The College Mathematics Journal

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A New Property of Repeating Decimals

Jane Arledge & Sarah Tekansik

To cite this article: Jane Arledge & Sarah Tekansik (2008) A New Property of
Repeating Decimals, The College Mathematics Journal, 39:2, 107-111, DOI:
10.1080/07468342.2008.11922283

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Published online: 28 Nov 2017.

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 A New Property of Repeating Decimals
Jane Arledge and Sarah Tekansik

Jane Arledge (arledge@mesastate.edu) received her B.S.

in mathematics and computer science from the University
of Texas of the Permian Basin, and her Ph.D. in
mathematics from the University of Colorado at Boulder
under the direction of David Grant. For the past 11 years
she has been at Mesa State College in Grand Junction,
Colorado, where she is a professor of mathematics. She
enjoys the ideal climate and magnificent scenery of
western Colorado.

Sarah Tekansik (stekansik@gmail.com) graduated from

Mesa State College in 2006 and is currently at the
University of Nebraska working on a master’s degree in
mathematics. Her recent interests include applied
mathematics and mathematical statistics, and upon
completion of her degree, she will pursue a career as a
mathematical statistician. When not doing mathematics or
statistics, she enjoys shopping and playing tennis.

For more than 200 years, mathematicians have been interested in the patterns found
in the repeating decimal expansions of rational numbers. A summary of this history is
included in [2]. In [4], it is noted that as early as 1802 mathematicians were writing
about the property of nines. A rational number has the property of nines if its decimal
expansion has a repeating period of even length, and the sum of the two halves of
the period is a string of nines. By period we mean the smallest repeating part of the
decimal expansion of a rational number. We note that if m is a positive integer and the
repeating decimal of 1/m has a period of length t, then its period is m1 (10t − 1). For
example, 17 (106 − 1) = 142857.142857 − .142857 = 142857. We see that 1/7 has the
property of nines since 142 + 857 = 999. More recent work with repeating decimals
includes [1], [4], and [5].
A surprising sum is obtained by splitting the period of the decimal expansion of
1/ p n+1 into appropriate blocks and adding them, where p is prime. Instead of the sum
being a string of nines, it is related to the period of 1/ p n . This result is connected to
the property of nines, as we will see.

A generalization of the property of nines

In [1], Brian Ginsberg extended the property of nines: if p is prime and if the length t
of the period of 1/ p is a multiple of 3, say 3d, then when this period is split into three
blocks of d-digits each and these are added, the sum is a string of d nines.
Extending Ginsberg’s methods, one can show that if t = kd, then the period of 1/ p
can be split into k blocks of length d, which when added yield a multiple of a d-
digit string of nines. For example, 1/7 = .142857 and 1 + 4 + 2 + 8 + 5 + 7 = 27 =
3 · 9. Similarly, 1/17 = .0588235294117647, and 05 + 88 + 23 + 52 + 94 + 11 +
76 + 47 = 396 = 4 · 99; also, 0588 + 2352 + 9411 + 7647 = 19998 = 2 · 9999.

VOL. 39, NO. 2, MARCH 2008 THE COLLEGE MATHEMATICS JOURNAL 107
 Not only are the sums 27, 396, and 19998 multiples of strings of nines, but if you
add the highest-order digit to the number that is the sum with this digit removed, you
get 2 + 7 = 9 and 96 + 3 = 99, and 9998 + 1 = 9999, in each case resulting in a
string of d nines. We can see, by shifting the decimal representation of 1/17 two digits
at a time and adding these numbers, that this strange property represents the carrying
of digits in each sum:
 1
0 . 05 88 23 52 94 11 76 47 . . . =
 17 
05 . 88 23 52 94 11 76 47 05 . . . = 17
1
(102 )
 1 
05 88 . 23 52 94 11 76 47 05 88 . . . = 17 (104 )
..
.  1 
+ 05 88 23 52 94 11 76 . 47 05 88 23 52 94 11 76 . . . = 17 (1014 )
05 94 17 70 64 76 52 . 99 99 99 99 99 99 99 99 . . .
Each column to the right of the decimal represents the blocks to be added, and the sum
of each column is 396, which with carrying gives the string of nines. Since 171
(1014 +
1012 + · · · + 102 + 1) is an integer, the prime 17 divides the sum of powers of 10. We
show this holds for any prime greater than 5.

Proposition 1. Let p be a prime greater than 5, and let t be the length of the period
of 1/ p. Let k be a positive integer greater than 1 that divides t, say t = kd. Then p
divides (10d )k−1 + (10d )k−2 + · · · + 10d + 1.

Proof. By a known result (see [2, Th. 1], t is the smallest positive integer for
which 10t ≡ 1 (mod p). Thus p divides 10t − 1 = 10kd − 1, which factors as
(10d − 1)((10d )k−1 + (10d )k−2 + · · · + 10d + 1), so p must divide one of the fac-
tors. Since d < t, p cannot divide the first factor. Hence p divides the sum.

It turns out that not only does p divide the sum of appropriate powers of 10, but p 2
divides this sum less p. This property is key to our main theorem, in which we relate
the periods of 1/ p and 1/ p 2 .

Reciprocals of consecutive prime powers

Let p be prime, and let the periods of 1/ p and 1/ p 2 have period lengths t and t2
respectively. When the period of 1/ p 2 is split into blocks of length t and the blocks
are added, then we almost get the period of 1/ p. For example, let p = 7, so p12 =
1
49
= .020408163265306122448979591836734693877551. If we split this period into
seven 6-digit blocks and add them, the sum is 3142854, which is the period of 1/7
plus a multiple of a 6-digit string of nines, and is also a multiple of the period of 1/7:
3142854 = 3 · 999999 + 142857 = 22 · 142857. Moreover, if we add the high-order
digit to the low-order one, we get the period of 1/7.
In generalizing this result, we use the concept of the order of 10 modulo m, defined
to be the smallest positive integer α such that 10α ≡ 1 (mod m). The following three
facts are easy to show using basic elementary number theory (a good general reference
is [3]):
Fact 1. The length of the period of the rational number 1/m is the order of 10
modulo m [2, Th. 1]. (We already used this fact in Proposition 1.)

108 
c THE MATHEMATICAL ASSOCIATION OF AMERICA
 Fact 2. If α is the order of 10 modulo m and if 10β ≡ 1 (mod m), then α divides
β.
Fact 3. If m and n are positive integers such that m divides n, then the order of 10
modulo m divides the order of 10 modulo n.
In order to split the period of 1/ p 2 into blocks the same length as the period of 1/ p,
we first need to know that t2 is divisible by t. In fact, t2 is either t or pt. A similar
result can be found in [2]; we sketch a proof below.

Lemma 1. Let p be a prime other than 2 or 5. Let t and t2 be the lengths of the
periods of 1/ p and 1/ p 2 , respectively. Then either t2 = t or t2 = t p.

Proof. By Fact 1, p divides 10t −1, so (10t ) p−1

 + · ·t· + 10 + t1 p−1
t
is congruent to 0
modulo p. Consequently, p divides (10 ) − 1 = (10 − 1)((10 )
2 t p
+ (10t ) p−2 +
· · · + 10t + 1), and hence 10t p ≡ 1 (mod p 2 ). Thus, by Fact 2, t2 |t p. Since p| p 2 , by
Fact 3 we also have t|t2 . The result follows.

It appears that more often than not, t2 = t p. In fact, we were able to find reference
in the literature ([2], [5]) to only three primes for which t = t2 : 3, 487, and 56,598,313,
and none for which tn = tn+1 when n > 1 [2, Th. 9].
When p = 7, then p12 ((10t ) p−1 + (10t ) p−2 + · · · + 10t + 1) = 49 1
(1036 + 1030 +
· · · + 106 + 1), which is the sum of the following numbers.

0 . 020408 163265 306122 448979 591836 734693 877551 020408 . . .

20408 . 163265 306122 448979 591836 734693 877551 020408 163265 . . .
20408 163265 . 306122 448979 591836 734693 877551 020408 163265 306122 . . .
20408 163265 306122 . 448979 591836 734693 877551 020408 163265 306122 448979 . . .
20408 163265 306122 448979 . 591836 734693 877551 020408 163265 306122 448979 591836 . . .
20408 163265 306122 448979 591836 . 734693 877551 020408 163265 306122 448979 591836 734693 . . .
+ 20408 163265 306122 448979 591836 734693 . 877551 020408 163265 306122 448979 591836 734693 877551 . . .
20408 183673 489795 938774 530612 265306 . 142857 142857 142857 142857 142857 142857 142857 142857 . . .

Each column to the right of the decimal represents a sum of the p blocks of the
period, with carrying occurring. Since the decimal part of the sum is the decimal
1
equivalent of 1/7, it appears that 49 (1036 + 1030 + · · · + 106 + 1) − 1/7 is an integer;
 36 
in other words, that 49 divides (10 + 1030 + · · · + 106 + 1) − 7 . The following
lemma says that p12 ((10t ) p−1 + (10t ) p−2 + · · · + 10t + 1) − 1p is indeed an integer.

Lemma 2. Let p be a prime other than 2 or 5. Let t and t2 be the lengths of the
periods of 1/ p and 1/ p 2 , respectively. Then p 2 divides (10t ) p−1 + (10t ) p−2 + · · · +
10t + 1 − p.

Proof. In the case that t2 = t, 10t ≡ 1 (mod p 2 ), and hence (10t ) p−1 + (10t ) p−2 +
· · · + 10t + 1 ≡ 1 + 1 + · · · + 1 ≡ p (mod p 2 ), giving us our result. In the more
likely case that t2 = t p, since 10t ≡ 1 (mod p) and p|(10t − 1), there exists a pos-
itive integer l such that pl + 1 = 10t . If we substitute and use the binomial theo-
rem to expand each term in ( pl + 1) p−1 + · · · + ( pl + 1) + 1, all terms contain a
power of p greater than or equal to 2, except for the last two terms in each expan-
sion. Thus, viewing modulo p 2 and reordering terms gives us (( p − 1)( pl) + 1) +
(( p − 2)( pl) + 1) + · · · + (2( pl) + 1) + ( pl + 1) + 1 ≡ ( pl)( p − 1 + p − 2 + p −
3 + · · · + 2 + 1) + 1 + · · · + 1 ≡ ( pl)( ( p−1)
2
p
) + 1 + · · · + 1 ≡ 1 + · · · + 1 ≡ p, and
hence p 2 | (10t ) p−1 + (10t ) p−2 + · · · + 10t + 1 − p .

VOL. 39, NO. 2, MARCH 2008 THE COLLEGE MATHEMATICS JOURNAL 109
 We now prove our main result: if we split the period of 1/ p 2 into blocks of the
proper size and add them, we get a multiple of the period of 1/ p. The periods of
1/ p n+1 and 1/ p n also share this property, which can be established using the same
ideas, but which we leave to the interested reader.

Theorem 1. Let p be a prime, not 2 or 5, and let t and t2 be the lengths of the
periods of 1/ p and 1/ p 2 , respectively. Let 1/ p 2 = .A1 A2 . . . A p , where the Ai are
blocks of t digits. Then A1 + A2 + · · · + A p is divisible by 1p (10t − 1).

Proof. By Lemma 1, we can represent 1/ p 2 with the t-digit blocks Ai , and there
are two possiblities to consider.
Case 1. t2 = t. Here, the period of 1/ p 2 has length t, and therefore each t-digit
block Ai , 1 ≤ i ≤ p, is simply the period of 1/ p 2 . In this case, A1 + A2 + · · · + A p is
p times the period of 1/ p 2 , which is p · p12 (10t2 − 1) = 1p (10t − 1). Our result follows.
Case 2. t2 = t p. Since A1 A2 . . . A p is the period of 1/ p 2 , and since each block Ai
is an integer with t digits, we can write p12 (10t p − 1) = A1 (10t ) p−1 + A2 (10t ) p−2 +
· · · + A p−1 10t + A p . Dividing both sides by 10t − 1 and subtracting 1/ p, we have an
integer on the left-hand side by Lemma 2. On the right, we have

A1 (10t ) p−2 + (A1 + A2 )(10t ) p−3 + (A1 + A2 + A3 )(10t ) p−4 + · · ·

+ (A1 + A2 + · · · + A p−2 )10t + (A1 + A2 + · · · + A p−1 )
A1 + A2 + · · · + A p 1
+ − .
10 − 1
t p
A1 +···+A p
Since all terms here are integers except the last two, 10t −1
− 1
p
= m, for some
integer m. Solving for the sum of the Ai yields

  1 + mp 1 t 
A1 + · · · + A p = 10t − 1 = 10 − 1 · l,
p p

where l = 1 + mp. Thus the period of 1/ p divides the sum of the blocks of the period
of 1/ p 2 .

As was true in our example, the sum of the Ai can be manipulated into the period
of 1/ p with the high-order digit(s) added to the lower-order ones; additionally, it can
be written as the period of 1/ p plus a multiple of a t-digit string of nines: A1 + · · · +
A p = m10t + 1p (10t − 1) − m = 1p (10t − 1) + m(10t − 1).

Future fun
This project was exceptionally rewarding since the conjectures were motivated by a
CAS program, and the mathematics is accessible to any undergraduate with some
knowledge of elementary number theory. Some interesting questions for future work
include the following: Are there similar results for reciprocals of powers of composite
numbers? Why is t2 = t so rarely? Can the relationship between 1/ p 2 and 1/ p (and
1/ p n+1 and 1/ p n ) be generalized to 1/ p m and 1/ p n for any n < m? Many different
perspectives regarding the decimal periods of rationals are included in the references.

110 
c THE MATHEMATICAL ASSOCIATION OF AMERICA
References

1. B. Ginsberg, Midy’s (nearly) secret theorem—an extension after 165 years, College Mathematics Journal 35
(2004) 26–30.
2. W.G. Leavitt, Repeating decimals, College Mathematics Journal 15 (1984) 299–308.
3. K.H. Rosen, Elementary Number Theory and its Applications, 5th ed., Pearson/Addison Wesley, 2005.
4. M. Shrader-Frechette, Complementary rational numbers, Mathematics Magazine 51 (1978) 90–98.
5. S. Yates, The mystique of repunits, Mathematics Magazine 51 (1978) 22–28.

Proof Without Words: Carnot’s Theorem for Acute Triangles

In an acute triangle ABC, the sum of the distances x, y, z from A
the circumcenter to the sides is equal to the sum of the inradius
r and circumradius R, that is, x + y + z = R + r . b
c
z
y
Lemma 1. ax + by + cz = r (a + b + c).
r x
Proof. B a C
A A

c b c b
z r r
y
x r
B a C B a C
1 1 1 1
ax + by + cz = r (a + b + c), ∴ ax + by + cz = r (a + b + c).
2 2 2 2

Lemma 2. cy + bz = a R; az + cx = bR; bx + ay = c R.

Proof. (that cy + bz = a R; the other two relationships are proved analogously).

A cy bz
α β α γ
c/2 b/2
R
γ β bc/2 cR bR bc/2
z y

β
x γ β γ
B a C aR

Proof of Carnot’s Theorem.

(a + b + c)(x + y + z)
= (ax + by + cz) + (cy + bz) + (az + cx) + (bx + ay)
= r (a + b + c) + (a + b + c)R
= (a + b + c)(R + r )
∴ x + y +z = R +r
Claudi Alsina (claudio.alsina@upc.edu)
Universitat Politècnica de Catalunya, Barcelona
Roger B. Nelsen (nelsen@lclark.edu)
Lewis & Clark College, Portland OR

VOL. 39, NO. 2, MARCH 2008 THE COLLEGE MATHEMATICS JOURNAL 111