Let [tex]n=abcba[/tex] be such a number. If 18 divides [tex]n[/tex], then both 2 and 9 divide [tex]n[/tex].
To be divisible by 2, we must have [tex]a\in\{2,4,6,8\}[/tex]. Meanwhile we can have [tex](b,c)\in\{0,1,2,\ldots,9\}^2[/tex].
To be divisible by 9, we the sum of the digits of [tex]n[/tex] must itself be divisible by 9, or
[tex]2a+2b+c=9k[/tex]
for some integer [tex]k[/tex].
The largest value of [tex]2a+2b+c[/tex] is 2•8 + 2•9 + 9 = 43, so we must have [tex]k\in\{1,2,3,4\}[/tex].
I'm not sure what the best way to get the final count may be, but there are 44 such numbers. It's rather tedious to do by hand.
• If [tex]k=1[/tex], then [tex]2a+2b+c=9[/tex], and we can do this in 4 ways.
For example,
2•2 + 2•0 + 5 = 9 [tex](n = 20502)[/tex]
• If [tex]k=2[/tex], then [tex]2a+2b+c=18[/tex] and can be done in 16 ways.
2•2 + 2•3 + 8 = 18 [tex](n = 23832)[/tex]
• If [tex]k=3[/tex], then [tex]2a+2b+c=27[/tex] and can be done in 18 ways.
2•2 + 2•7 + 9 = 27 [tex](n = 27972)[/tex]
• If [tex]k=4[/tex], then [tex]2a+2b+c=36[/tex] and can be done in 6 ways.
2•6 + 2•8 + 8 = 36 [tex](n = 68886)[/tex]
