Find the highest number by which 101 and 137 can be divided so as to leave a remainder 5 in each case.

To solve this problem, we need to find the highest number \( d \) by which both 101 and 137 can be divided to leave a remainder of 5.

### Step 1: Understanding the problem
We need to find the highest number \( d \) such that:
- When 101 is divided by \( d \), the remainder is 5.
- When 137 is divided by \( d \), the remainder is 5.

This means that both 101 and 137 can be written as:
- \( 101 = d \times q_1 + 5 \) (for some quotient \( q_1 \))
- \( 137 = d \times q_2 + 5 \) (for some quotient \( q_2 \))

### Step 2: Simplify the equations
We can subtract 5 from both 101 and 137 to get:
- \( 101 - 5 = 96 \)
- \( 137 - 5 = 132 \)

Now, the problem is reduced to finding the highest divisor of both 96 and 132, i.e., the greatest common divisor (GCD) of 96 and 132.

### Step 3: Find the GCD of 96 and 132
We can use the Euclidean algorithm to find the GCD of 96 and 132:

1. Divide 132 by 96:
\( 132 \div 96 = 1 \) (quotient), remainder \( 132 - 96 = 36 \).

2. Divide 96 by 36:
\( 96 \div 36 = 2 \) (quotient), remainder \( 96 - 2 \times 36 = 24 \).

3. Divide 36 by 24:
\( 36 \div 24 = 1 \) (quotient), remainder \( 36 - 24 = 12 \).

4. Divide 24 by 12:
\( 24 \div 12 = 2 \) (quotient), remainder \( 24 - 2 \times 12 = 0 \).

Since the remainder is now 0, the GCD of 96 and 132 is 12.

### Step 4: Conclusion
The highest number \( d \) that divides both 101 and 137 with a remainder of 5 is 12.