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Let $$n = 2^8\times 3^5\times 5^3\times 7^3\times 11^2\times 13^2\times 17\times 19\times 23\times 29\times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73\times 79 \times 83 \times 89 \times 97 \times 101 \times 103\times 107 \times 109 \times 113 \times 127 \times 131 \times 137 \times 139 \times 149 \times 151 \times 157 \times 163 \times 167 \times 173 \times 179 \times 181 \times 191 \times 193 \times 199 \times 211\times 233 \times 239 \times 241 \times 251 \times 257 \times 263 \times 269 \times 271\times 277 \times 281 \times 283\times 293\times 307\times 311 \times 313 \times 331 \times 347 \times 359 \times 367 \times 373 \times 397 \times 409 \times 419 \times 431 \times 433\times 443\times 487 \times 491 \times 499\times 509 \times 577\times 593\times 619\times 641\times 653\times 659\times683\times719\times 743 \times 761\times 809 \times 911\times 953 \times 1013 \times 1019 \times 1031 \times 1049 \times 1103 \times 1223 \times 1229 \times 1301 $$

$n$ has 97 prime factors, and around $2.6\times10^{229}$. Let $$A=\{3 , 7 , 9, 19, 31, 39, 49, 63, 79, 99, 127, 159, 199, 249, 319, 399, 499, 511, 639, 999, 1023\}$$ Out of $42$ numbers of the form $10^n \pm k, k\in A$ which ones are prime?

My observation is $\forall x\in A$, $x+1$ is of the form $2^\alpha5^\theta$. However, I can't see a patern amongst the prime factors of $n$. I feel like Fermat's Little Theorem will help eliminate most of the numbers, but, ,I have no idea how to prove any of them are actually prime.

For example, as $\phi(7) | n $, $10^n\equiv 1\pmod7$. Thus, $10^n - 1023 \equiv 1-1023 \equiv 1022 \equiv 0$. Same argument works for 99 and 127 too.

Or in general, for any $x\in A\setminus \{3\}$, take an odd prime divisor $p$ of $x-1$, then, $\phi(p) = p-1 | n$, thus, $10^n - x$ is divisible by $p$.

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    $\begingroup$ If any number of these $42$ is actually a prime, it will very hard (or very long) to prove it. So, assuming that this is a prepared-for-solving problem, I think that the answer is probably "none". $\endgroup$ – ajotatxe Jan 23 '17 at 15:37
  • $\begingroup$ Could you eliminate some numbers with Fermat's little theorem ? If yes, please report us the current status! $\endgroup$ – Peter Jan 23 '17 at 17:24
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    $\begingroup$ Added an example. $\endgroup$ – Emre Jan 23 '17 at 17:46
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This is only a partial answer, just for some $10^n-k$

For $k=(7,19,31,49,79,127,199,319,499,511)$ the last four digits of $10^n-k$ are $$(9993,9981,9969,9951,9921,9873,9801,9681,9501,9489)$$

As the sum of the digits in each of these is divisible by 3, as are all the preceding 9s, $10^n-k$ is divisible by 3, and hence composite, for $$k=(7,19,31,49,79,127,199,319,499,511)$$

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  • $\begingroup$ This cases are actually dealt with in the original statement of the question. $\endgroup$ – Emre Jan 23 '17 at 20:25
  • $\begingroup$ Sorry about that, I was negligent in not the reading the question properly. However, I’ve up-voted this fascinating puzzle, one that I expect to have an elegant solution; just wish I could see it. $\endgroup$ – Old Peter Jan 24 '17 at 19:05

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