A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year 2003?
I need to find the smallest number of main courses, let's call it m, such that the number of possible dinners is at least 365, since 2003 is not a leap year. The number of possible dinners is the product of the number of appetizers, main courses, and desserts, which are 2m, m, and 3 respectively. So I have the inequality 2m * m * 3 >= 365, which simplifies to 6m^2 >= 365. I want to find the smallest integer value of m that satisfies this inequality. I could try to solve for m by taking the square root of both sides, but that would give me a decimal value, and I need an integer. Instead, I can try some values of m and see if they work. If m = 1, then 6m^2 = 6, which is too small. If m = 2, then 6m^2 = 24, which is also too small. If m = 3, then 6m^2 = 54, which is still too small. If m = 4, then 6m^2 = 96, which is getting closer, but not enough. If m = 5, then 6m^2 = 150, which is also not enough. If m = 6, then 6m^2 = 216, which is still not enough. If m = 7, then 6m^2 = 294, which is almost there, but not quite. If m = 8, then 6m^2 = 384, which is more than enough. So the smallest integer value of m that works is 8. Therefore, the restaurant should offer at least 8 main courses to ensure a different dinner each night in 2003. 