How can i use modular arithmetic to find all numbers that are divisible by 33? I have $33 = 3 \cdot 11$ Which digits sums are positive integers in base 10.
Can i use the euclidean algorithm to find all the postive integers that are divisible by 33 and then back-track with the extended euclidean algorithm to find all x or y but i'dont know which numbers i have to use in the Euclidean algorithm.
And $gcd(3,11) = 1$
Are i on the right trail to solve this problem?
If you want to use the Euclidean algorithm, which involves a lot of divisions, then you may as well just divide once by $33$ and look at the remainder: $33 \mid n$ if and only if the remainder when $n$ is divided by $33$ is $0$. That is not very interesting mathematically and I suspect it is not what you're looking for as a method.
Here is a much more interesting method to test if a positive integer $n$ is divisible by $33$. Write $n = 10n' + a$, so $a$ is the units digit of $n$. Then $n$ is divisible by $33$ if and only if $n' + 10a$ is divisible by $33$, and since $n' + 10a$ is typically smaller than $n$ we get a recursive procedure that will bring us down to a small number that we can directly check for divisibility by $33$.
Example. Let $n = 24354$. Then $n' = 2435 + 10\cdot 4 = 2475$. We continue this process: $$ 24354 \leadsto 2475 \leadsto 247 + 10\cdot 5 = 297 \leadsto 29 + 10\cdot 7 = 99, $$ and $99$ is divisible by $33$, so the original number is divisible by $33$. In fact, $24354 = 33 \cdot 738$.
Example. Let $n = 5823$. Then we look at $$ 5823 \leadsto 582 + 30 = 612 \leadsto 61 + 20 = 81 \leadsto 8+10=18, $$ and $18$ is not divisible by $33$, so the original number $5823$ is not divisible by $33$.
This method does not appear to involve modular arithmetic, but it does behind the scenes: $33 \mid n$ if and only if $33 \mid (n'+ 10a)$ because $10^2 \equiv 1 \bmod 33$: $$ n \equiv 0 \bmod 33 \Longleftrightarrow 10n' + a \equiv 0 \bmod 33 \Longleftrightarrow n' + 10a \equiv 0 \bmod 33, $$ where we multiplied through by $10$ (an invertible number modulo $33$) in the last equivalence.
This method is a special case of what I call a "universal divisibility test" for divisibility by numbers $m$ that are relatively prime to $10$. It is based on using the inverse of $10 \bmod m$, and when $m = 33$ the number $10$ happens to be its own inverse mod $33$. For more information and examples of this approach, which is based on modular arithmetic for its justification but does not involve modular arithmetic when carrying out the method, look here.
