A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 13. What is the value of the divisor?

Let the number be N and the divisor be D.
According to the question:

N=qD+24
where q is the quotient.

When twice the original number is divided by D:

2N=2(qD+24)=2qD+48
When 2N is divided by D, the term 2qD is completely divisible, so the remainder depends on dividing 48 by D.
Given that the actual remainder is 13, we can set up the equation:

48=kD+13
where k is an integer. Since N left a remainder of 24, the divisor D must be greater than 24.

48−13=kD
35=kD
Since D>24, the only possible integer factor of 35 that satisfies this condition is 35 itself (with k=1).
Therefore, the divisor is 35.