The integers from 1 through 10 (inclusive) are divided into three groups, each containing at least one number. These groups satisfy the additional property that if x is in a group and 2x≤10, then
2x is in the same group. How many
different ways are there to create the groups?
To solve this problem, let's consider the possible ways to divide the integers from 1 through 10 into three groups.
We start by creating an empty group, and then we can distribute the integers one by one. Since each group must contain at least one number, let's distribute the numbers to the groups one at a time.
Starting with the first number, "1", we have three options: we can put it in the first group, second group, or third group.
For the next number, "2", we need to consider the additional property mentioned in the problem. If 2x ≤ 10, where x is a number already in a group, then 2x should also be in the same group.
Since "1" is already in a group, 2(1) = 2 is also in the same group. This means we can only put "2" in the same group as "1".
For the next number, "3", we apply the same logic. If 2x ≤ 10, where x is any of the numbers already in a group, then 2x should also be in the same group.
Since "1" is already in a group, 2(1) = 2 is in the same group, and "2" is also in a group, 2(2) = 4 is in the same group. This means we can only put "3" in the same group as "1" and "2".
We continue this process for each number, ensuring that if any existing number, x, is in a group, then 2x is also in the same group.
By the time we reach the last number, "10", we will have distributed all the numbers into the three groups.
To determine the number of different ways to create the groups, we need to count all the possible distribution combinations.
Since we have only one option for each number based on the properties mentioned, there is only one possible way to distribute the numbers into the groups.
Therefore, there is only "1" different way to create the groups.