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

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:

1/142857 = 0.000007000007…
1/766923 = 0.000013000013…

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.


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