Is there any certain formula to figure out a problem like 9_8_7_ is divisible by 44 but not by 4? This is not a real question just similar to one I have seen. I realize that anything divisible by 44 will be divisible by 4, I do not need to find an answer to the problem I am trying to see if there is a faster way than trying all numbers in all situations! Thank you!
closed as not a real question by Brandon Carter, Kannappan Sampath, Asaf Karagila, J. M., t.b. Mar 24 '12 at 16:05
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
Every number divisible by $44$ is also divisible by $4$. To see this, consider a number $n$ divisible by $44$. This means that we can write $n$ as: $$ n = 44k$$ where $k$ is some integer. Now factor $44$ further, we get: $$ n = 4\times 11\times k = 2\times 2\times 11\times k. $$ Obviously, $n$ is divisible by $4$ (also by $2$, $11$, and $22$).
I think you are looking for "modular arithmetic." If you want to determine if a number is divisible by $n$, then you want to determine if that number is congruent to $0$ mod $n$. Methods for doing this depend on $n$. Your question is far to general to have any sort of formula or specific trick.
In fact, it is impossible to tell if $4$ divides a number of the form 9_8_7_ without knowing the last digit, but it will be divisible by $4$ iff the last digit is $2$ or $6$. This is because $4$ divides $20$ and in turn $100$, so first of all we only need to look at the last two digits, and secondly we only need to consider that last two digits mod $20$. Specifically, 9_8_7x is congruent to $1x$ mod 20, and the only numbers divisible by $4$ in the teens are $12$ and $16$.