New Light Math

Choosing subsets without consecutive integers (in one form or another) is a problem that comes up quite frequently in harder problems. This approach to solving those problems is both simple and useful. To find a possible subset of {1, 2, 3, ..., x} that does not contain consecutive integers, we can first note that integers of a subset must have a "space" or integer between them. Next, the key to this approach is to think of all x integers in the set as dots . By ordering specific dots in a sequence, we can map the chosen (or red) dots to the chosen integers in the subset. Now we have that chosen dots are red, so let the "non-chosen" dots be green. But all the orderings include consecutive integers, so we need to remove some "buffer" dots that can later be added between the chosen dots. So suppose we need to find the total number of 3 integer subsets of {1, 2, 3, ..., 7} such that no two integers are consecutive. Then we have 2 "buffer dots" to be...

2006 AMC 12A #8 Problem: How many sets of two or more consecutive positive integers have a sum of 15? Solution: It is simple to see there are 3 total ways: {1, 2, 3, 4, 5} | {4, 5, 6} | {7, 8}. But why? Is there a way to generalize this? Yes! The number of sets of two or more positive integers that have a sum of k is equal to the number of positive odd divisors of k excluding 1. Let's break this down with an example. Take 369 = 3^2 * 41. From above, we have 5 ways. To find such sets, we take the factors of 369 and let them be our median of the set. Then, (regardless if any of the integers in the set are negative) we have a feasible set. First, 123 will be our median. Our set is: {122, 123, 124}. This set has 3 numbers and needs no manipulation. Next, our median equals 41. Our set becomes longer: {37, 38, 39, 40, 41, 42, 43, 44, 45}. This also doesn't require manipulation. Now, we have 9 as our median. Our set is then: {-11, -10 , -9, ... 9, 10, 11, ... 29}. However, this set ...