When in each box 5 or 6 dozens of apples were packed,three dozens were left.Therefore,bigger boxes were taken to pack 8 or 9 dozens of apples. However, still three dozens of apples remained. What was the least number of dozens of apples to be packed?

Options

A. 363 dozens
B. 315 dozens
C. 345 dozens
D. 335 dozens

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the problem
The problem states that when packing apples in boxes of either 5 or 6 dozens, there are always 3 dozens left over. This means that the total number of dozens of apples can be expressed in terms of 5 or 6 dozens plus 3.

### Step 2: Set up the equations
Let the total number of dozens of apples be represented as \( N \). According to the problem:
- When packed in boxes of 5 dozens:
\[ N \equiv 3 \ (\text{mod} \ 5) \]
- When packed in boxes of 6 dozens:
\[ N \equiv 3 \ (\text{mod} \ 6) \]

### Step 3: Solve the congruences
From the above equations, we can see that \( N - 3 \) must be a multiple of both 5 and 6. Therefore, we need to find the least common multiple (LCM) of 5 and 6.

### Step 4: Calculate the LCM
The prime factorization of 5 is \( 5^1 \) and for 6 is \( 2^1 \times 3^1 \). The LCM is found by taking the highest power of each prime:
\[ \text{LCM}(5, 6) = 2^1 \times 3^1 \times 5^1 = 30 \]

### Step 5: Express \( N \)
Since \( N - 3 \) is a multiple of 30, we can express \( N \) as:
\[ N = 30k + 3 \]
for some integer \( k \).

### Step 6: Consider the bigger boxes
Next, the problem states that when using bigger boxes (8 or 9 dozens), there are still 3 dozens left over. Thus:
- For 8 dozens:
\[ N \equiv 3 \ (\text{mod} \ 8) \]
- For 9 dozens:
\[ N \equiv 3 \ (\text{mod} \ 9) \]

### Step 7: Solve the new congruences
Now, we need to find \( N \) such that:
- \( N - 3 \) is a multiple of 8 and 9.

Calculating the LCM of 8 and 9:
- The prime factorization of 8 is \( 2^3 \) and for 9 is \( 3^2 \).
\[ \text{LCM}(8, 9) = 2^3 \times 3^2 = 72 \]

### Step 8: Combine the conditions
Now we have two conditions:
1. \( N - 3 \) is a multiple of 30.
2. \( N - 3 \) is a multiple of 72.

Thus, \( N - 3 \) must be a multiple of the LCM of 30 and 72.

### Step 9: Calculate the LCM of 30 and 72
The prime factorization of 30 is \( 2^1 \times 3^1 \times 5^1 \) and for 72 is \( 2^3 \times 3^2 \).
\[ \text{LCM}(30, 72) = 2^3 \times 3^2 \times 5^1 = 360 \]

### Step 10: Express \( N \) again
Thus, we can express \( N - 3 \) as:
\[ N - 3 = 360m \]
for some integer \( m \). Therefore,
\[ N = 360m + 3 \]

### Step 11: Find the least value of \( N \)
To find the least number of dozens of apples, we set \( m = 1 \):
\[ N = 360 \times 1 + 3 = 363 \]

### Step 12: Conclusion
The least number of dozens of apples to be packed is:
\[ \boxed{363} \]