To solve the problem step by step, we will denote the divisor as \( D \), and the two numbers as \( N_1 \) and \( N_2 \).
### Step 1: Set up the equations based on the problem statement
When \( N_1 \) is divided by \( D \), it leaves a remainder of 55. This can be expressed as:
\[ N_1 = k_1 D + 55 \]
for some integer \( k_1 \).
When \( N_2 \) is divided by \( D \), it leaves a remainder of 98. This can be expressed as:
\[ N_2 = k_2 D + 98 \]
for some integer \( k_2 \).
### Step 2: Find the sum of the two numbers
Now, let's find the sum \( N_1 + N_2 \):
\[
N_1 + N_2 = (k_1 D + 55) + (k_2 D + 98) = (k_1 + k_2) D + (55 + 98)
\]
This simplifies to:
\[
N_1 + N_2 = (k_1 + k_2) D + 153
\]
### Step 3: Analyze the remainder when the sum is divided by \( D \)
According to the problem, when \( N_1 + N_2 \) is divided by \( D \), it leaves a remainder of 45. Therefore, we can express this as:
\[
N_1 + N_2 \equiv 45 \mod D
\]
This means:
\[
(k_1 + k_2) D + 153 \equiv 45 \mod D
\]
Since \( (k_1 + k_2) D \) is divisible by \( D \), we can ignore it in the modulo operation:
\[
153 \equiv 45 \mod D
\]
### Step 4: Set up the equation from the modulo operation
From the above equation, we can write:
\[
153 - 45 = nD
\]
for some integer \( n \). This simplifies to:
\[
108 = nD
\]
### Step 5: Find the possible values for \( D \)
To find the divisor \( D \), we need to find the divisors of 108. The divisors of 108 are:
1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108.
### Step 6: Determine the valid divisor
Since we know that the remainders (55 and 98) must be less than the divisor \( D \), we can eliminate any divisors that are less than or equal to 55. The possible divisors greater than 55 are:
- 108
Thus, the only valid divisor \( D \) that satisfies all conditions is:
\[
D = 108
\]
### Final Answer
The divisor is \( D = 108 \).