An important problem-solving skill - especially for the AMC - is figuring out how to frame (or reframe) a situation to make it easier to solve. This course will **guide you through the thought processes** needed to excel at this problem-solving art.

Consider the problem of finding the sum of all positive factors of 72. Typically, factors are found by creating pairs (or using a factor tree), which would be fairly inefficient since we would need to manually find and add all of the factors.

However, since \(72=2^3 \cdot 3^2\) has two prime factors, we can think about all of its factors using a grid instead:

Using this reframing, what is the sum of all of the positive factors of \(72?\)

To find the sum of the factors of a number with more than two prime factors, we can't make a grid of factors, but we *can* extend the same idea to more dimensions.

What is the sum of the positive factors of \(360 = 2^3 \cdot 3^2 \cdot 5?\)

The art of reframing a problem is present across all topics. In the next few problems, we'll look at an application in Combinatorics: the branch of Discrete Math concerned with counting.

How many ways are there to divide 5 indistinguishable candies among 3 distinguishable friends?

Note that all candies must be distributed, and people can receive 0 candies.

In combinatorics problems, a reframing can reveal an equivalent set of possibilities that is easier to count.

Consider the previous problem of distributing 5 indistinguishable candies among 3 distinguishable people. This was relatively tedious to count, especially because the people were distinguishable (i.e., 5-0-0 gives 3 possible distributions, but 4-1-0 gives 6).

However, we can reframe this problem: if we line up 5 candies and two indistinguishable dividers, **the distinct permutations of indistinguishable candies and dividers will give precisely the possible distributions of candies**.

For example, this corresponds to the distribution of 2-1-2.

In a fish tank, there are 4 distinct fish. You spread 10 identical food pellets into the tank.

How many different distributions of food pellets among the fish are possible, given that fish can receive 0 pellets and every pellet goes to some fish?

We can generalize the answer from the previous problems. When distributing \(n\) identical objects to \(r\) distinguishable bins, it's equivalent to arranging \(n\) identical objects and \(r-1\) identical dividers, so it can be done in \[\binom{n+r-1}{r-1}\] ways. However, in problem-solving, it's important to **understand the method rather than to memorize the formula**.

By doing so, you'll be able to tackle problems that you encounter that you haven't seen before by relating them to what you already know, and making the necessary modifications.

In the next problem, we'll see a slight modification on the previous fish feeding problem.

In a fish tank, there are 4 distinct fish. You spread 10 identical food pellets into the tank. However, you don't want any fish to go hungry, so **each fish must receive at least 1 pellet**.

How many different distributions of food pellets among the fish are possible?

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