2. A winemaker has three kinds of wine. He has 400 litres of chenin blane, 440 litres chardonnay and 460 litres of sauvignon blanc. Calculate the smallest possible number of barrels with equal volumes which can be fillec with the different wines without mixing them so that no wine is left.

To solve this problem, we need to determine the greatest common divisor (GCD) of the volumes of the wines: 400 litres, 440 litres, and 460 litres. The GCD will give us the largest volume of a barrel that can be filled evenly without leaving any wine behind. The prime factorization of each volume is as follows: - 400 = 2^4 × 5^2 - 440 = 2^3 × 5 × 11 - 460 = 2 × 5 × 23 The GCD takes the lowest power of all prime factors common to the three numbers. For our case: - The lowest power of 2 is 2^1. - The lowest power of 5 is 5^1. Thus, the GCD = 2^1 × 5^1 = 10 litres. Now we can find the smallest possible number of barrels: - For Chenin Blanc: 400 litres ÷ 10 litres = 40 barrels - For Chardonnay: 440 litres ÷ 10 litres = 44 barrels - For Sauvignon Blanc: 460 litres ÷ 10 litres = 46 barrels In total, the smallest possible number of barrels needed is 40 + 44 + 46 = 130 barrels.