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A329914
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Numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k.
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9
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21, 23, 27, 29, 33, 39, 57, 59, 69, 71, 83, 87, 99, 101, 107
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OFFSET
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1,1
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COMMENTS
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The idea of this sequence comes from the 21-digit integer 112359550561797732809 in Penguin dictionary (see reference) with this property: "The smallest number which, when 1 is placed at both ends, the number is multiplied by 99". The terms of this sequence are the other numbers k that have the same property than 99 and the corresponding smallest numbers M in each set {M_k} are in A329915 (see link).
The Diophantine equation to solve is 1M1 = k * M with M that has q digits, this is equivalent to 10^(q+1) + 1 = (k-10) * M, with number of zeros in 10^(q+1) + 1 = q also.
Some results coming from this Diophantine equation:
q >= 2 and 21 <= k <= 110, so this sequence is finite. The integers (k-10) end with 1, 3, 7 or 9, hence k also.
Integer (k-10) must be a divisor of 10^(q+1)+1 = A000533(q+1).
For k = 21, there is 21 * 91 = 1[91]1 but also 21 * 9091 = 1[9091]1; hence, 91 and 9091 are terms of M_21.
Since 10^(q+1)+1 mod (k-10) is periodic and the period length cannot exceed k-10, it is easy to check that the sequence is indeed full. - Giovanni Resta, Nov 26 2019
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REFERENCES
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D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.
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LINKS
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EXAMPLE
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23 * 77 = 1[77]1, so k = 23 is a term and 13 * 77 = 1001; remark: number M = 77 has 2 digits and 10^3+1 has 2 zeros also.
29 * 52631579 = 1[52631579]1, so 29 is a term et 19 * 52631579 = 10^9 + 1 = 1000000001.
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MATHEMATICA
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Select[Range[21, 110], GCD[10, # - 10] == 1 && MemberQ[Mod[10^Range[#] + 1, # - 10], 0] &] (* Giovanni Resta, Nov 26 2019 *)
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CROSSREFS
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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STATUS
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approved
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