I am currently familiar with the method of checking if a number is divisible by $2$, $3$, $4$, $5$, $6$, $8$, $9$, $10$, $11$. However is there a way to check if a no is divisible by $23$ ?
I read something at this link regarding this matter but couldn't really figure out what the author was trying to say. Any help in this regard would be appreciated. Thanks.
Best Answer
Here is the explanation to that link:
We know that $10 \cdot 7 = 70 \equiv 1 \mod 23$ as $3 \cdot 23 = 69$.
Thus, if we consider $10a + b \mod 23$ and we note that $7$ is coprime to $23$, then $10a + b \equiv 0 \mod 23 \iff 7(10a + b) = 70a + 7b \equiv a + 7b \equiv 0 \mod 23$
And that is how he got that $10a + b$ is divisible by $23$ iff $a + 7b$ is divisible by $23$. And this lets that blogger get rid of a digit, more or less, at each step to easily verify divisibility.