Getting a specific digit from a ratio expansion in any base (nth digit of x/y)

Question

Is there an algorithm that can calculate the digits of a repeating-decimal ratio without starting at the beginning?

I\'m looking for a solution that doesn\'t use ar

Answer 1

edit: (I'm leaving post here for posterity. But please don't upvote it anymore: it may be theoretically useful but it's not really practical. I have posted another answer which is much more useful from a practical point of view, doesn't require any factoring, and doesn't require the use of bignums.)

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@Daniel Bruckner has the right approach, I think. (with a few additional twists required)

Maybe there's a simpler method, but the following will always work:

Let's use the examples q = x/y = 33/57820 and 44/65 in addition to 33/59, for reasons that may become clear shortly.

Step 1: Factor the denominator (specifically factor out 2's and 5's)

Write q = x/y = x/(2^a₂5^a₅z). Factors of 2 and 5 in the denominator do not cause repeated decimals. So the remaining factor z is coprime to 10. In fact, the next step requires factoring z, so you might as well factor the whole thing.

Calculate a₁₀ = max(a₂, a₅) which is the smallest exponent of 10 that is a multiple of the factors of 2 and 5 in y.

In our example 57820 = 2 * 2 * 5 * 7 * 7 * 59, so a₂ = 2, a₅ = 1, a₁₀ = 2, z = 7 * 7 * 59 = 2891.

In our example 33/59, 59 is a prime and contains no factors of 2 or 5, so a₂ = a₅ = a₁₀ = 0.

In our example 44/65, 65 = 5*13, and a₂ = 0, a₅ = a₁₀ = 1.

Just for reference I found a good online factoring calculator here. (even does totients which is important for the next step)

Step 2: Use Euler's Theorem or Carmichael's Theorem.

What we want is a number n such that 10ⁿ - 1 is divisible by z, or in other words, 10ⁿ ≡ 1 mod z. Euler's function φ(z) and Carmichael's function λ(z) will both give you valid values for n, with λ(z) giving you the smaller number and φ(z) being perhaps a little easier to calculate. This isn't too hard, it just means factoring z and doing a little math.

φ(2891) = 7 * 6 * 58 = 2436

λ(2891) = lcm(7*6, 58) = 1218

This means that 10²⁴³⁶ ≡ 10¹²¹⁸ ≡ 1 (mod 2891).

For the simpler fraction 33/59, φ(59) = λ(59) = 58, so 10⁵⁸ ≡ 1 (mod 59).

For 44/65 = 44/(5*13), φ(13) = λ(13) = 12.

So what? Well, the period of the repeating decimal must divide both φ(z) and λ(z), so they effectively give you upper bounds on the period of the repeating decimal.

Step 3: More number crunching

Let's use n = λ(z). If we subtract Q' = 10^a₁₀x/y from Q'' = 10^{(a₁₀ + n)}x/y, we get:

m = 10^a₁₀(10ⁿ - 1)x/y

which is an integer because 10^a₁₀ is a multiple of the factors of 2 and 5 of y, and 10ⁿ-1 is a multiple of the remaining factors of y.

What we've done here is to shift left the original number q by a₁₀ places to get Q', and shift left q by a₁₀ + n places to get Q'', which are repeating decimals, but the difference between them is an integer we can calculate.

Then we can rewrite x/y as m / 10^a₁₀ / (10ⁿ - 1).

Consider the example q = 44/65 = 44/(5*13)

a₁₀ = 1, and λ(13) = 12, so Q' = 10¹q and Q'' = 10¹²⁺¹q.

m = Q'' - Q' = (10¹² - 1) * 10¹ * (44/65) = 153846153846*44 = 6769230769224

so q = 6769230769224 / 10 / (10¹² - 1).

The other fractions 33/57820 and 33/59 lead to larger fractions.

Step 4: Find the nonrepeating and repeating decimal parts.

Notice that for k between 1 and 9, k/9 = 0.kkkkkkkkkkkkk...

Similarly note that a 2-digit number kl between 1 and 99, k/99 = 0.klklklklklkl...

This generalizes: for k-digit patterns abc...ij, this number abc...ij/(10^k-1) = 0.abc...ijabc...ijabc...ij...

If you follow the pattern, you'll see that what we have to do is to take this (potentially) huge integer m we got in the previous step, and write it as m = s*(10ⁿ-1) + r, where 1 ≤ r < 10ⁿ-1.

This leads to the final answer:

• s is the non-repeating part
• r is the repeating part (zero-padded on the left if necessary to ensure that it is n digits)
• with a₁₀ = 0, the decimal point is between the nonrepeating and repeating part; if a₁₀ > 0 then it is located a₁₀ places to the left of the junction between s and r.

For 44/65, we get 6769230769224 = 6 * (10¹²-1) + 769230769230

s = 6, r = 769230769230, and 44/65 = 0.6769230769230 where the underline here designates the repeated part.

You can make the numbers smaller by finding the smallest value of n in step 2, by starting with the Carmichael function λ(z) and seeing if any of its factors lead to values of n such that 10ⁿ ≡ 1 (mod z).

update: For the curious, the Python interpeter seems to be the easiest way to calculate with bignums. (pow(x,y) calculates x^y, and // and % are integer division and remainder, respectively.) Here's an example:

>>> N = pow(10,12)-1
>>> m = N*pow(10,1)*44//65
>>> m
6769230769224
>>> r=m%N
>>> r
769230769230
>>> s=m//N
>>> s
6
>>> 44/65
0.67692307692307696

>>> N = pow(10,58)-1
>>> m=N*33//59
>>> m
5593220338983050847457627118644067796610169491525423728813
>>> r=m%N
>>> r
5593220338983050847457627118644067796610169491525423728813
>>> s=m//N
>>> s
0
>>> 33/59
0.55932203389830504

>>> N = pow(10,1218)-1
>>> m = N*pow(10,2)*33//57820
>>> m
57073676928398478035281909373919059149083362158422691110342442061570390868211691
45624351435489450017295053614666205465236942234520927014873746108612936700103770
32168799723279142165340712556208924247665167762020062262193012798339674852992044
27533725354548599100657212037357315807679003804911795226565202352127291594603943
27222414389484607402282947077135939121411276374956762365963334486336907644413697
68246281563472846765824974057419578000691802144586648218609477689380837080594949
84434451746800415081286751988931165686613628502248356969906606710480802490487720
51193358699411968177101349014181943964026288481494292632307160152196471809062608
09408509166378415773088896575579384296091317883085437564856451054998270494638533
37945347630577654790729851262538913870632998962296783120027672085783465928744379
10757523348322379799377378069872016603251470079557246627464545140089934278796264
26841923209961950882047734347976478727084053960567277758561051539259771705292286
40608785887236250432376340366655136630923555863023175371843652715323417502594258
04219993081978554133517813905223106191629194050501556554825319958491871324801106
88343133863714977516430300933932895191975095122794880664130058803182289865098581
80560359737115185
>>> r=m%N
>>> r
57073676928398478035281909373919059149083362158422691110342442061570390868211691
45624351435489450017295053614666205465236942234520927014873746108612936700103770
32168799723279142165340712556208924247665167762020062262193012798339674852992044
27533725354548599100657212037357315807679003804911795226565202352127291594603943
27222414389484607402282947077135939121411276374956762365963334486336907644413697
68246281563472846765824974057419578000691802144586648218609477689380837080594949
84434451746800415081286751988931165686613628502248356969906606710480802490487720
51193358699411968177101349014181943964026288481494292632307160152196471809062608
09408509166378415773088896575579384296091317883085437564856451054998270494638533
37945347630577654790729851262538913870632998962296783120027672085783465928744379
10757523348322379799377378069872016603251470079557246627464545140089934278796264
26841923209961950882047734347976478727084053960567277758561051539259771705292286
40608785887236250432376340366655136630923555863023175371843652715323417502594258
04219993081978554133517813905223106191629194050501556554825319958491871324801106
88343133863714977516430300933932895191975095122794880664130058803182289865098581
80560359737115185
>>> s=m//N
>>> s
0
>>> 33/57820
0.00057073676928398479

with the overloaded Python % string operator usable for zero-padding, to see the full set of repeated digits:

>>> "%01218d" % r
'0570736769283984780352819093739190591490833621584226911103424420615703908682116
91456243514354894500172950536146662054652369422345209270148737461086129367001037
70321687997232791421653407125562089242476651677620200622621930127983396748529920
44275337253545485991006572120373573158076790038049117952265652023521272915946039
43272224143894846074022829470771359391214112763749567623659633344863369076444136
97682462815634728467658249740574195780006918021445866482186094776893808370805949
49844344517468004150812867519889311656866136285022483569699066067104808024904877
20511933586994119681771013490141819439640262884814942926323071601521964718090626
08094085091663784157730888965755793842960913178830854375648564510549982704946385
33379453476305776547907298512625389138706329989622967831200276720857834659287443
79107575233483223797993773780698720166032514700795572466274645451400899342787962
64268419232099619508820477343479764787270840539605672777585610515392597717052922
86406087858872362504323763403666551366309235558630231753718436527153234175025942
58042199930819785541335178139052231061916291940505015565548253199584918713248011
06883431338637149775164303009339328951919750951227948806641300588031822898650985
8180560359737115185'