It is asked:
What is the remainder when $25$ divides a number where natural ones are arranged in order from $1$ to $35$ as a single number $1234567...333435$?
Before showing the way I did it, let's consider this: since $12=6\cdot 2$ it divides $9216$, coming from divisibility tests for $6$ and $2$ which the number $9216$ satisfies.
As I remember, when evaluating if $25$ divides $123...35$, I didn't use this method (firstly), instead, just picked the last 2 digits, $35$ and divided it into $25$: $35=1\cdot 25+10.$ So I concluded that when $123...35$ is divided by $25$, it will also have the remainder $10$, which is correct. Another neat way to show it is by writing $123..35=123...333425+10$. Here, the number $123...25$ divides into $25$ evenly, since the last 2 digits, $25$ also divides into $25$ evenly, so $10$ above indicates the remainder.
And here is the origin of confusion: $25=5\cdot 5$ so, to see if $25$ divides $123...35$ evenly, the latter huge number ($123...35$) must divide into $5$ evenly, which it does, and one can conclude that $123...35$ divided by $25$ won't have any remainder, which is untrue. By the way, it implies as well, that the case with $12$ further above in the first paragraph is not always applicable in other cases (not universal).
But I also don't trust my method completely in that it is universal, consider this apparent proof: when evaluating if $16$ divides $114664$ evenly, picking the last digits, $64$ and dividing it by $16$ or by $4,$ (because $16=4^2$) doesn't produce any remainder. But concluding $114664$ divided by $16$ either doesn't have a remainder is erroneous.
Finally, my questions are:
1) what is your thoughts on my way of solving the problem?
2) what is the correct (and universal) way of solving the problem?
3) is not using the method which is used in evaluating if $12$ divides $9216$ evenly when solving like problems saying evaluate if $x$ divides into $y$ evenly universal? (Though I believe it is not universal, I want to hear from you.)