divisibility rule by 11 reverse the decimal representation a-b+a-b+a=0 to be smallest $3a= 2b$ which a and b are integgers a= 2 b=3 23232

StarLex1 wrote: divisibility rule by 11 reverse the decimal representation a-b+a-b+a=0 to be smallest $3a= 2b$ which a and b are integgers a= 2 b=3 23232 ...and the largest would be $69696$ @below: fair, though the sum of the digits can be zero. If $S_n$ is the sum of the digits of a number $n$, then $11 | S_n$ iff $S_n \equiv 0 \pmod{11}$.

What about 98989 and 17171? I'm probably wrong - but isn't the divisibility rule by 11 that the alternating sum of digits has to be a multiple of 11 - not 0?

Ging wrote: What about 98989 and 17171? I'm probably wrong - but isn't the divisibility rule by 11 that the alternating sum of digits has to be a multiple of 11 - not 0? Nope, you are right. The smallest is actually $17171$ and the largest is $98989$.