Find the greatest number which, when divided by 122 and 243 leaves remainders 2 and 3 respectively.

Options

A. 60
B. 120
C. 180
D. 240

Answer

To find the greatest number which, when divided by 122 and 243, leaves remainders 2 and 3 respectively, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 122 = 2 \)
- \( N \mod 243 = 3 \)

This means that:
- \( N = 122k + 2 \) for some integer \( k \)
- \( N = 243m + 3 \) for some integer \( m \)

### Step 2: Rewrite the equations
From the first equation, we can express \( N \) as:
\[ N - 2 = 122k \]
This implies that \( N - 2 \) is a multiple of 122.

From the second equation, we can express \( N \) as:
\[ N - 3 = 243m \]
This implies that \( N - 3 \) is a multiple of 243.

### Step 3: Set the equations equal
We can set the two expressions for \( N \) equal to each other:
\[ 122k + 2 = 243m + 3 \]
Rearranging gives:
\[ 122k - 243m = 1 \]

### Step 4: Solve for \( k \) and \( m \)
We need to find integer solutions for the equation \( 122k - 243m = 1 \). This is a linear Diophantine equation.

### Step 5: Find the HCF
To solve this, we first find the highest common factor (HCF) of 122 and 243 using the Euclidean algorithm:
1. \( 243 = 122 \times 1 + 121 \)
2. \( 122 = 121 \times 1 + 1 \)
3. \( 121 = 1 \times 121 + 0 \)

Thus, the HCF is 1, which means there are integer solutions for \( k \) and \( m \).

### Step 6: Find particular solutions
Using the Extended Euclidean Algorithm, we can express 1 as a linear combination of 122 and 243:
1. From \( 1 = 122 - 1 \times 121 \)
2. Substitute \( 121 = 243 - 1 \times 122 \):
\[ 1 = 122 - 1 \times (243 - 1 \times 122) \]
\[ 1 = 2 \times 122 - 1 \times 243 \]

Thus, one particular solution is \( k = 2 \) and \( m = 1 \).

### Step 7: General solution
The general solution can be expressed as:
\[ k = 2 + 243n \]
\[ m = 1 + 122n \]
for any integer \( n \).

### Step 8: Substitute back to find \( N \)
Substituting \( k \) back into the equation for \( N \):
\[ N = 122(2 + 243n) + 2 \]
\[ N = 244 + 29526n + 2 \]
\[ N = 246 + 29526n \]

### Step 9: Find the greatest \( N \)
To find the greatest number, we set \( n = 0 \):
\[ N = 246 \]

### Final Answer
The greatest number which, when divided by 122 and 243, leaves remainders 2 and 3 respectively is:
**246**

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