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 ...

2019 AMC 10B #1 Problem: Alicia had two containers. The first was $\frac{5}{6}$  full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\frac{3}{4}$  full of water. What is the ratio of the volume of the first container to the volume of the second container? Solution: Let the volume of the first container be equal to X. Similarly, define Y to be the volume of the second container. From the problem, we see that  $\frac{5}{6}$ X = $\frac{3}{4}$Y. Solving, we get X/Y = 9/10 . 2019 AMC 10B #12 Problem: What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than 2019? Solution:  We see that the largest digit in any base-seven representation of a positive number is 6, so we can maximize the number of 6's in the base-seven representation. We know that $666_7 = 6*(7^0+7^1+7^2) = 342_10$. So, we have 2019-342 = 1677 fo...

2020 AMC 10A #22 Problem: For how many positive integers $n \leq 1000$  is $$\lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor$$ not divisible by 3 ? (Recall that $\lfloor x \rfloor$  is the greatest integer less than or equal to $x$ .) Solution: Observe that the 3 values $\lfloor{\frac{998}{n}}\rfloor$,  $\lfloor \frac{999}{n} \rfloor$ , and  $\lfloor{\frac{1000}{n}}\rfloor$ must have exactly 1 of them that is not equal to the others, to satisfy the condition that the expression  $$\lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor$$ is not divisible by 3. Note that the "turning points", or values of n that produce different values for $\lfloor{\frac{998}{n}}\rfloor$,  $\lfloor \frac{999}{n} \rfloor$ , and  $\lfloor{\frac{1000}{n}}\rfloor$ are the factors of 999 and 1000 . We realize that 998 has 4 factors: 1, 2, 499, 998. Note that when $n=1$, 499...