series sum whose value is equal to 1/(1−1/10m). Therefore

#### 1/n = (K/10m)[1/(1−1/10m).] and when the denominators on the RHS are multiplied together 1/n = K/(10m−1) nK = 10m−1

The term (10m−1) is just m 9's in a row.

The concatenation block is arbitrary to a degree. For example, the concatenation block for 0.333333… could be chosen as 3, 33, 333 or any number of 3's in a row. The theorem holds for any such choice.

## The Set of Integers such that their decimal representationsreciprocals have the concatenated block form

The product nK is equal to 99…99. This means that n must be a factor of 99…99. But 99…99 equal to 9(11…11) so n must be a factor of 9 or 11…11. For m equal to 1, n must be a factor of 9. The factors of 9 are 1, 3 and 9. So there cannot be any other single digit integers other than 1, 3 and 9 whose reciprocal has a decimal representation of the concatenated block form.

for m equal to 2 the only integers with the property being considered are the divisors of 99 which are 1, 3, 9, 11, 33 and 99.

For m=3 the integers 1, 3 and 9 have the property but also the divisors of 111. The factorization of 111 is 3*37 so its divisors are 1, 3, 37, 111, which means that they have the property but also 333 and 999.

The factorization of 1111 is 11*101 which means its divisors are 1, 11, 101 and 1111. So the set of numbers with the property is {1, 3, 9, 11, 33, 99, 101, 1111}.

The factorization of 111111 is 3 * 7 * 11 * 13 * 37, so its divisors include 1, 3, 7, 11, 13, 21, 33, 37, 39, 77, 91, 111, 143, 231, 259, 407 , 429, 481, 777, 1001, 1221, 1443, 2849, 3003, 3367, 5291, 8547, 10101, 15873, 37037 and 111111. It becomes tedious to list all the divisors of 111111 . but it may include more besides the ones shown. The divisors of 999999 include all of the divisiors of 111111 and their multiples by 9.