A merchant has three kinds of wine of the first kinds 403 gallons, of the second 527 gallons and of the third 589 gallons. What is the least number of full casks of equal size in which this can be stored without mixing ?

Options

A. 21
B. 29
C. 33
D. 31

Answer

To solve the problem of how many full casks of equal size are needed to store 403 gallons, 527 gallons, and 589 gallons of wine without mixing, we need to find the highest common factor (HCF) of these three quantities. Here’s a step-by-step solution:

### Step 1: Factorization of Each Quantity
We will first factor each of the quantities of wine.

1. **403 gallons**:
- To factor 403, we can check for divisibility by prime numbers.
- 403 is divisible by 13 (403 ÷ 13 = 31).
- So, the prime factorization of 403 is:
\[ 403 = 13 \times 31 \]

2. **527 gallons**:
- To factor 527, we check for divisibility.
- 527 is divisible by 17 (527 ÷ 17 = 31).
- So, the prime factorization of 527 is:
\[ 527 = 17 \times 31 \]

3. **589 gallons**:
- To factor 589, we check for divisibility.
- 589 is divisible by 19 (589 ÷ 19 = 31).
- So, the prime factorization of 589 is:
\[ 589 = 19 \times 31 \]

### Step 2: Identify the Common Factors
Now that we have the prime factorizations:
- 403 = \( 13 \times 31 \)
- 527 = \( 17 \times 31 \)
- 589 = \( 19 \times 31 \)

We can see that the common factor among all three quantities is **31**.

### Step 3: Calculate the HCF
The highest common factor (HCF) of the three quantities is:
\[ HCF = 31 \]

### Step 4: Determine the Number of Casks
To find the least number of full casks of equal size, we divide each quantity by the HCF:
- For 403 gallons: \( \frac{403}{31} = 13 \)
- For 527 gallons: \( \frac{527}{31} = 17 \)
- For 589 gallons: \( \frac{589}{31} = 19 \)

### Step 5: Total Number of Casks
The total number of casks needed is the sum of the individual casks:
\[ 13 + 17 + 19 = 49 \]

### Final Answer
The least number of full casks of equal size required to store the wine without mixing is **49**.

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