At least one person lives in each house. Any two neighboring houses together are inhabited by at most six people. What is the largest number of people that could be living on Jump Street? Here I have drawn a model of Jump Street and I have nine houses here in a row. I'll begin by making an example where each house contains exactly three inhabitants. And with nine times three, we have a total of 27 people living on Jump Street. So that allows us to discount choices A and B. And what we are doing here is we're making sure that neighboring houses contain the maximum of six inhabitants. We've used three and three. Now let's try four and two. So if I have four people in the first house, the second house would have to have two. And then we will alternate with fours and twos until the very end. So this calculation here, if we were to add up all these numbers, would give us four times five, which is 20, and four times two for a total of 28 inhabitants. So we have increased the population once again. And the most extreme case would be when we have the minimum of one person living in each house and then five people living in some of the houses. That still makes a total of six. So five plus one plus five plus one, and we'll alternate like this. Does that in fact give us a sum that is larger than 28? And calculating this, we have five times five, 25, plus four ones. And that increases the total to 29. Now we cannot increase our total past in this number. That would require leaving some houses uninhabited. So 29 here is indeed the maximum for this problem. And our answer is D, 29 people is the most that can live on Jump Street.

Jump Street

maximum population

neighboring houses

inhabitants arrangement

population optimization