Find the greatest number that divides 338 and 59 and leaves remainders of 2 and 5 respectively.

Answer

To find the greatest number that divides 338 and 59 and leaves remainders of 2 and 5 respectively, we can follow these steps:

### Step 1: Set up the equations based on the problem statement
Let the greatest number be denoted as \( \alpha \). According to the problem, we can express the conditions as follows:
- For 338: \( 338 = \alpha \cdot Q_1 + 2 \)
- For 59: \( 59 = \alpha \cdot Q_2 + 5 \)

### Step 2: Rearrange the equations
From the first equation, we can rearrange it to find:
\[ 338 - 2 = \alpha \cdot Q_1 \]
\[ 336 = \alpha \cdot Q_1 \]

From the second equation, we can rearrange it to find:
\[ 59 - 5 = \alpha \cdot Q_2 \]
\[ 54 = \alpha \cdot Q_2 \]

### Step 3: Identify the relationship
From the above equations, we see that \( \alpha \) must be a common divisor of both 336 and 54. Therefore, we need to find the highest common factor (HCF) of 336 and 54.

### Step 4: Find the prime factorization of 336 and 54
- For 54:
\[ 54 = 2 \times 27 = 2 \times 3^3 \]
So, the prime factorization of 54 is \( 2^1 \times 3^3 \).

- For 336:
\[ 336 = 2 \times 168 = 2 \times 2 \times 84 = 2^2 \times 2 \times 42 = 2^3 \times 21 = 2^3 \times 3 \times 7 \]
So, the prime factorization of 336 is \( 2^4 \times 3^1 \times 7^1 \).

### Step 5: Determine the HCF
To find the HCF, we take the lowest power of each prime factor present in both factorizations:
- For \( 2 \): The minimum power is \( 2^1 \) (from 54).
- For \( 3 \): The minimum power is \( 3^1 \) (from 336).
- \( 7 \) is not present in 54, so we do not include it.

Thus, the HCF is:
\[ HCF = 2^1 \times 3^1 = 2 \times 3 = 6 \]

### Step 6: Conclusion
The greatest number \( \alpha \) that divides both 338 and 59, leaving remainders of 2 and 5 respectively, is:
\[ \alpha = 6 \]