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Repeating Decimals: A Period Piece Author(s): Kenneth A. Ross Reviewed work(s): Source: Mathematics Magazine, Vol. 83, No. 1 (February 2010), pp. 33-45 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/10.4169/002557010X479974 . Accessed: 29/02/2012 09:22
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Repeating Decimals: A Period PieceAuthor(s): Kenneth A. RossReviewed work(s):Source:
Mathematics Magazine,
Vol. 83, No. 1 (February 2010), pp. 33-45Published by:
Mathematical Association of America
Stable URL:
http://www.jstor.org/stable/10.4169/002557010X479974.
Accessed: 29/02/2012 09:22
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at
.
http://www.jstor.org/page/info/about/policies/terms.jspJSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.
Mathematical Association of America
is collaborating with JSTOR to digitize, preserve and extend access to
Mathematics Magazine.
http://www.jstor.org
VOL. 83, NO. 1, FEBRUARY 2010
33
Repeating Decimals: A Period Piece
KENNETH A. ROSS
University of OregonEugene, OR 97403rossmath@pacinfo.com
Since retiring I have been working with younger people, mostly ages 9–12. They tendto ﬁndlong division boring, soI showthem that long division is interesting bystudyingrepeating decimals. Some good questions from the kids led me to look into the resultsin this paper. Most of the results are known—as noted in the history section at the endof this article—but they have not been readily accessible in a general form.The ﬁrst interesting repeating decimal is the decimal expansion for
17
=
0
.
142857.I have known forever that the repeating portions of
27
,
37
,
47
,
57
, and
67
are all cyclic per-mutations of 142857, but I only recently stumbled upon another property of 142857:If you break its set of digits into 2 strings of equal length, then the numbers add to999: 142
+
857
=
999. I will call this the 2-block property. The number 142857 alsohas the 3-block property: 14
+
28
+
57
=
99. If we look at
37
=
0
.
428571, then again428
+
571
=
999 but in this case 42
+
85
+
71
=
198. That is not quite as nice as99, but 198 is twice 99 and we regard this as close enough to say that 3
/
7 has the 3-block property. With this understanding, 1
/
7 and 3
/
7 also satisfy the 6-block property,because the sums of their digits are multiples of 9.These nice block properties are surprisingly common. In this paper, we give twosimple theorems that explain nearly all appearances of this phenomenon. First, weillustrate with special cases, stating the theorems and a corollary along the way. Proofsfollow the examples, and help to clarify what is going on.For any fraction, the length of the smallest repeating portion of its decimal is calledthe
period,
for which we always write
. Thus the periods of
t
7
for
t
=
1, 2, 3, 4, 5,6 are all 6. We say that the fraction satisﬁes the
m
-block property if
m
divides
and,when we break the repeating portion into
m
blocks of equal length
k
=
/
m
, the sumof the
m
blocks is a string of
k
nines or an integer multiple of a string of
k
nines.T
ABLE
1 contains some examples where the denominators are primes.These illustrate the following corollary, which is a corollary to both Theorems 1 and2 below. As we will see in the history section, the prime 487 is especially interestingin our story.C
OROLLARY
.
Consider a prime p
≥
7
, and let
be the period of t
/
p where
1
≤
t
<
p. (Then
≤
p
−
1
.) If m divides
where m
>
1
, then the m-block property holds for t
/
p.
For
m
=
2, the sum of the two blocks,
A
and
B
, is exactly a string of nines. Justadd
A
and
B
in the standard way, working from right to left, and you will see that eachcolumn adds to 9—no carrying. This property is sometimes referred to as the “ninesproperty.” In this case, the sum is exactly a string of nines. This is also true for 1
/
p
for
m
=
3; the numerator being 1 is crucial here as a glance at 3
/
7 shows. We explain thisafter proving Theorem 2.The
m
-block property holds for 1
/
n
in many cases when
n
is not prime. SeeT
ABLE
2 below. You’re invited to momentarily ignore the third column and verifysome of the claims in the table about
m
-block properties. The challenge is to see whysome
m
-block properties hold, while others don’t.
Math. Mag.
83
(2010) 33–45. doi:
10.4169/002557010X479974
.c
Mathematical Association of America
34
MATHEMATICS MAGAZINETABLE 1
denominator
fraction
m
-block property holds for
7 6 1
/
7
=
0
.
142857
m
=
2
,
3
,
67 6 3
/
7
=
0
.
428571
m
=
2
,
3
,
613 6 1
/
13
=
0
.
076923
m
=
2
,
3
,
613 6 11
/
13
=
0
.
846153
m
=
2
,
3
,
617 16 1
/
17
=
0
.
0588235294117647
m
=
2
,
4
,
8
,
1619 18 1
/
19
=
0
.
052631578947368421
m
=
2
,
3
,
6
,
9
,
1831 15 1
/
31
=
0
.
032258064516129
m
=
3
,
5
,
1531 15 11
/
31
=
0
.
354838709677419
m
=
3
,
5
,
1573 8 17
/
73
=
0
.
23287671
m
=
2
,
4
,
8487 486 1
/
487
=
0
.
0020533
···
m
=
2
,
3
,
6
,
9
,
18
,
27
,
54
,
81
,
162
,
243
,
486
Do you see a pattern for when the
m
-block property holds? The key is to look at
k
=
/
m
and see whether it is a multiple of any of the periods of the reciprocals of the prime factors in the denominator. Those periods are listed in the third column of T
ABLE
2. For example, for 77
=
7
·
11, the periods of 1
/
7 and 1
/
11 are 6 and 2,respectively. For
m
=
2
,
3 and 6, the corresponding values of
k
are 3, 2 and 1. Of these
k
’s, only 2 is a multiple of 6 or 2, and the
m
-block property only fails for
m
=
3 where
k
=
2. Here is the general theorem, which is Theorem 3 in Harold Martin’s 2007 paper[
24
].T
HEOREM
1.
Let n
=
p
a
1
1
···
p
a
r
r
where the primes p
j
≥
7
are distinct. Let
bethe period of t
/
n, where
0
<
t
<
n and t is relatively prime to n, so that the fractiont
/
n is reduced. For each prime p
j
, we write
(
p
j
)
for the period of
1
/
p
j
. If
=
mk,where m
>
1
and k is an integer, and if none of the periods
(
p
j
)
divides k, then t
/
nhas the m-block property.
TABLE 2
m
-block propertydenominator
(
p
j
)
s
fraction holds for fails for
77
=
7
·
11 6 6, 2 1
/
77
=
0
.
012987
m
=
2
,
6
m
=
391
=
7
·
13 6 6, 6 1
/
91
=
0
.
010989
m
=
2
,
3
,
6 no
m
143
=
11
·
13 6 2, 6 19
/
143
=
0
.
132867
m
=
2
,
6
m
=
3259
=
7
·
37 6 6, 3 19
/
259
=
0
.
073359
m
=
3
,
6
m
=
2407
=
11
·
37 6 2, 3 19
/
407
=
0
.
046683
m
=
6
m
=
2
,
31001
=
7
·
11
·
13 6 6, 2, 6 151
/
1001
=
0
.
150849
m
=
2
,
6
m
=
3803
=
11
·
73 8 2, 8 1
/
803
=
0
.
00124533
m
=
8
m
=
2
,
4451
=
11
·
41 10 2, 5 1
/
451
=
0
.
0022172949
m
=
10
m
=
2
,
51147
=
31
·
37 15 15, 3 1
/
1147
=
0
.
000871839581517
m
=
3
,
15
m
=
51241
=
17
·
73 16 16, 8 1
/
1241
=
0
.
0008058017727639
m
=
4
,
8
,
16
m
=
2
VOL. 83, NO. 1, FEBRUARY 2010
35
All of the examples in T
ABLE
2 are easy to verify directly. All of the
m
-block properties that hold, hold by Theorem 1. Each failure in T
ABLE
2 occurs when atleast one of the
(
p
j
)
’s divides
k
. Nevertheless, we are morally obligated to check the failures, because the converse of Theorem 1 is not true: The
m
-block propertycan hold, even if some period
(
p
j
)
divides
k
. Here is the simplest example: For
n
=
253
=
11
·
23 and
=
22
=
mk
where
m
=
11 and
k
=
2, we have1253
=
0
.
0039525691699604743083
,
the 11-block sum is 594
=
6
·
99, and yet
k
=
2 is a multiple of
(
11
)
=
2. A class of examples follows the proof of Theorem 2.
Whydowe restrictprimestobebigger than 5?
First,avoiding factors of2and5inthe denominators is a major simpliﬁcation, because they are also factors of 10. Second,we can still get to these cases indirectly. If
n
is divisible by 2 or 5, then the repeatingportion of
t
/
n
also occurs as the repeating portion of another fraction
t
∗
/
n
∗
where 2and 5 are not factors of
n
∗
. Consider, for example, 17
/
280. Since 280
=
2
3
·
5
·
7, toeliminate the 2’s and 5 in the denominator, we multiply the fraction by 10
3
and obtain10
3
·
17280
,
which reduces to4257
=
60
+
57
=
60
.
714285
.
Therefore
17280
=
0
.
060714285, and the repeating portion of the decimal expansion isthe same as for 5
/
7.
For these reasons, we assume that
n
is relatively prime to 10
.We also avoid allowing 3 as a factor of the denominator, because the block propertyinvolving nines rarely holds in this case, essentially because 3 divides 9. Moreover, if 3 is a factor of
n
in Theorem 1, the hypotheses never hold:
k
is always a multiple of
(
3
)
=
1. There are some patterns, though, if the denominator has at most two factorsof 3; consider23117
=
233
2
·
13
=
0
.
196581 and65219
=
653
·
73
=
0
.
29680365
.
But this is another story.
We do not allow multiples of 3 in the denominator
, sincethe extra complications tend to obscure the main ideas of this article.We are now ready to address denominators that are powers of a prime. We avoidedthese fractions in T
ABLE
2, in part because they have long periods. For prime powers,the following theorem gives easily-checked conditions that are
equivalent
to the
m
-block property.T
HEOREM
2.
Consider a prime power p
a
where p
≥
7
, and consider an integer t where
0
<
t
<
p
a
and t is relatively prime to p. Let
be the period of t
/
p
a
and suppose
=
mk where m
>
1
and k is an integer. The following are equivalent:
(a)
The m-block property holds for t
/
p
a
.
(b)
m is not a power of p.
(c)
The period
(
p
)
of
1
/
p does not divide k.
T
ABLE
3 gives examples where the denominators are powers of primes and theblock property fails. Note that, in each case, neither (b) nor (c) in Theorem 2 holds,since
m
is a power of
p
and
(
p
)
divides
k
.

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