Integer, no more than 50,000. Having no multipliers of 2 and 5. 

Fraction period 
Period numbers 
In this article, we will consider a topic from number theory: simple fractions and their properties. The topic is quite interesting, not fully studied and I think it will be interesting to look at it again.
What is it about?
As you know, any fraction of the form \(\cfrac{1}{N}\), where N is a natural (integer and positive) number, and does not have the numbers 2 and 5 in the multipliers, has such a property as a period.
For example in numbers
\(\cfrac{1}{13}=0.07692307692307692307692307692308\)
\(\cfrac{1}{41}=0.02439024390243902439024390243902\)
\(\cfrac{1}{101}=0.00990099009900990099009900990099\)
\(\cfrac{1}{91}=0.01098901098901098901098901098901\)
it is easy to notice that the numbers are repeated on the right side
for the number \(13\) it is \(076923\), for \(41\) it is \(02439\), for \(101\) it is \(0099\), for \(91\) it is \(010989\)
This is the period of the number. The main task associated with finding the period of fractions is that they are not subject to some wellknown rule.
That is, the fraction of the number 17 has a period equal to 16, and the period of the number 11 is already equal to 2. The number 7 has a period of 6, and \(7^2\) is 42.
How can you determine the period of a simple fraction?
We use the statement that for any prime number p, except 2 and 5, the statement
is true
\(10^{p1}mod p=1\)
That is, taking the remainder of ten to the 18th power when dividing it by 19, we get one (1)
This works always and on all numbers mutually prime with ten. In general, this is called Fermat's small theorem, and we have only considered a special case where the base is 10.
So the period of a simple fraction is always! a multiple of \({p1}\)
Judge for yourself
Number 11 \(p1=10\) and the period is 2
the number is 41 \(p1=40\) and the period is 5
number 101 \(p1=100\) and the period is 4
number 19 \(p1=18\) and the period is 18
the number is 6 \(p1=6\) and the period is 6
In addition, by denoting the period with the letter \(T\), we can easily prove that \(10^{T}mod p=1\)
Thus, raising 10 to the power of 1,2,3..... and dividing it by a given prime number, we get the residuals.
As soon as the remainder is 1  that's it, we found the period of a simple fraction.
Simple brute force works well on relatively small numbers, but what to do when a simple fraction has a number of tens of digits in its denominator?
It is necessary to decompose \(p1\) into multipliers and sort through all combinations of multipliers in order to substitute the value to a power.
We will stop at a complete search, just in order to calculate not only the period, but also to determine all the numbers in this period.
How do I do it?
Imagine \(10^{p1}mod p=1\) as \(10*M=1+p*K\), where \(K\) and \(M\) are some integers.
We obtained a linear diophantine equation with two variables
Let's take the number 41 as an example right away.
So the first step
\(10M41P=1\) Replace \(P\) with \(P\), so it would get rid of the minus sign.
Solving the equation \(10M41P=1\) we get that \(M=37+41R\)
Knowing that \(M\) is a power of ten, we can represent it as \(M=10W\)
that is, \(10W=37+41R\), where \(R\) and \(W\) are some integers.
That is, we got the Diophantine equation again, which we also solve.
Without going through each step, I will write to you what numbers are obtained 1,37,16,18,10,1,37....
Have you noticed that after doing 5 iterations, we returned to the original unit? So  this quantity is the period of a simple fraction.
In principle, nothing complicated yet, moreover, the reader will say: "and what's the point, I can also raise ten to a power and take the remainder, and it will turn out even faster"
I completely agree, but the material is about something a little different  about a broader understanding of simple fractions and their periods.
But since they wanted to take the leftovers...
The Diophantine equation is certainly not very convenient to solve at each stage, but how can it be simplified?
Having calculated the value of the remainder 37 for the first time, we do not need to solve the Diophantine equation in principle. We just take the remainder of the number 37 to the power of,2,3,4.... when dividing it by 41
\(37^{0}mod 41=1\)
\(37^{1}mod 41=37\)
\(37^{2}mod 41=16\)
\(37^{3}mod 41=18\)
\(37^{4}mod 41=10\)
Such a scheme is already comparable to a simple search as suggested by an unknown reader.
What else can be extracted from the algorithm that uses Diophantine equations?
For example, we can find out all the numbers in a period.
\(10*M=1+p*K\) solving this equation, we found what is equal to \(M\) but did not find what is equal to \(K\)
at \(p=41\)
\(K=9\)
Now multiplying each remainder \(1, 37, 16, 18, 10\) by \(K\) and taking the last digit, we get the value of the period, which is "inverted", that is, it is not read from right to left, but from left to right.
in our example it will be \(9, 3, 4, 2, 0\)
that is, the period of the fraction is 0.(02439)
Now consider the example of the fraction \(\cfrac{1}{117}\)
1. Solving the diophantine equation \(10*M+117*K=1\), we get a pair of values \(M=82, K=7\). The last digit in the period will be 7
2. \(82^{2} mod 117=55\) , So \(M=55\) and the penultimate digit in the period will be \(82*7mod 10=4\) 4(four)
3. \(82^{3} mod 117=64\) , So \(M=64\) and the previous digit in the period will be \(55*7mod 10=5\) 5(five)
3. \(82^{4} mod 117=100\) , So \(M=100\) and the previous digit in the period will be \(64*7mod 10=8\) 8(eight)
3. \(82^{4} mod 117=10\) , So \(M=10\) and the previous digit in the period will be \(100*7mod 10=0\) 0(zero)
3. \(82^{5} mod 117=1\) , So \(M=1\) a and the first digit in the period will be \(10*7mod 10=0\) 0(zero)
Once we have reached \(M=1\) at step 5, therefore our period is 6.
And our answer is \(\cfrac{1}{117}=0.008547008547005847008547.......\)
We offer for acquaintance a table of fractions of prime numbers up to 1500, and their periods
Number  Period  Number  Period  Number  Period 
3  1  421  140  947  473 
7  6  431  215  953  952 
11  2  433  432  967  322 
13  6  439  219  971  970 
17  16  443  221  977  976 
19  18  449  32  983  982 
23  22  457  152  991  495 
29  28  461  460  997  166 
31  15  463  154  1009  252 
37  3  467  233  1013  253 
41  5  479  239  1019  1018 
43  21  487  486  1021  1020 
47  46  491  490  1031  103 
53  13  499  498  1033  1032 
59  58  503  502  1039  519 
61  60  509  508  1049  524 
67  33  521  52  1051  1050 
71  35  523  261  1061  212 
73  8  541  540  1063  1062 
79  13  547  91  1069  1068 
83  41  557  278  1087  1086 
89  44  563  281  1091  1090 
97  96  569  284  1093  273 
101  4  571  570  1097  1096 
103  34  577  576  1103  1102 
107  53  587  293  1109  1108 
109  108  593  592  1117  558 
113  112  599  299  1123  561 
127  42  601  300  1129  564 
131  130  607  202  1151  575 
137  8  613  51  1153  1152 
139  46  617  88  1163  581 
149  148  619  618  1171  1170 
151  75  631  315  1181  1180 
157  78  641  32  1187  593 
163  81  643  107  1193  1192 
167  166  647  646  1201  200 
173  43  653  326  1213  202 
179  178  659  658  1217  1216 
181  180  661  220  1223  1222 
191  95  673  224  1229  1228 
193  192  677  338  1231  41 
197  98  683  341  1237  206 
199  99  691  230  1249  208 
211  30  701  700  1259  1258 
223  222  709  708  1277  638 
227  113  719  359  1279  639 
229  228  727  726  1283  641 
233  232  733  61  1289  92 
239  7  739  246  1291  1290 
241  30  743  742  1297  1296 
251  50  751  125  1301  1300 
257  256  757  27  1303  1302 
263  262  761  380  1307  653 
269  268  769  192  1319  659 
271  5  773  193  1321  55 
277  69  787  393  1327  1326 
281  28  797  199  1361  680 
283  141  809  202  1367  1366 
293  146  811  810  1373  686 
307  153  821  820  1381  1380 
311  155  823  822  1399  699 
313  312  827  413  1409  32 
317  79  829  276  1423  158 
331  110  839  419  1427  713 
337  336  853  213  1429  1428 
347  173  857  856  1433  1432 
349  116  859  26  1439  719 
353  32  863  862  1447  1446 
359  179  877  438  1451  290 
367  366  881  440  1453  726 
373  186  883  441  1459  162 
379  378  887  886  1471  735 
383  382  907  151  1481  740 
389  388  911  455  1483  247 
397  99  919  459  1487  1486 
401  200  929  464  1489  248 
409  204  937  936  1493  373 
419  418  941  940  1499  214 