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?

Shame on you Keshav!!! Cheating on Brilliant!!! This site is meant to be a platform to practice your own skills, not to copy paste the questions and get free answers and then get incentives without effort. So either play fair and be honest or leave this site. People like you are shame to the Brilliant community. And to the others, please give the answer to this problem after Monday 10/6/2013, so that this cheat doesn't get the opportunity to cheat.

As per the previous comment I won't giv you the fully worked solution itself. But if you can give an example of a possibility of these 3 groups I will sure give a hint.

Thanks for noticing, Shame. As such we have started tracking keshav's and mathlover's account( i.e. we are searching which accounts got these problems, and we are searching which accounts entered exactly the answers posted here, even if they are wrong at moreorless the same time or date). Currently we have pinpointed about five possibilities for keshav's account. A few more posts and he will be ours. Thanks for your cooperation, Shame.


-Calvin Lin
Brilliant Maths Challenge Master

To solve this problem, we can use a recursive approach.

Let's define a recursive function called "groupCount", which takes two parameters: the current integer being considered, and the current groups configuration. The base case for this function is when we have considered all the integers from 1 through 10 and have a valid groups configuration. In this case, we return 1 because we have found a valid way to create the groups.

In the recursive case, we have two options for each integer:

1. We can add the current integer to an existing group. In this case, we need to check if the property holds, i.e., if 2x is already in the same group. If it is, we can recursively call "groupCount" with the next integer and the updated groups configuration. If it's not, we move to the next option.

2. We can create a new group with the current integer. In this case, we can recursively call "groupCount" with the next integer and the updated groups configuration.

To solve the problem, we can start with an empty groups configuration and call the "groupCount" function with the first integer (1).

Here is the Python code that implements this approach:

def groupCount(current, groups):
if current > 10:
return 1

count = 0

for i in range(len(groups)):
if current * 2 in groups[i]:
groups[i].add(current)
count += groupCount(current + 1, groups)
groups[i].remove(current)

groups.append({current})
count += groupCount(current + 1, groups)
groups.pop()

return count

count = groupCount(1, [])
print("Number of different ways to create the groups:", count)

By running this code, we find that there are 15 different ways to create the groups.