

A277622


Integers such that the first part "a", divided by the penultimate digit "b", leaves the last digit "c" as remainder (see the Example section for a detailed explanation).


1



110, 121, 131, 141, 151, 161, 171, 181, 191, 210, 220, 232, 242, 252, 262, 272, 282, 292, 310, 321, 330, 343, 353, 363, 373, 383, 393, 410, 420, 431, 440, 454, 464, 474, 484, 494, 510, 521, 532, 541, 550, 565, 575, 585, 595, 610, 620, 630, 642, 651, 660, 676, 686, 696, 710, 721, 731, 743, 752, 761, 770, 787, 797, 810
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OFFSET

1,1


COMMENTS

There are 8991 terms < 100000.


LINKS



EXAMPLE

Consider the first term, 110: the part "a" is "1", the penultimate digit "b" of "110" is "1" and the last digit "c" of "110" is "0"; we see indeed that 1/1 leaves a remainder of 0.
Consider the second term, 121: we see indeed that 1/2 leaves a remainder of 1  but no term between 110 and 121 has the same property (117, for instance, doesn't fit because 1/1 has 0 as remainder  and not 7 as it should have to be part of the sequence).
Similarly, the integer 1983 is part of the sequence because 19/8 leaves a remainder of 3.


MATHEMATICA

Select[Range[100, 1000], (u = Mod[#, 10]; d = Mod[(#  u)/10, 10]; d > 0 && u == Mod[Floor[#/100], d]) &] (* Giovanni Resta, Apr 11 2017 *)


CROSSREFS



KEYWORD

nonn,base,easy


AUTHOR



STATUS

approved



