OFFSET
0,2
COMMENTS
Every integer n has a multiple of the form 99...9900...00. To see that n has a multiple that's a palindrome (allowing 0's on the left) with even digits, let 9n divide 99...9900...00; then n divides 22...2200...00. - Dean Hickerson, Jun 29 2001
a(81), if it exists, is greater than 5 million. - Harvey P. Dale, Dec 19 2021
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..80
P. De Geest, Smallest multipliers to make a number palindromic.
EXAMPLE
a(12) = 5 since 5*12 = 60 (i.e., "060") is a palindrome.
MATHEMATICA
a[n_] := For[k = 1, True, k++, id = IntegerDigits[k*n]; If[AllTrue[id, EvenQ], rid = Reverse[id]; If[id == rid || (id //. {d__, 0} :> {d}) == (rid //. {0, d__} :> {d}), Return[k]]]]; a[0] = 1; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Apr 01 2016 *)
skpal[n_]:=Module[{k=1}, While[Count[IntegerDigits[k n], _?OddQ]>0 || (!PalindromeQ[(k n)/10^IntegerExponent[n k]]), k++]; k]; Array[skpal, 70, 0] (* Harvey P. Dale, Dec 19 2021 *)
PROG
(ARIBAS): stop := 500000; for n := 0 to 75 do k := 1; test := true; while test and k < stop do m := omit_trailzeros(n*k); if test := not all_even(m) or m <> int_reverse(m) then inc(k); end; end; if k < stop then write(k, " "); else write(-1, " "); end; end;
(Haskell)
a061797 0 = 1
a061797 n = head [k | k <- [1..], let x = k * n,
all (`elem` "02468") $ show x, a136522 (a004151 x) == 1]
-- Reinhard Zumkeller, Feb 01 2012
CROSSREFS
KEYWORD
nonn,base,easy,nice
AUTHOR
Amarnath Murthy, Jun 17 2001
EXTENSIONS
More terms from Klaus Brockhaus, Jun 27 2001
STATUS
approved