Transposable integer
Some specific numbers permute permute or shift cyclically when it is multiplied with a number n. Examples are:
- 142857 × 3 = 428571 (shifts cyclically one place left)
- 142857 × 5 = 714285 (shifts cyclically one place right)
- 128205 × 4 = 512820 (shifts cyclically one place right)
- 076923 × 9 = 692307 (shifts cyclically two place left)
These numbers are the repeating digits of the representative repeating decimals. The rules for finding these numbers are:
- A number X shift right cyclically by k positions when it is multiplied with an integer n. X is then the repeating digits of 1⁄F, whereby F = n × 10k - 1; or a factor of it; excluding any values for which 1⁄F has a period less than k + 1; and F must be coprime to 10. Most often it is convenient to choose the smallest F that fits the above.
- A number X shift left cyclically by k positions when it is multiplied with an integer n. X is then the repeating digits of 1⁄F, whereby F is 10k - n, or a factor of it; excluding any values for which 1⁄F has a period less than k + 1; and F must be coprime to 10. Most often it is convenient to choose the smallest F that fits the above.
- A number X shift left cyclically by k positions when it is multiplied with a fraction r⁄s. X is then the repeating digits of s⁄F, whereby F is (s × 10k - r), or a factor of it; excluding any values for which 1⁄F has a period less than k + 1; and F must be coprime to 10.
Sample proofs are provided below.
Performing cyclic permutation by multiplication
A long division of 1 by 7 gives:
0.142857...
7 ) 1.000000
.7
3
28
2
14
6
56
4
35
5
49
1
At the last step, 1 reappears as the remainder. The cyclic remainders are {1, 3, 2, 6, 4, 5}. We rewrite the quotients with the corresponding remainders below them at all the steps:
Quotients 1 4 2 8 5 7
Remainders 1 3 2 6 4 5
and also note that:
- 1⁄7 = 0.142857...
- 3⁄7 = 0.428571...
- 2⁄7 = 0.285714...
- 6⁄7 = 0.857142...
- 4⁄7 = 0.571428...
- 5⁄7 = 0.714285...
By observing the remainders at each step, we can thus perform a desired cyclic permutation by multiplication. E.g.,
- The number 142857, corresponding to remainder 1, can be made to permute to 428571 by multiplying with 3, the corresponding remainder of the latter.
- The number 142857, corresponding to remainder 1, can be made to permute to 857142 by multiplying with 6, the corresponding remainder of the latter.
- The number 857142, corresponding to remainder 6, can be made to permute to 571428 by multiplying with 5⁄6; i.e. dividing by 6 and then multiplying with 5, the corresponding remainder of the latter.
In this manner, cyclical shift of left or right of any position can be performed.
Less importantly, this technique can be applied to any number to cyclically shifting it right or left by any given number of places for the following reason:
- Every repeating decimal can be expressed as a rational number (fraction).
- Every number, when added with a decimal point in front and concatenated with itself infinite times, can be converted to a fraction, e.g. we can transform 123456 in this manner to 0.123456123456..., which can thus be converted to fraction 123456⁄999999. This fraction can be further simplified but it will not be done here.
- To permute the number 123456 to 234561, all one needs to do is to multiply 123456 by 234561⁄123456. This looks like cheating but if 234561⁄123456 is a whole number (in this case it is not), the mission is completed.
Shifting numbers cyclically
We are now ready to derive specific numbers that shift cyclically left or right by any number of positions when they are multiplied by n. These permutations can be:
- Shifting right cyclically by single position (parasitic numbers);
- Shifting right cyclically by double positions;
- Shifting right cyclically by any number positions;
- Shifting left cyclically by single position;
- Shifting left cyclically by double positions; and
- Shifting left cyclically by any number positions
Parasitic numbers
When a parasitic number is multiplied by n, not only it exhibits the cyclic behavior but the permutation is such that the last digit of the parasitic number now becomes the first digit of the multiple. For example, 102564 x 4 = 410156. Note that 102564 is the repeating digits of 4⁄39 and 410156 the repeating digits of 16⁄39.
Shifting right cyclically by double positions
A number X shift right cyclically by double positions when it is multiplied with an integer n. X is then the repeating digits of 1⁄F, whereby F = n × 102 - 1; or a factor of it; but excluding values for which 1⁄F has a period less than 3; and F must be coprime to 10.
Most often it is convenient to choose the smallest F that fits the above.
Proof
First recognize that X is the repeating digits of a repeating decimal, which always possesses a cyclic behavior in multiplications. Number X and its multiple then will have the following relationship:
- The number X is the repeating digits of the fraction 1⁄F, say cdxxx...ab, where a, b, c, d each represents a digit and x represents an unknown digit.
- The multiple is thus the repeating digits of the fraction n⁄F, say abcdxxx....
- F must be coprime to 10 so that when 1⁄F is expressed in decimal there is no preceding non-repeating digits.
- F should not be 3 or 9 for which 1⁄F has a period of 1, or any other value for which 1⁄F has a period less than 3.
- If the first remainder is taken to be n then 1 shall be the third remainder in the long division for n⁄F in order for this cyclic permutation to take place.
- In order that n × 102 = 1 (mod F) then F shall be either (n × 102 - 1), or a factor of it; but excluding any value for which 1⁄F has a period less than 3; and must be coprime to 10 as deduced in items 3) and 4) above.
This completes the proof.
Summary of results
The following multiplication moves the last two digits of each original number to the first two digits and shift every other digits to the right:
- 1⁄199 x 2 = 2⁄199; and the like, such as multiplication of 2 with 2⁄199, 3⁄199, ..., 99⁄199
- When expressed in decimal, the repeating decimal of 1⁄199 has a period of 99, i.e. it has 99 repeating digits.
- 0033444816 0535117056 8561872909 6989966555 1839464882 9431438127 090301 x 3, which is 1⁄299 x 3 = 3⁄299; and the like, such as multiplication of 3 with 2⁄299, 3⁄299, ..., 99⁄299.
- These fractions have a period of 66. Take note that 299 = 13 x 23, and the period is accurately predicted by the formula, LCM(6, 22) = 66, in above sections.
- Note that the solutions listed above have included:
- 076923 x 3 = 230769, which is 1⁄13 x 3 = 3⁄13; and the like, such as multiplication of 3 with 2⁄13, 3⁄13, and 4⁄13
- 0434782608 6956521739 13 x 3 = 1304347826 0869565217 39, which is 1⁄23 x 3 = 3⁄23; and the like, such as multiplication of 3 with 2⁄23, 3⁄23, .... 7⁄23
- 0025062656 64160401 x 4 = 0100250626 56641604, which is 1⁄399 x 4 = 4⁄399; and the like, such as multiplication of 4 with 2⁄399, 3⁄399, ..., 99⁄399.
- These fractions have a period of 18. Take note that 399 = 3 x 7 x 19, and the period is accurately predicted by the formula, LCM(1, 6, 18) = 18, in above sections.
- Note that the solutions listed above have included:
- 142857 x 4 = 571428, which is 1⁄7 x 4 = 4⁄7
- 0526315789 47368421 x 4 = 2105263157 89473684, which is 1⁄19 x 4 = 4⁄19; and the like, such as multiplication of 4 with 2⁄19, 3⁄19, and 4⁄19
- 1⁄499 x 5 = 5⁄499; and the like, such as multiplication of 5 with 2⁄499, 3⁄499, ..., 99⁄499.
- The number 499 is a full reptend prime. The fraction 1⁄499 is a cyclic number which has a period of 498.
There are many other possibilities.
Shifting left cyclically by single position
Problem: A number X shift left cyclically by single position when it is multiplied with 3. Find X.
Solution: First recognize that X is the repeating digits of a repeating decimal, which always possesses some interesting cyclic behavior in multiplications. Number X and its multiple then will have the following relationship:
- The number X is the repeating digits of the fraction 1⁄F, say abxxx... .
- The multiple is thus the repeating digits of the fraction 3⁄F, say bxxx...a.
- In order for this cyclic permutation to take place, then 3 shall be the next remainder in the long division for 1⁄F. Thus F shall be 7 as 1 × 10 ÷ 7 gives remainder 3.
This yields the results that:
- X = the repeating digits of 1⁄7
- =142857, and
- the multiple = 142857 × 3 = 428571, the repeating digits of 3⁄7
The other solution is represented by 2⁄7 x 3 = 6⁄7:
- 285714 x 3 = 857142
No solution can be found by this method for a multiplier of other positive integer n, except for the trivial solution of 1, because:
- Integer n must be the subsequent remainder in a long division of a fraction 1⁄F. Given that n = 10 - F, and F is coprime to 10 in order for 1⁄F to be a repeating decimal, then n shall be less than 10.
- For n to be 2, F must be 10 - 2 = 8. However 1⁄8 does not generate a repeating decimal.
- For n to be 7, F must be 10 - 7 = 3. However 7 > 3 and 7⁄3 = 2.333 > 1 and does not fit the purpose.
- Similarly there is no solution for any other number of n less than 10 except n = 3.
However, if the multiplier is not restricted to be an integer (though ugly), there are many other solutions from this method. E.g., if a number X shift right cyclically by single position when it is multiplied with 3⁄2, then 3 shall be the next remainder after 2 in a long division of a fraction 2⁄F. This deduces that F = 2 x 10 - 3 = 17, giving X as the repeating digits of 2⁄17, i.e. 1176470588235294, and its multiple is 1764705882352941.
The following summarizes some of the results found in this manner:
- 105263157894736842 x 0.5 = 052631578947368421, which is 2⁄19, a 2-parasitic number (and other 2-parasitic numbers)
- 1176470588235294 x 3⁄2 = 1764705882352941, i.e. 2⁄17 x 3⁄2 = 3⁄17; and the like, such as multiplication of 3⁄2 with 4⁄17, 6⁄17, 8⁄17 and 10⁄17
- 153846 x 3.5 = 538461, i.e. 2⁄13 x 7⁄2 = 7⁄13
- 18 x 4.5 = 81, i.e. 2⁄11 x 9⁄2 = 9⁄11
- 190476 x 4.75 = 904761, i.e. 4⁄21 x 19⁄4 = 19⁄21
- 1304347826086956521739 x 7⁄3 = 3043478260869565217391, i.e. 3⁄23 x 7⁄3 = 7⁄23; and the like, such as multiplication of 7⁄3 with 6⁄23, 9⁄23, 12⁄23, 15⁄23, 18⁄23 and21⁄23
Shifting left cyclically by double positions
A number X shift left cyclically by double positions when it is multiplied with an integer n. X is then the repeating digits of 1⁄F, whereby F is R = 102 - n, or a factor of R; excluding values of F for which 1⁄F has a period less than 3; and must be coprime to 10.
Most often it is convenient to choose the smallest F that fits the above.
Proof
First recognize that X is the repeating digits of a repeating decimal, which always possesses a cyclic behavior in multiplications. Number X and its multiple then will have the following relationship:
- The number X is the repeating digits of the fraction 1⁄F, say abcdxxx... .
- The multiple is thus the the repeating digits of the fraction n⁄F, say cdxxx...ab.
- F must be coprime to 10 so that 1⁄F has no preceding non-repeating digits.
- F should not be 3 or 9 for which 1⁄F has a period of 1, or any other value for which 1⁄F has a period less than 3.
- If the first remainder is taken to be 1 then n shall be the third remainder in the long division for 1⁄F in order for this cyclic permutation to take place.
- In order that 1 × 102 = n (mode F ) then F shall be either (102 - n), or a factor of it; but excluding any value for which 1⁄F has a period less than 3; and F must be coprime to 10 as deduced in items 3) and 4) above.
This completes the proof.
Summary of results
The following summarizes some of the results obtained in this manner:
- 142857 x 2 = 285714
- 285714 x 2 = 571428
- 428571 x 2 = 857142
And thus others listed below, where the white spaces between the digits divide the digits into 10-digit groups solely for the purpose of counting the digits:
- 0103092783 5051546391 7525773195 8762886597 9381443298 9690721649 4845360824 7422680412 3711340206 1855670103 0927835051 5463917525 7731958762 8865979381 4432989690 7216494845 3608247422 6804123711 3402061855 67 x 3 = ... , , i.e. 1⁄97 x 3 = 3⁄97; and the like, such as multiplication of 3 with 2⁄97, 3⁄97, 4⁄97, 5⁄97, ...., 31⁄97, 32⁄97
- No solution for n = 4
- 0526315789 47368421 x 5 = 2631578947 36842105, i.e. 1⁄19 x 5 = 5⁄19; and the like, such as multiplication of 5 with 2⁄19, and 3⁄19
- 0212765957 4468085106 3829787234 0425531914 893617 x 6 = 1276595744 6808510638 2978723404 2553191489 361702, i.e. 1⁄47 x 6 = 6⁄47; and the like, such as multiplication of 6 with 2⁄47, 3⁄47, 4⁄47, 5⁄47, 6⁄47, and 7⁄47
- 0322580645 16129 x 7 = 2258064516 12903, i.e. 1⁄31 x 7 = 7⁄31; and the like, such as multiplication of 7 with 2⁄31, 3⁄31, and 4⁄31; and such as multiplication of 7 with 1⁄93, 2⁄93, 4⁄93, 5⁄93, 7⁄93, 8⁄93, 10⁄93, 11⁄93, and 13⁄93
- 0434782608 6956521739 13 x 8 = 3478260869 5652173913 04, i.e. 1⁄23 x 8 = 8⁄23; and 2⁄23 x 6 = 16⁄23
- 076923 x 9 = 692307, i.e. 1⁄13 x 9 = 9⁄13
- No solution for n = 10
- 0112359550 5617977528 0898876404 4943820224 7191 x 11 = 1235955056 1797752808 9887640449 4382022471 9101, i.e. 1⁄89 x 11 = 11⁄89; and the like, such as multiplication of 11 with 2⁄89, 3⁄89, 4⁄89, 5⁄89, 6⁄89, 7⁄89, and 8⁄89
- No solution for n = 12
- 0344827586 2068965517 24137931 x 13 = 4482758620 6896551724 13793103, i.e. 1⁄29 x 13 = 13⁄29; and 2⁄29 x 13 = 26⁄29; and the like, such as multiplication of 13 with 1⁄87, 2⁄87, 4⁄87, and 5⁄87
- 0232558139 5348837209 3 x 14 = 325581395348837209302, i.e. 1⁄43 x 14 = 14⁄43; and the like, such as multiplication of 14 with 2⁄43, and 3⁄43
- 0588235294117647 x 15 = 8823529411764705, i.e. 1⁄17 x 15 = 15⁄17
See also
References
- C. A. Pickover, Wonders of Numbers, Chapter 28, Oxford University Press UK, 2000.
- Sequence OEIS: A092697 in the On-Line Encyclopedia of Integer Sequences.
- Gardner, Martin. Mathematical Circus: More Puzzles, Games, Paradoxes and Other Mathematical Entertainments From Scientific American. New York: The Mathematical Association of America, 1979. pp. 111-122.
- Kalman, Dan; 'Fractions with Cycling Digit Patterns' The College Mathematics Journal, Vol. 27, No. 2. (Mar., 1996), pp. 109-115.
- Leslie, John. "The Philosophy of Arithmetic: Exhibiting a Progressive View of the Theory and Practice of ....", Longman, Hurst, Rees, Orme, and Brown, 1820, ISBN 1-4020-1546-1
- Wells, David; "The Penguin Dictionary of Curious and Interesting Numbers", Penguin Press. ISBN 0-14-008029-5