The Campers Problem (Part 4) A counselor has 311 + 2 campers with her at ajunction in a hiking trail (where n, is a natural number}. She knows their camp is twenty minutes down one of four possible paths. It wiii be dark in one hour and the group must find their camp before dark. n of the 3n + 2 campers sometimes lie, and unfortunately the counselor doesn't know which 11 they are. The counselors problem is that she needs to find the camp in one hour by sending campers down 3 paths (the counselor checks one path herself). |n parts 173 we showed that the counselor can make a code to solve her problem. In part 4 we will prove that 371 + 2 is the minimum number of campers the counselor needs to solve her problem To prove this, show that with only 311 + 1 campers the counselor cannot solve her problem. Start by using the pigeonhole principle as depicted below. Pi Ieons Cam ers Holes [Pathsl @ 3n+1 ' / 3 @ Hint: Look at the proof from the notes that showed with only 7 campers and n : 2 liars the counselor cannot solve her problem. A very similar proof that worked in the case of n : 2 will also work in this general case. Note: After you have compieted this proof you can use the fact that 311, + 2 is the minimum number of campers to quickly solve the campers' problem in the future for n liars, n 2 1