A very large number is formed by writing consecutive digits this way:
$$12345678910111213…$$
Find the last two digits of this number when it is divisible by $72$ the first time.
Could anyone advise me on how to approach this problem? Hints will suffice, thank you.
Best Answer
If you let $a_k$ be formed by concatenating $1,2,3,...,k$, the residues of the $a_k$ $\pmod{9}$ repeat cyclically: $1,3,6,1,6,3,1,0,0$.
So $a_k$ will be divisible by $9$ if and only if $k$ is congruent to $8$ or $9 \pmod{9}$. This should cut down your search field quite a bit.