Consider the sequence of numbers $$24, 2534, 253534, 25353534, \ldots$$ Let $$N$$ be the first number in the sequence that is divisible by 99. Find the number of digits in the base 10 representation of $$N$$.

The term with $$k$$ blocks of 53 is the digit string 2 followed by $$k$$ copies of 53 followed by 4, and it has $$2k+2$$ digits. Splitting the even length string into two digit groups gives $$25, 35, 35, \ldots, 35, 34$$, and a number is divisible by 99 exactly when these groups add to a multiple of 99, that is $$59 + 35(k-1) \equiv 0 \pmod{99}$$. Since $$35 \times 17 \equiv 1 \pmod{99}$$, this gives $$k - 1 \equiv 17 \times 40 \equiv 86$$, so $$k = 87$$ and the number of digits is $$2 \times 87 + 2 = 176$$.