bingboy

First, we can use the given functional equation to show that f(0) = 1. We have f(0) + f(0) = f(0 + 0) = f(0) = 2. This equation has two solutions: f(0) = 0 or f(0) = 1. If f(0) = 0, then f(x) = 0 for all x, since f(x) = f(x + 0) = x + f(0) = x + 0 = x. However, we know that f(0) = 2, so f(0) must be 1.

Now, we can use the given functional equation again to show that f(2012) = 2013. We have f(2012) + f(0) = f(2012 + 0) = f(2012) = 2012 + f(0) = 2012 + 1 = 2013. This equation has only one solution: f(2012) = 2013.

Therefore, the answer is 2013.

There are two cases to consider:

Case 1: One child sits next to two adults, and the other two children sit next to one adult each.

Case 2: Each child sits next to one adult.

In Case 1, we can choose which child sits next to two adults in 3 ways. Once we have chosen that child, we can seat the other two children and the three adults in a row in 6! ways. However, we have overcounted the number of arrangements by a factor of 2, since we have counted each arrangement as if the children were distinguishable.

In Case 2, we can seat the three adults in a row in 3! ways. Then, we can seat the three children between the adults in 3! ways. However, we have overcounted the number of arrangements by a factor of 3, since we have counted each arrangement as if the children were distinguishable.

Therefore, the total number of ways to seat the three adults and three children at a circular table if each child must be next to at least one adult is:

(3)(6!)/2 + (3!)(3!)/3 = 2! (6 + 3!) = \boxed{24}.

When the balls are indistinguishable, there is only one way to place them in the boxes. We can simply put one ball in each box.

However, the boxes are also indistinguishable, so there are 8! ways to arrange the boxes.

Therefore, there are 8!​ ways to place 4 balls in 8 boxes if neither the balls nor the boxes are distinguishable.

There are 5 vowels: a, e, i, o, and u. There are 4 possible first digits: 1, 2, 4, and 6. For each of the three letters, there are 23 possible letters that can be used (since no letter can be repeated). For each of the three digits, there are 9 possible digits that can be used (since no digit can be repeated).

Therefore, the total number of possible license plates is 5 * 4 * 23 * 23 * 23 * 9 * 9 * 9 = 3000.

There are two cases to consider:

Case 1: Two children sit next to one adult, and one child sits between two adults.

To seat the three adults, there are 2! ways, since rotations of the same arrangement are considered the same.

To seat the three children, there are two possible arrangements:

The child sitting between the two adults can be chosen in 3 ways. The other two children can then be seated in 2! ways.

The two children sitting next to the same adult can be chosen in 3 ways. The other child can then be seated in 2 ways.

Therefore, the total number of ways to seat the adults and children in Case 1 is 2! * (3 * 2! + 3 * 2) = 24.

Case 2: One child sits next to each adult.

To seat the adults, there are 2! ways.

To seat the three children, there are 3! ways.

Therefore, the total number of ways to seat the adults and children in Case 2 is 2! * 3! = 12.

Therefore, the total number of ways to seat the three adults and three children at the circular table if each child must be next to two adults is 24 + 12 = 36.

Let f be the number of female members and m be the number of male members. We know the following:

40f + 25m = 30(f + m)

Simplifying this equation, we get:

15f = 5m

Therefore, the ratio of female to male members is f/m=1/3.

There are 2 ways to start with the letter A: either from the first A in the array, or from the second A. To reach the second A, we must move down once. Once we have reached the first or second A, we can move in any direction to reach the letter R. There are 3 ways to do this: we can move down, right, or down and then right. Once we have reached the letter R, we can again move in any direction to reach the letter C. There are 3 ways to do this: we can move down, left, or left and then down. Once we have reached the letter C, the last step is to move up to reach the letter H, and there is only 1 way to do this. Therefore there are 2×3×3×1=18​ ways to spell the word "ARCH" in the array provided.

If there are 21 yellow tiles, then there are 21 tiles along one diagonal. Since the tiles are square, the total number of tiles along one side of the square is 21 + 1 = 22. Therefore, the total number of tiles in the square is 22 * 22 = 484. Since there are 21 yellow tiles, there are 484 - 21 = 463 gray tiles. So the answer is 463