6 Digit Mayhem

There are 24 numbers on that list, not 22. you forgot to take off 4 and 8, since we are looking for 2 digit numbers.

An alternate proof:

The prime factorization of 148 is $4\times 37$ . Thus, we are looking for all numbers of the form $\overline{ababab}$ that have both 4 and 37 as prime factors. We consider 37 first. Note that $\begin{aligned} 1 &\equiv 1 &\mod 37 \\ 10 &\equiv 10 &\mod 37 \\ 100 &\equiv -11 &\mod 37 \\ 1000 &\equiv 1 &\mod 37 \\ 10000 &\equiv 10 &\mod 37 \\ 100000 &\equiv -11 &\mod 37 \end{aligned}$

Thus, $\overline{ababab} = a(100000 + 1000 + 10) + b(10000 + 100 + 1) \equiv_{37} a(-11 + 1 + 10) + b(10 -11 + 1) \equiv_{37} 0$ . So, in fact, all numbers of the form $\overline{ababab}$ are divisible by 37.

To see if our number is divisible by 4, consider $\overline{ababab} = \overline{abab00} + \overline{ab} = \overline{abab}\times 100 + \overline{ab}$ . Since $\overline{abab}\times 100$ will always be divisible by 4, our number will be divisible by 4 precisely when $\overline{ab}$ is divisible by 4. There are 25 two digit numbers divisible by 4, but three of them (00, 04, 08) have a leading zero (which would lead to our original number having only 5 digits) and two more of them (44 and 88) have $a=b$ .

Thus there are $\mbox{20}$ numbers of the form $\overline{ababab}$ that are divisible by 148.