Mr. Black has three kinds of wine. Of the first kind 403 litres, of the second kind 434 litres of the third kind 465 litres. What is the least number of full carks of equal size in which this can be stored without mixing?

Options

A. 31
B. 39
C. 42
D. 51

Answer

To find the least number of full casks of equal size in which Mr. Black can store his wines without mixing, we need to determine the greatest common divisor (GCD) of the three quantities of wine: 403 liters, 434 liters, and 465 liters. The size of each cask will be equal to the GCD, and the number of casks will be the total volume of wine divided by the GCD.

### Step 1: Find the GCD of the three quantities.

1. **Prime Factorization**:
- **403**:
- 403 is not divisible by 2 (it's odd).
- Check divisibility by 3: 4 + 0 + 3 = 7 (not divisible).
- Check divisibility by 5: last digit is not 0 or 5.
- Check divisibility by 7: 403 ÷ 7 = 57.57 (not divisible).
- Check divisibility by 11: 403 ÷ 11 = 36.64 (not divisible).
- Check divisibility by 13: 403 ÷ 13 = 31 (divisible).
- So, 403 = 13 × 31.

- **434**:
- 434 is even, so divide by 2: 434 ÷ 2 = 217.
- Now factor 217:
- Check divisibility by 3: 2 + 1 + 7 = 10 (not divisible).
- Check divisibility by 7: 217 ÷ 7 = 31 (divisible).
- So, 434 = 2 × 7 × 31.

- **465**:
- 465 is odd, so check divisibility by 3: 4 + 6 + 5 = 15 (divisible).
- Divide by 3: 465 ÷ 3 = 155.
- Now factor 155:
- Check divisibility by 5: last digit is 5 (divisible).
- 155 ÷ 5 = 31.
- So, 465 = 3 × 5 × 31.

2. **Common Factors**:
- The common prime factor among 403, 434, and 465 is **31**.

3. **GCD**:
- Therefore, GCD(403, 434, 465) = 31.

### Step 2: Calculate the number of casks.

1. **Total Volume of Wine**:
- Total = 403 + 434 + 465 = 1302 liters.

2. **Number of Casks**:
- Number of casks = Total volume ÷ GCD = 1302 ÷ 31 = 42.

### Conclusion:
The least number of full casks of equal size in which Mr. Black can store the wine without mixing is **42**.

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