It's like binary but not really

Why it must start and end with "1" is left as an exercise for the reader.

Since it starts and ends with "1", we can write the number as $\displaystyle\sum_{i=0}^k100^i$ .

If k is positive and even, then we have: $\displaystyle\sum_{i=0}^k100^i=10^0+10^2+\cdots+10^{2k}\equiv[10^0+10^1+\cdots+10^k][(-10)^0+(-10)^1+\cdots+(-10)^k]$ (This result can be proven by induction.)

If k is an odd number greater than 1, then we have: $\displaystyle\sum_{i=0}^k100^i=10^0+10^2+\cdots+10^{2k}\equiv[10^0+10^2][10^0+10^4+\cdots+10^{2k-2}]$ (This result can be proven by induction.)

For example, when $k=3$ (which corresponds to $1010101$ ): $1010101=10^0+10^2+10^4+10^6=(10^0+10^2)(10^0+10^4)=101\times10001$

For example, when $k=4$ (which corresponds to $101010101$ ): $101010101=10^0+10^2+10^4+10^6+10^8=(10^0+10^1+10^2+10^3+10^4)(10^0-10^1+10^2-10^3+10^4)=11111\times9091$

Therefore the only solution is $k=1$ which corresponds to $\mbox{101}$ .