Private Equity puzzles, solved step by step

He proposes 98 for himself, 0, 1, 0 and 1. Solve from the end. With two left, partner 4 takes everything. With three, partner 3 buys partner 5 for one unit. With four, partner 2 buys partner 4 for one. With five, partner 1 needs two votes besides his own and buys the two partners who would get nothing in the next round, 3 and 5, for one unit each.

Why start from the end rather than the beginning?

Think of planning a train journey with a fixed arrival time: you work back from when you must arrive to when you must leave. Each partner's vote depends on what they would get if the current proposer were removed, so you can only value a vote once you know the next round, and the only round you can solve directly is the last one. With two partners left, partner 4 proposes 100 for himself and 0 for partner 5. His own vote is one of two, which is half, and half is not fewer than half, so it passes.

With three left, partner 3 needs two yes votes. Partner 5 gets nothing if partner 3 is removed, so one unit buys him: 99, 0, 1. With four left, partner 2 needs two votes. Partner 4 gets nothing in the three-partner round, so one unit buys him: 99, 0, 1, 0.

Which votes does the most senior partner buy, and why so cheaply?

With five partners he needs three yes votes: his own and two more. Look at the four-partner row. Partners 3 and 5 get nothing there, while partner 4 gets one unit and partner 2 gets 99. The cheapest votes always belong to whoever would be worst off in the next round, and here that is partners 3 and 5, so one unit each beats their alternative of zero. He keeps 100 less 2, which is 98. The answer feels unfair, and that is the point: in this game power comes from position in the sequence, not from any idea of a fair share.

Which assumptions does the answer rest on?

Three, and naming them is part of the answer. The tie rule, the tie-breaking preference and pure self-interest each change the split if you alter them. If a plan needed a strict majority, the proposer would need more votes. If an indifferent partner voted yes, the bought votes would cost zero instead of one, and the senior partner would keep 100. And real partners care about fairness and future rounds of carry, which is why actual carry allocations are set by negotiation and track record rather than by this logic.

Ten lockers are open: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100, the perfect squares. Locker n is toggled once by each person whose number divides n, so its final state depends on how many divisors n has. Divisors come in pairs that multiply to n, which makes the count even, except for a perfect square, where one divisor pairs with itself. An odd count leaves the locker open.

What decides whether one locker ends open?

Think of a light switch flicked by everyone who walks past: if an odd number of people flick it, the light ends on. Locker n is flicked by person d exactly when d divides n, so the number of toggles is the number of divisors of n, and the locker ends open only if that number is odd. The whole puzzle reduces to one question: which numbers have an odd number of divisors? Do not simulate a hundred people; reason about one locker.

Why do only perfect squares have an odd number of divisors?

Pair each divisor with its partner: for 12, 1 goes with 12, 2 with 6, 3 with 4. Every divisor has a different partner, so the count is even and the locker closes. For a perfect square, one divisor is its own partner, 6 x 6 for 36, so it is counted once and the total turns odd. There are ten perfect squares from 1 to 100, so ten lockers stay open. The answer for any number of lockers N is the whole part of the square root of N.