Talk:Palindromic gapful numbers: Difference between revisions

 

:Having calculated an outer 101, the 101 rem 11 == 2 determines permitted inners, in this case only 2, since 2*10 rem 11 is the needed 9. That idea needs generalising to d{0}d*10^k rem n11, ideally passing the outer remainder, needed inner length and trailing zeroes, then we don't actually have to perform any potentially expensive inner remainders at all, and maybe not even on the way in. A few small snippets from a quick experiment yields some clear patterns, both horizontal and vertical:

i=6, j=1, k=3: 33r66=33, 330r66=0, 3300r66=0, 33000r66=0, 330000r66=0, 3300000r66=0, 33000000r66=0, 330000000r66=0, 3300000000r66=0,

i=6, j=2, k=3: 303r66=39, 3030r66=60, 30300r66=6, 303000r66=60, 3030000r66=6, 30300000r66=60, 303000000r66=6, 3030000000r66=60, 30300000000r66=6,

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:Obviously there is a fair bit of 30+36=66 and 60+6=66 going on there, however as well as the full dividend it can be say 22, in this case a third of it

i=6, j=8, k=1: 100000001r66=35, 1000000010r66=20, 10000000100r66=2, 100000001000r66=20, 1000000010000r66=2, 10000000100000r66=20, 100000001000000r66=2, 1000000010000000r66=20, 10000000100000000r66=2,

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:However, there are also some longer cycles:

i=7, j=1, k=1: 11r77=11, 110r77=33, 1100r77=22, 11000r77=66, 110000r77=44, 1100000r77=55, 11000000r77=11, 110000000r77=33, 1100000000r77=22,

i=7, j=2, k=1: 101r77=24, 1010r77=9, 10100r77=13, 101000r77=53, 1010000r77=68, 10100000r77=64, 101000000r77=24, 1010000000r77=9, 10100000000r77=13,