Problem:
A number when divided by a divisor left a remainder of 23. When twice the number was divided by the same divisor, the remainder was 11. Find the divisor.

Solution:
Let's assume the number to be divided is 'x' and the divisor is 'd'. We are given the following information:

- When 'x' is divided by 'd', the remainder is 23.
- When 2 times 'x' is divided by 'd', the remainder is 11.

Step 1: Setting up the equations
Using the given information, we can set up the following equations:

Equation 1: x = q1 * d + 23 (where q1 is the quotient when 'x' is divided by 'd')
Equation 2: 2x = q2 * d + 11 (where q2 is the quotient when 2 times 'x' is divided by 'd')

Step 2: Solving the equations
To find the divisor, we need to solve these equations simultaneously. Let's eliminate 'x' from the equations:

2x = q2 * d + 11
2(q1 * d + 23) = q2 * d + 11
2q1 * d + 46 = q2 * d + 11
2q1 * d - q2 * d = 11 - 46
d(2q1 - q2) = -35

Simplifying further, we get:
d = -35 / (2q1 - q2)

Step 3: Finding the possible values for 'd'
Since 'd' is a divisor, it must be a positive integer. So, we need to find the possible values of 'd' for which the above equation holds true.

The divisor 'd' can be any positive integer that divides -35. The positive divisors of 35 are 1, 5, 7, and 35. However, we need to check if any of these values satisfy the equation.

Step 4: Checking the values of 'd'
Substitute each possible value of 'd' and check if it satisfies the equation:

For d = 1: d = -35 / (2q1 - q2) => 1 = -35 / (2q1 - q2) => -35 = 2q1 - q2 (Since 'd' cannot be negative)
This equation does not have any integer solutions for q1 and q2.

For d = 5: d = -35 / (2q1 - q2) => 5 = -35 / (2q1 - q2) => -35 = 10q1 - 5q2 => 7 = 2q1 - q2
This equation has an integer solution when q1 = 2 and q2 = 3.

For d = 7: d = -35 / (2q1 - q2) => 7 = -35 / (2q1 - q2) => -35 = 14q1 - 7q2 => 5 = 2q1 - q2
This equation has an integer solution when q1 = 3 and q2 = 1.

For d =