Let the original number be '$a$'
Let the divisor be '$d$'
Let the quotient of the division of $a$ by $d$ be '$x$'
Therefore, we can write the relation as $a/d = x$ and the remainder is 24.
i.e., $a = dx + 24$
When twice the original number is divided by $d, 2a$ is divided by $d$.
We know that $a = dx + 24$. Therefore, $2a = 2dx + 48$
The problem states that $(2dx + 48)/d$ leaves a remainder of 11.
$2dx$ is perfectly divisible by $d$ and will therefore, not leave a remainder.
The remainder of 11 was obtained by dividing 48 by $d$.
When 48 is divided by 37, the remainder that one will obtain is 11.
Hence, the divisor is 37.