Disprove that for all primes $p$ if $a_{p-2}a_{p-3}...a_{2}a_{1}a_{0}$ is a multiple of $p$ then $a_{p-3}a_{p-4}...a_{2}a_{1}a_{0}a_{p-2}$ is also a multiple of $p$.
Assume all digits are non-zero such that $1\le{a_{i}}\le9$ for each $i$.
Need some help, thanks!