1) start from number = 3 Loop while(number <= 100) 2) display number 3) number = number+3, display number 4) number = number+6 Loop

put those integers into an array, pick every third element, out of which discard every third element.

I'm assuming the question wants us to find integers that are divisible by 3 but not by 9. This can be easily obtained using a mod function inside the following if statement: if(number % 3 == 0 && number % 9 != 0) Here is a short program I wrote in c++ to show how to solve this problem. Instead of returning the set of integer, I just printed them out on the screen: #include #include using namespace std; int main(int argc, char** argv) { int i = 0; vector list; vector::iterator it; for(i = 1; i <= 100; i++) { if(i%3 == 0 && i%9 != 0) { list.push_back(i); } } for(it = list.begin(); it != list.end(); it++) { cout << *it << endl; } return 0; } If I missed anything, please let me know. Happy coding and problem solving!

A variation on Matt's answer: (1..100).select { |n| n % 3 == 0 }.reject { |n| n % 9 == 0 }

(1..100).map { |i| (i % 3).zero? && !(i % 9).zero? ? i : nil }.compact

That'll certainly work, Tyler, but the OP indicated he was interviewing for a Ruby On Rails - not C++ - gig.

The requirement doesn't say if the input has to be a Range. If it doesn't have to be, then we don't need to traverse each element but to simply calculate it. def get_nums_by_3_not_by_9(max) arr = [] x = max.to_i / 3 x.times do |i| next if i % 3 == 0 arr << i * 3 end return arr end

(1..100).select do |n| n%3 ==0 and n%9 != 0 end

(1..100).to_a.delete_if{|x| !(x%3==0 && x%9>0)} or (1..100).to_a.select{|x| x%3==0 && x%9>0} or (1..100).to_a.map{|x| x%3==0 && x%9>0 ? x : nil}.compact or (1..100).to_a.reject{|x| !(x%3==0 && x%9>0)}

python [x for x in range(0,100) if x % 3 == 0 and x % 9 != 0]

matt has the best answer