%I #19 Dec 01 2019 22:34:05
%S 110,121,131,141,151,161,171,181,191,210,220,232,242,252,262,272,282,
%T 292,310,321,330,343,353,363,373,383,393,410,420,431,440,454,464,474,
%U 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
%N 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).
%C There are 8991 terms < 100000.
%H Jean-Marc Falcoz, <a href="/A277622/b277622.txt">Table of n, a(n) for n = 1..8991</a>
%e 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.
%e 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).
%e Similarly, the integer 1983 is part of the sequence because 19/8 leaves a remainder of 3.
%t Select[Range[100, 1000], (u = Mod[#, 10]; d = Mod[(# - u)/10, 10]; d > 0 && u == Mod[Floor[#/100], d]) &] (* _Giovanni Resta_, Apr 11 2017 *)
%K nonn,base,easy
%O 1,1
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Apr 11 2017