Last two digits of $12345678910...$ when it is divisible by $72$ the first time. 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.
 A: 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. 
A: Well a natural number can be written as $$N=9k+S$$ where S is sum of digits. So it is divisible by 9 if there  sum is. And for 8 ($2^3$) we need last 3 digits divisible by 8.And simply we need both conditions for divisibility by 72. 
A: Let $a_n=123\cdots n$. Modulo $9$, a number leaves the same remainder as the sum of its digits, so $a_n$ is divisible by $9$ if and only if $1+2+\cdots +n=\frac{n(n+1)}{2}$ is. Thus, either $n$ or $n+1$ is a multiple of $9$. Modulo $8$, the last three digits $a_n$ leave the same remainder as $a_n$ (since $8$ divides $1000$) so in particular $n$ is even.
The only option when $n$ has 1 digit is $n=8$, which does not work since $678$ leaves a remainder of $2$ modulo $4$. Thus $n$ has at least $2$ digits, so by considering the last two digits of $a_n$ we see that $n$ must be a multiple of $4$. After $8$, the least multiple of $4$ that is either a multiple of $9$, or $1$ less than a multiple of $9$ is $n=36$, which works since $536$ is divisible by $8$.
A: Hope its help.
you can add $1+2+3+4+5+...+n=\frac{n(n+1)}{2}$
$$1+2+3+4+5+...+n=\frac{n(n+1)}{2}=9k \to n(n+1)=18k$$check the possible numbers.
for example $8*9=18k$ but $12345678\neq 8q\\$.
next possiblities $9*10 \to 123456789\neq 8q $
...
$17*18\to 1234567891011121314151617\neq 8q\\18*19...$
